1. In this activity, you will be looking for a relationship between the mass of the cart and the acceleration of the cart.

In which scenario will the two objects have the least gravitational force between them?

yawa3891 [41]

Answer:

A

Explanation:

7 0

3 years ago

Read 2 more answers

An elevator carrying a person of mass m is moving upward and slowing down. How does the magnitude f of the force exerted on the

V125BC [204]

The force f of the elevator on the man keeps reducing as the elevator keeps going up while the gravitational force mg keeps increasing moving upwards.

<h3>What is an elevator?</h3>

An elevator is an electrical device that lifts people up and down a tall building or structure.

for the elevator to go up, f > mg.

for the elevator to come down mg > f.

Analysis

since the force on the man is f = ma

where a is the acceleration of the elevator, then it means when a increases, f will increase and when it decreases, f would decrease. slowing down means a, is decreasing going up and this reduces the force as the elevator keeps going up.

on the other hand, gravity acts faster on bodies that are slower in motion so since g, increases going up, mg would also increase.

Learn more about forces in an elevator : brainly.com/question/13526583

#SPJ1

6 0

2 years ago

What was the greatest discovery by galileo during his inclined-plane experiments?

Sergeeva-Olga [200]

Galileo discovered during his inclined-plane experiments that a ball rolling down an incline and onto a horizontal surface would roll indefinitely.

3 0

3 years ago

Miswer

quester [9]

Answer: 4.0

Explanation:

5 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago