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aniked [119]
2 years ago
14

Think about Newton's 2nd Law, Force equals mass times acceleration. Liz puts a 1 kg weight and a 10 kg weight on identical wagon

s. She then applies a 10N force to each wagon. What will be true about the acceleration of the bottom wagon that has more mass?
Question 16 options:
1 The acceleration of the bottom wagon will be greater.
2 The acceleration of the bottom wagon will be smaller.
3 The bottom wagon will accelerate the same amount as the top one.
Physics
2 answers:
Sonja [21]2 years ago
7 0

The acceleration of the bottom wagon will be greater. hope it helps TR3way

Alexeev081 [22]2 years ago
5 0
More force is needed for more mass. Therefore, if the mass is greater and the force is not enough then the object will less likely accelerate
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8 months ago
Which situation is work not being done? A) A bookcase is slid across carpeting. B) A stack of books is carried at waist level ac
Nastasia [14]

Answer:

B) A stack of books is carried at waist level across a room

Explanation:

Work is defined as:

W=Fd cos \theta

where

F is the force applied

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

From the formula, we see that the work done is zero when the force and the displacement are perpendicular to each other. Let's now analyze each situation:

A) A bookcase is slid across carpeting. --> work is done, because the force that pushes the bookcase is in the same direction of the displacement

B) A stack of books is carried at waist level across a room. --> no work is done, because the force to carry the book is vertical, while the displacement of the books is horizontal

C) A chair is lifted vertically with respect to the floor. --> work is done, because the force that lifts the chair is vertical, and the displacement is vertical as well

D) A table is dropped onto the ground. --> work is done, because the force of gravity (that makes the table falling down) is vertical and the displacement of the table is also vertical.

5 0
3 years ago
When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____
valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0
3 years ago
A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0
Kazeer [188]

Answer:

\Delta K = 52J

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: \Delta K=K_f-K_i

We know that the formula for the kinetic energy for an object is:

K=\frac{mv^2}{2}

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}

Which for our values is:

\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J

3 0
3 years ago
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