Home

# In the above question, the acceleration of mass m is

2. In the above question, what is her centripetal acceleration in m/s2? a. 0.04 b. 0.24 c. 0.4 d. There's no centripetal acceleration because speed is constant 3. A pendulum of mass 0.5 kg is released from a position 10 cm above the lowest point of its swing. What is the speed of the pendulum as it passes the lowest point? a a = F / m. Force, Mass, and Acceleration Units. There are countless unit types that can be used to measure force, mass, and acceleration, but the most common ones (and those used by this calculator) are shown below: Metric force → N (Newtons) mass → kg (kilograms) acceleration → m/s 2 (meters per second squared) Imperia

### Solved: 2. In The Above Question, What Is Her Centripetal ..

force = mass x acceleration. mass = force x acceleration. acceleration = force x mass. Tags: Question 11. SURVEY. 120 seconds. Q. A 10kg object is at rest, it accelerates because a 50N force is applied Problem # 2. Two blocks of mass m and M are hanging off a single pulley, as shown. Determine the acceleration of the blocks. Ignore the mass of the pulley. Hint and answer. Problem # 3. Two blocks of mass m and M are connected via pulley with a configuration as shown Determine the acceleration of the masses and the tension in the string. (i) When unequal masses m 1 and m 2 are suspended from a pulley (m 1 > m 2) m 1 g - T = m 1 a, and T - m 2 g = m 2 a. On solving equations, we get. (ii) When a body of mass m 2 is placed on a frictionless horizontal surface, then. Mass Pulley System acceleration, a = What is the acceleration of the system if m 1 = 5 kg and m 2 = 2kg? Answer: 1. When the body m 2 moves in down ward direction. m 2 g - T = m 1 a T = m 2 g - m 1 a. 2. New tension can be found from the relation m 1 g - T = m 2 a T = m 1 g - m 2 a. 3. Acceleration of system, a. Question 3. The collision of two ice hockey players are shown.

### Force Mass Acceleration Calculator calculatordonke

You can express acceleration by standard acceleration, due to gravity near the surface of the Earth which is defined as g = 31.17405 ft/s² = 9.80665 m/s². For example, if you say that an elevator is moving upwards with the acceleration of 0.2g , it means that it accelerates with about 6.2 ft/s² or 2 m/s² (i.e., 0.2*g ) This Short Answer question also works as a part of the AP Physics C: Mechanics curriculum. A wooden wheel of mass M, consisting of a rim with spokes, rolls down a ramp that makes an angle θ with the horizontal, as shown above. The ramp exerts a force of static friction on the wheel so that the wheel rolls without slipping. (a) i

let's now tackle Part C so they tell us block three of mass M sub three so that's right over here is added to the system is shown below there is no friction between block three and the table all right indicate whether the magnitude of the acceleration of block two is now larger smaller or the same as in the original two block system explain how you arrived at your answer so let's just think. mass (m) is measured in kilograms (kg) acceleration (a) is measured in metres per second squared (m/s²) The equation shows that the acceleration of an object is: proportional to the resultant force.. Angular acceleration α is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: α = Δω Δt α = Δ ω Δ t, where Δ ω is the change in angular velocity and Δ t is the change in time. The units of angular acceleration are (rad/s)/s, or rad/s 2. If ω increases, then α is positive In the above question, the force acting on the object is [CPMT 1971] In the above question, the acceleration of the car will be [CPMT 1971] A) A cart of mass M is tied by one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M

Mass Radius A) 1/2 m 1/2R B) m R C) m 2R D) 2m R 17. A student who weighs 500 N on Earth travels to a planet whose mass and radius are twice that of Earth. His weight on that planet is about A) 1000 N B) 500 / √2 N C) 500 N D) 250 N 18. Mars has 1/10 the mass of Earth and 1/2 its diameter C. start moving and continue to increase its acceleration D. not move E. start moving and then slow to a stop 11. A 25 N box is pulled across a rough surface by an applied force of 22 N. The coefficient of kinetic friction between the box and the surface is 0.3. Find the acceleration of the box. Use g = 10 m/s2. A. 2.9 m/s2 B. 4.8 m/s2 C. 5.8 m/s 6 m/s 23. An object of mass 2 kg increases in speed from 2 m/s to 4 m/s in 3 s. What was the total work performed on the object during this time interval? A. 4 J B. 6 J C. 12 J D. 24 J E. 36 J 24. The figure above shows the forces acting on an object of mass 2 kg. What is the object's acceleration? A. 2 m/s 2 B. 2.5 m/s 2 C. 3 m/s 2 D. 3.5 m.

A bead of mass m is located on a parabollic wire with its axis vertical and vertex at the origin as shown and whose equation is x 2 = 4 a y. the wire is fixed and the bead can slide on it without friction.it is released from the point y= 4A on the wire frame from rest. the tangential acceleration of the bead reaches to position given by y=a is Question 11. The value of g at a particular point is 9.8 m/sec² suppose the earth suddenly shrink uniformly to half its present size without losing any mass. The value of g at the same point (assuming that the distance of the point from the centre of the earth does not shrink) will become (a) 9.8 m/sec² (b) 4.9 m/sec² (c) 19.6 m/sec² (d) 2. Click here������to get an answer to your question ️ An object with a mass 10kg moves at a constant velocity of 10m/s. A constant force then acts for 4 seconds on the object and gives it a final speed of 2m/s inopposite direction. The acceleration produced in it is (a) 3m/s? (b)-3m/s? (c) 0.3m/s² (d) -.3m/s2 In the above question, the force acting on the object is (a)30N (b)- 30N (c)3N (d)-3N.

Question. An electron moves in circular motion in a uniform magnetic field. The velocity of the electron at point P is 6.8 × 10 5 m s -1 in the direction shown. The magnitude of the magnetic field is 8.5 T. a. State the direction of the magnetic field The formula for the m.o.i. of a pulley is 1/2mr^2, where m is the mass and r is the radius. So the m.o.i. of your pulley would be I=1/2*5kg*.25m^2=.156kg*m^2. The product of the m.o.i. and the angular velocity is going to equal the torque (rotational force) on the pulley - question from Steven Kelly. This question goes back to basic principles of the physics of motion as first discovered by Galileo and Sir Isaac Newton. You are correct that the force of gravity on Earth creates an acceleration of 32 ft/s², or about 10 m/s² 1) The direction of the force vector corresponds directly to the direction of acceleration because of Newton's Second Law's equation: F = m * a. The mass of an object is a one-dimensional quantity and only changes the magnitude of the result, not its direction 30 seconds. Q. One Newton is the force needed to cause a. answer choices. 1-kg object to accelerate at 1 m/s 2 . 1-g object to accelerate at 1 cm/s 2. 1-kg object to accelerate at 1 km/s 2. none of the above

### Force= Mass x Acceleration Science Quiz - Quiziz

acceleration net force mass In briefest form, where a is acceleration, F is net force, and m is mass, a F m The acceleration is equal to the net force divided by the mass. From this relationship we see that doubling the net force acting on an object doubles its acceleration. Suppose instead that the mass is doubled. Then acceleration will be. Q. A solid cylinder of mass m and radius R has a string wound around it. A person holding the string pulls it vertically upward, as shown above, such that the cylinder is suspended in midair for a brief time interval Δt and its center of mass does not move. The tension in the string is T, and the rotational inertia of the cylinder about its axis is 1/2 mR A2. This question is about kinematics. Lucy stands on the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves her hand with a speed of 15 m s-1 at the instant her hand is 80 m above the surface of the sea. Air resistance is negligible and the acceleration of free fall is 10 m s-2 [G = 6.67 × 1O-11 N.m 2 /kg 2, mass or the earth =6 × 10 24 kg, radius or the earthe 6.4 × 10 6 m] Question 1. A satellite of mass 1000 kg revolves around the earth in a circular path. If the distance between the satellite and the centre of the earth is 40000 km, find the gravitational force exerted on the satellite by the earth. Answer: 250.

### Pulley Problem

Two stars, each of mass M, form a binary system.The stars orbit about a point a distance R from the center of each star, as shown in the diagram above. The stars themselves each have radius r. 5. In terms of each star's tangential speed v, what is the centripetal acceleration of each star? Answe The centripetal acceleration of a satellite of mass 1000 kg orbiting at 1600 km above the earth's surface is (take, radius of earth = 6.4 x 10 6 m and g = 9.8 ms -2 ) (a) 6.272 ms − ### Pulley Mass System Definition, Examples in Physics

A less massive object should have a smaller percentage difference. The acceleration that I used for the question above was the 50 grams mass acceleration, and if the mass was greater than this, the percentage difference would increase, not lower. My measurements agree with this statement, however, lowering the mass too much will also increase the percent difference, as seen with the 20 grams. A ball with a mass of 10 kg is barreling downhill with the total force of 50 N. Assuming the direction of the total force is perpendicular to the hill's slope, what's the ball's acceleration? From that question, we can conclude that: F = 50N. m = 10kg. Now, we're going to input that data into the equation. a = F / m. a = 50 / 10. a = 5 m/s2. Question 3 . SURVEY . 30 seconds . Q. A seesaw with mass x is perfectly balanced with a fulcrum in the center. A force F is acting on an object of mass m, giving it an acceleration of a. What is the acceleration if the mass is halved and the force quadrupled? In the diagram above, two masses, one with mass m, the other with mass M = 2m.

In this case, the mass is already in kilograms, and therefore it remains m=90 Kg. II) Identify the gravitational acceleration. In our case above, the space in question is the earth, which ah a gravitational acceleration of 9.8 m/s2. III) Finally, after identifying the 'g' and 'm,' substitute these values to equation [ii] above. F=m*g =9.8 * 90. Consider the two collisions shown above. In both cases a solid disk of mass M, radius R, and initial angular velocity ω0 is dropped onto an initially stationary second disk having the same radius. In Case 2 the mass of the bottom disk is twice as big as in Case 1 Question. An object is at rest at time t=0 t = 0. The variation with t t of the acceleration a a of the object is shown from t=0 t = 0 to t=20 s t = 20 s. What is the speed of the object when t=15 s t = 15 s? A. 25 ms−1 25 m s − 1. B. 50 ms−1 50 m s − 1. C. 75 ms−1 75 m s − 1. D. 100 ms−1 100 m s − 1 The magnitude of the acceleration of block A after release is nearly 9.81 m/s2. Briefly explain your reasoning without deriving or using equations. In the absence of block A, block B would be in free fall and have an acceleration of 9.81 m/s 2 down mass will experience a smaller acceleration. d. the object with greater mass wil l experience a small acceleration and the object with less mass will experience an even smaller acceleration. ____ 22. Two perpendicular forces, one of 45.0 N directed upward and the second of 60.0 N directed to the right, act simultaneously on an object with a.

c) Acceleration due to gravity 'g' depends on the distance T from the centre of the earth. Draw a graph showing the variation of 'g' with 'r'. Answer: a) Yes. Moon attracts everybody towards its centre. Question 4. Imagine a point mass 'm' maintained at the centre of a shell of uniform density having mass 'M' Question 9. SURVEY. 60 seconds. Q. Acceleration is defined as the CHANGE in. answer choices. time it takes for one place to move to another place. velocity of an object. distance divided by the time interval. velocity divided by the time interval

### Plus One Physics Chapter Wise Questions and Answers

AP Physics C Mechanics: Unit - 7 - Gravitation Practice Test Question 1 A satellite of mass m orbits 3.8x10 ⁸ m above the Earth. If the force due to gravity of the Earth on the satellite is 7.99N, what is the mass of the satellite? Use: mE=5.97x10 24 kg; r E=6.37x10 5 kg; G=6.67x10−11 m The correct answer to above question is based on [NCERT 1975] A) Acceleration due to A body of mass m is taken to the bottom of a deep mine. Then [NCERT 1982] Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If \[{{R}_{e}}\] is the maximum. The force applied to the cart in the above Question by spring scale FA is still 10.5N. The cart now moves toward the right with an acceleration also toward the right of 1.75m/s^2. If the combined force is ∑F, the mass is m and the acceleration is a, write a mathematical relationship which relates these three physical quantities. ∑F = ma Examples of correct answers: g or 9.8 m/s2 or 10 m/s2 (or just 9.8 or 10) For a correct answer and correct justification 1 point Examples: Nearly equal to g. Because block B is almost in free fall. 10 m/s2, because block A has negligible mass and the tension in the string is nearly zero. Claim: The acceleration of the blocks is close to g

