A solid yellow line on your side of the center stripe means pass with care.<br> A. True<br> B. False

The following graph shows the force exerted on and the displacement of object being pulled

Tomtit [17]

The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.

<h3>Work done by the applied force</h3>

The area under force versus displacement graph is work done.

The total work done by pulling the object 7 m, can be grouped into two areas;

First area, A1 = area of triangle from 0 m to 2.0 m
Second area, A2 = area of trapezium, from 2.0 m to 7.0 m

A1 = ¹/₂ bh

A1 = ¹/₂ x (2) x (20)

A1 = 20 J

A2 = ¹/₂(large base + small base) x height

A2 = ¹/₂[(7 - 2) + (7-3)] x 50

A2 = ¹/₂(5 + 4) x 50

A2 = 225 J

<h3>Total work done </h3>

W = A1 + A2

W = 20 J + 225 J

W = 245 J

Learn more about work done here: brainly.com/question/8119756

3 0

2 years ago

A 2000kg suv accelerates from rest at a rate of 3.00m/s^2. The total amount of force resisting its motion 1500N. How much force

choli [55]

The total force that the SUV exerts is:

F = 2000 kg * 3 m/s^2

F = 6000 N

Since a resisting force amounting to 1500 N is exerted, then the force exerted by the SUV tires is:

F tire = 6000 N – 1500 N

F tire = 4500 N

7 0

3 years ago

A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel

Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0

3 years ago

For a ∆x of 0.1mm, what is ∆px, the uncertainty in the transverse momentum of a photon passing through a slit (where uncertainty

Colt1911 [192]

Answer:

$0.53\times 10^{-30}kgms^{-1}$

Explanation:

Uncertainty principle say that the position and momentum can not be measured simultaneously except one relation which is described below,

$\Delta x\Delta p=\frac{h}{4\pi }$

Given that the uncertainty in x is 0.1 mm.

Therefore,

$\Delta p=\frac{6.626\times 10^{-34} }{4\times 3.14\times 1\times 10^{-4}m }\\\Delta p=0.53\times 10^{-30}kgms^{-1}$

Therefore, uncertainty in the transverse momentum of photon is $0.53\times 10^{-30}kgms^{-1}$

6 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?