The expression for the acceleration of the cart for the experiment you did in la

Question

The expression for the acceleration of the cart for the experiment you did in lab is a = (m1g ? f) (m1 + m2) where m1 + m2 is kept constant throughout. (a) Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a = 1

If you now double the hanging mass m1, (so m = 2m1), how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a = 3

Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart? The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2.
(b) Consider the situation when friction force is non-zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a = 5

If you now double the hanging mass m1, (so m = 2m1) how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a = 7

Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart? The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2. The expression for the acceleration of the cart for the experiment you did in lab is a = (m1g ? f) (m1 + m2) where m1 + m2 is kept constant throughout. (a) Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a = 1

If you now double the hanging mass m1, (so m = 2m1), how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a = 3

Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart? The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2.
(b) Consider the situation when friction force is non-zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a = 5

If you now double the hanging mass m1, (so m = 2m1) how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a = 7

Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart? The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2. a = (m1g ? f) (m1 + m2) m1 + m2 (a) Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a = 1

If you now double the hanging mass m1, (so m = 2m1), how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a = 3

Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart? The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2.
(b) Consider the situation when friction force is non-zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a = 5

If you now double the hanging mass m1, (so m = 2m1) how will the value of the denominator for the expression for the acceleration change? The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a = 7

Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart? The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2. 1 The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment. 3 The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2. 5 The denominator will double in value. The denominator value will be 2m1 + m2. Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment. 7 The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled. The cart's acceleration will remain the same as the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of 2. (m1g ? f) (m1 + m2)

Explanation / Answer

1)a = (m1g)/(m1+m2)

The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled.

2)a = (m1g-f)/(m1+m2)

Doubling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.

3)The cart's acceleration will double as the numerator in the expression for the acceleration will double when m1 is doubled.

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