### Acceleration Calculator Definition Formul

• A ball of mass M swings in a horizontal circle at the end of a string of radius R at a constant tangential speed v0. A student gradually pulls the string inward such that the radius of the circle decreases while keeping the tangential speed v0 of the ball constant, as shown above
• MCQ Questions for Class 9 Science: Ch 9 Force and Laws of Motion. 1. A plate, a ball and a child all have the same mass. The one having more inertia is the. 2. The inertia of an object tends to cause the object. 3. The rate of change of momentum with respect to time is measured in. 4
• And that second acceleration, 6 m/s 2, is twice the first acceleration, 3 m/s 2: a 2 = 2(a 1) 6 m/s 2 = 2(3 m/s 2) 6 m/s 2 = 6 m/s 2. Clearly, both the acceleration and the force on this object change by the same factor, 2. These matching factor changes would occur for any factor change and for any constant mass
• e the mass of the object in question. This law can be summed up with the equation F = ma, where F is the force, m is the mass of the object, and a is acceleration. you can still use the above equation, F grav = (GM earth m)/d 2 to deter
• A)5.0 m/s2 to the right B)5.0 m/s2 to the left C) 3.0 m/s2 to the right D)3.0 m/s2 to the left 3.Two forces are applied to a 2.0-kilogram block on a frictionless horizontal surface, as shown in the diagram below. The acceleration of the block is A) 1,000 N B)10 N C)0.1 N D)0.001 N 4.An object with a mass of 0.5 kilogram starts from res
1. Newton's law can be represented by the equation F net = m x a, where F net is the total force acting on the object, m is the object's mass, and a is the acceleration of the object. When using this equation, keep your units in the metric system. Use kilograms (kg) for mass, newtons (N) for force, and meters per second squared (m/s 2) for.
2. From the above equation, it is clear that acceleration due to gravity is more at poles and less at the equator. So if a person moves from the equator to poles his weight decreases as the value of g decreases. Variation of g due to Rotation of Earth. Consider a test mass (m) is on a latitude making an angle with the equator
3. He drops an object from a height of 1.6 m above the surface of the moon and the object takes 1.4 s to fall. Use this data to show that the acceleration due to gravity on the surface of the moon is 1.6 m s-2. (vi)The astronaut has a mass of 120 kg. Calculate his weight on the surface of the moon
4. 8. A satellite with mass m=617kg is placed into a circular orbit 1.00×107mabove the surface of the earth, which has a mass M=5.98×1024kg and a radius r=6.38×106m. a. Calculate the linear speed of the satellite as it orbits the earth at this altitude. b. Calculate the radial acceleration of the satellite
5. For example, the equation above gives the acceleration at 9.820 m/s 2, when GM = 3.986×10 14 m 3 /s 2, and R=6.371×10 6 m. The centripetal radius is r = R cos( φ ) , and the centripetal time unit is approximately ( day / 2 π ), reduces this, for r = 5×10 6 metres, to 9.79379 m/s 2 , which is closer to the observed value

### Question 1c: 2015 AP Physics 1 free response (video

• We can also relate the linear acceleration of the mass in that the linear acceleration is the angular acceleration times the length of the rod (d). If we take these two substitutions and put them into the original F = m a equation, we can wind up with an equation that relates the moment and the angular acceleration for our scenario
• This force accelerates the chair with a mass of 45 kg by 0.6435 m/s^2. The acceleration of the chair is 0.6435 m/s^2 and the normal force is 523.59 N. Ask a Question Ask a Question
• ute, its speed will be A) 50 m/s. B) 500 m/s. C) 3000 m/s. D) 3600 m/s. Answer: C Diff: 1 Topic: Linear Motion 16) Drop a rock from a 5-m height and it accelerates at 10 m/s2 and strikes the ground 1 s later. Drop the same rock from a height of 2.5 m and its acceleration of fall is about A) half as much. B) the same amount
• On substitution using the above equation, it will be seen that the acceleration due to gravity on the 5 kg mass is 1.25G m/s2 while that on the 3 kg mass is 0.75G m/s2. This proves that mass directly affects the gravitational acceleration experienced on a body, i.e, the heavier the mass, the higher the gravitational acceleration felt ### Newton's Second Law - Forces, acceleration and Newton's

• The acceleration vector is pointed towards the center of the mass producing the acceleration. The mass of the body being accelerated isn't used in this equation. Use the mass of the body producing the acceleration. To find the acceleration of objects on Earth use Earth's mass. The simulation below shows a vector field. Each vector shows the.
• A long, thin, rod of mass M = 0.500kg and length L = 1.00 m is free to pivot about a fixed pin located at L/4. The rod is held in a horizontal position as shown above by a thread attached to the far right end. a. Given that the moment of inertia about an axis of rotation oriented perpendicular to the rod and passing through its center of mass.
• M = Mass of object + mass of block = 1 + 5 = 6 kg Let V be the velocity of the combined object after the impact. Movement of combined objects = Total momentum after the impact i.e. 6 V = 10 V = 10/6= 1.67 m/s. Question 14. An object starting from rest travels 20m in the first 2 sec. and 160 m in the next 4 sec
• 11.Repeat steps 2-10, for hanging mass values of m H = 25 g, 35 g, and 45 g. For each value of the hanging weight, measure the acceleration of the cart five times. Be sure to measure the actual mass of the hanging weight for each trial. 12.For each value of the hanging mass, calculate the average value of th
• Problem 1. A block of mass 5 Kg is suspended by a string to a ceiling and is at rest. Find the force Fc exerted by the ceiling on the string. Assume the mass of the string to be negligible. Solution. a) The free body diagram below shows the weight W and the tension T 1 acting on the block. Tension T 2 acting on the ceiling and F c the reaction.
• mass m is at rest. If the blocks had instead stuck together after the collision, with what speed would they move if m = 12 kg ? A. 2.0 m/s B. 2.7 m/s C. 3.2 m/s D. 4.0 m/s E. 4.6 m/s 18. The figure above shows two positively charged particles. The +Q charge is fixed in position, and the +q charge is brought close to +Q and released from rest.
• e (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Answer: Question 14. 10

A simple pendulum of length L and mass (bob) M is oscillating in a plane about a vertical line between angular limit - 0 and +0. For an angular displacement (lol 0), the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions : (1986 - 2 Marks) (a) T cos = Mg A 1.0-kg block slides down a vertical wall while you push on it with a force of 12 N upward at an angle of 30 degrees above the horizontal. What is the magnitude of the block's acceleration (in m. ### Angular Acceleration Physic

For a correct equation for the acceleration 1 point a ABe Bt For multiplying the equation above by the mass of the object 1 point F ma m ABe mABe BtBt Fee .012 1.18 5 0.071 55tt Note: Credit given for using A, B, and m or plugging in the given value A hanging mass m = 4.0 kg is connected by a light cord over a pulley to a mass M = 6.0 kg, which slides on a smooth horizontal surface. The pulley rotates about a frictionless axle and has a radius..

### Question Bank for JEE Main & Advanced Physics NLM

1. particle B of mass 2 m, which hangs freely, vertically below P. The system is released from rest, with the string taut, when A is 1.3 m from P and B is 1 m above the horizontal floor, as shown in Figure 3. Given that B hits the floor 2 s after release and does not rebound, (a) find the acceleration of A during the first two seconds, (2
2. ed by its mass (m) and the acceleration due to gravity (g) affecting that object: w = mg. On Earth, g = 9.8 m/s C
3. e the speed of each object when the two pass each other. b) Deter
4. Use the steps outlined above to find the magnitude of the acceleration a of a chair and the magnitude of the normal force FN acting on the chair: Yusef pushes a chair of mass m = 55.0 kg across a carpeted floor with a force F⃗ p (the subscript 'p' here is lowercase and throughout the question) of magnitude Fp = 160 N directed at θ = 35.0.
5. 2. A pulley of mass ml=M and radius R is mounted on frictionless bearings about a fixed axis through O. A block of equal mass m2=M, suspended by a cord wrapped around the pulley as shown above, is released at time t = 0. The acceleration of the block is measured to be (2/3)g in an experiment using a computer-controlled motion sensor. a
6. Question: In the common setup shown in Figure 1, the hanging cylinder of mass m is released from rest. If the friction between the cart of mass M and the horizontal track is negligible, find the acceleration of the cart and the tension in the rope

### A bead of mass m is located on a parabolic wire with its

1. Doubling the acceleration to 9.72 m/s2 isn't possible simply by suspending more mass because all objects, regardless of their mass, fall freely at 9.8 m/s2 near Earth's surface. If the puck were struck in the same way by an astronaut on a patch of ice on Mars, where the acceleration of gravity is 0.35 g, so that the puck left the hockey stick.
2. 6.7 m/s2. A ball is thrown straight up in the air. When the ball reaches its highest point, which of the following is true? None of the other answers are true. The figure above shows an object of mass 0.4 kg that is suspended from a scale and submerged in a liquid
3. e the mass of the block. The block is now placed on a rough horizontal surface having a coefficient of static friction μs = 0.2, and a coefficient of kinetic (sliding) friction μk = 0.1
4. For a mass m= kg, the elevator must support its weight = mg = Newtons to hold it up at rest. If the acceleration is a= m/s² then a net force= Newtons is required to accelerate the mass. This requires a support force of F= Newtons. Note that the support force is equal to the weight only if the acceleration is zero, and that if the acceleration is negative (downward), the support force is less.
5. A constant horizontal force F is applied to m 1. What is the acceleration of each mass? In one sense, we can (almost) solve this example intuitively -- in our head. A force F is applied to an object whose mass is m = m 1 + m 2. So its acceleration must be. a = F / m. or. a = F / (m 1 + m 2) That's the right answer

### MCQ Questions for Class 11 Physics Chapter 8 Gravitation

1. below. Model it as a sti rod of negligible mass, d = 3.20 m long, joining particles of mass m 1 = 0.130 kg and m 2 = 58.0 kg at its ends. It can turn on a frictionless, horizontal axle perpendicular to the rod and 15.0 cm from the large-mass particle. The operator releases the trebuchet from rest in a horizontal orientation
2. D: 8.9 m/s 2. Answer. 9.8 m/s 2. Q.5> The value of gravitational acceleration is - A: increases as height increase form the earth. B: decreases as height increase from the earth. C: remains constant D: None of the above. Answer. Increases as height increase from the earth. Q.6> Where the value of gravitational acceleration is less due to the.
3. what I want to do in this video is think a little bit about what happens to some type of projectile maybe a ball a ball or rock if I were to throw it up straight up into the air so to do that and what I want to do is on a plot its distance relative to time so there's a few things that I'm going to tell you about my throwing of the rock in the air well I'll have an initial velocity I'll have an.
4. Laws of Motion Class 11 MCQs Questions with Answers. Question 1. Question 2. A spherical ball of mass 10-6 kg hits a wall 1000 times per second normally with a velocity of 1000 m/s and rebounds with same velocity along the initial direction. The force experienced by the wall is
5. Find an expression for P(θ) for the case where (a) the mass m 1 and mass m 2 move as a single unit, and (b) there is relative slipping between mass m 1 and mass m 2. Download solution Problem # N-7: In the pulley system shown, m A = 10 kg, m B = 20 kg, m C = 8 kg, and m D = 5 kg
6. A high diver of mass 68.2 kg jumps off a board 10.4 m above the water. The acceleration of gravity is 9.8 m/s2 . If his downward motion is stopped 3.58 s after he enters the water, what average upward force did the water exert on him? I already calculated that his velocity before hitting the water is 14.277 m/s

### An object with a mass 10kg moves at a constant velocity of

1. The acceleration is produced by the gravitational force that the earth exerts on the object. Applying Newton's second law to an object in free fall gives. W = mg, an equation that relates the mass and weight of an object. Figure 1 gives the free-body force diagram for an object sliding down a frictionless incline that is at an angle, θ, above.
2. Acceleration is the rate of change of velocity with time. Since velocity is a vector, this definition means acceleration is also a vector. When it comes to vectors, direction matters as much as size. In a simple one-dimensional problem like this one, directions are indicated by algebraic sign
3. Mass is a property of an object. It depends on the amount of stuff in it. But weight is a force that depends on the strength of gravity. That is why objects weigh about 1/6 as much on the moon as they do on earth. On earth, the weight of an object is given by W = Mg, where M is its mass and g is the acceleration due to earth's gravity
4. The average acceleration is the total change in velocity divided by the total time. This can be found using the equation a = Δv ÷ Δt. For example, if the velocity of an object changes from 20 m/s to 50 m/s over the course of 5 seconds the average acceleration would be: a = (50 m/s - 20 m/s) ÷ 5s. a = 30 m/s ÷ 5s. a = 6 m/s 2
5. Question Answers; Chapter 10 Gravitation MCQ Test 1 Science Class 9 Science. June 23, Both mass and weight (d) None of the above. 6. Find the gravitational force between two spheres of mass m and m/2 having radius x are in contact with each other. 10. The acceleration due to gravity varies on earth with (a) Distance from equator (b.
6. f. 80.0 m/s 5. The acceleration of an object at a time, t, during a trip of duration, T, is defined to be : All of the above remarks (a .through d) are true of the Ce~ter of Mass Point..y None of the above answers IS true and correct. X 17. A car initially traveling north at 2 rn/s has a constant acceleration of 0.5 rn/s2 northward Question 1: You push with a 20-N horizontal force on a 2-kg mass resting on a horizontal surface against a horizontal frictional force of 12N. What is the acceleration of the mass? 1. Cannot tell from this information. 2. 10 m/s2 3. 6.0 m/s2 4. 4.0 m/s2 a=(F-Ffr)/m=4m/s Substituting this into the above equations we find that a P = 2.2 m/s 2 (centripetal acceleration) Therefore, N 1 = m(g − 2.2) N 2 = m(g + 2.2) With g = 9.8 m/s 2, N 1 = 7.6m N 2 = 12m At the top of the Ferris wheel the passengers experience 0.78g (they feel lighter). At the bottom of the Ferris wheel the passengers experience 1.2g (they feel. Which would require a greater force — accelerating a 2 kg mass at 5 m s-2 or a 5 kg mass at 3 ms-2? Answer: F 1 = m 1 x a 1 =2 x 5 = 10 N, F 2 = m 2 x a 2 = 5 x 3 = 15 N ∴ F 2 > F 1. Question 44. If the mass of a body is doubled, what will happen to its acceleration, if the applied force remains constant ? Answer: a = F/m Question 22. An object of mass m has angular acceleration a = 0.2 rad s2. What is the angular displacement covered by the object after 3 second? (Assume that the object started with angle zero with zero angular velocity). Answer: Given, Angular acceleration = α = 0.2 rad s-2 Time = 3s Initial velocity = 0 Solution   Question 4. One of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10 8 m. Show that the mass of Jupiter is about one-thousandth that of the sun. Answer: For the satellite around Jupiter. T 10 = 1.769 days. R 10 = 4.22 × 10 8 m. For earth around sun Two masses are hung are connected by a light cord and hung from a frictionless pulley of negligible mad are shown. Mass m1=3.00kg and mass m2=2.00kg. When the two masses are released from rest, the resulting acceleration of the . physics. THree blocks of masses 3m, 2m and m are connected to strings A,B, and C Instructions: Once students have worked on deriving expression (1) above, the instructor may proceed with the demonstration as follows. If the accelerometer is used as the mass on the pendulum, the reading at the lowest point of the swing is exactly the value of centripetal acceleration Use the steps outlined above to find the magnitude of the acceleration {eq}a {/eq} of a chair and the magnitude of the normal force {eq}F_N {/eq} acting on the chair  The acceleration due to gravity of that planet whose mass and radius are half those of earth, will be (g is acceleration due to gravity at earth's surface) (a)2g (b)g (c)g/2 (d)g/4 Answer is: (a)2g. Asked by निकोदिमुस 5th December 2017 6:58 PM. Answered by Expert A mass m=0.7 kg is released from rest at the origin O. The mass falls under the influence of gravity. When the mass reaches point A, it is distance x below the origin O; When the mass reaches the p.. Since F grav = F net, the above expressions for centripetal force and gravitational force are equal. Thus, (M sat * v 2) / R = (G* M sat * M Central ) / R 2. Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by M sat