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3 years ago

Which of these variables affect the kinetic energy of a body? Select all that apply.

1 answer:

5 0

Speed, mass, and acceleration. Acceleration affects the kinetic energy, due to it affecting speed; therefore, the more acceleration, the more of an increase/decrease in kinetic energy.

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A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>

Mass of the box, m = 3.1 kg
Distance moved by the box, d = 2.5 m
Coefficient of friction, = 0.35
Inclination of the force, θ = 30⁰

Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

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

8 0

2 years ago

Help me please this is for physics

Yuri [45]

<h2>Hello there! :)</h2>

It's a pleasure to be helping you today with your<u> physics question!</u>

Answer:

$23.1m/s$

Explanation:

We want to find the initial speed of the ball.

To do this, we have to apply the formula for the time of flight of a projectile:

$T=\frac{2_{v0~sin 0} }{g}$

where θ = angle of flight

g = acceleration due to gravity

v0 = initial speed

Therefore, substituting the given values into the formula, we have that:

$\boxed{4.2=\frac{2~x~_{v0~sin63} }{9.8}}$

⇒ 2 × $_{v0}$ ×0.8910= 9.8 × 4.2

⇒ $\boxed{{v0}=\frac{9.8~times~4.2}{2~times~0.8910}}$

$\boxed{{v0} =23.1m/s}$

That is the initial speed of the ball.

<em />

<em>I hope this helps you!</em>

<em>Good Luck with your Assignment!</em>

3 0

3 years ago

A jetliner can fly 4.9 hours on a full load of fuel. Without any wind it flies at a speed of 1.88 x 102 m/s. The plane is to mak

PSYCHO15rus [73]

Answer:

1371.26 Km

Explanation:

First of all, we need to find the velocity of the plane relative to the ground since the air has a velocity of 78.2 m/s due east and without any wind, it flies at a velocity of 188 m/s.

Thus, during the west trip, the velocity will be;

Vw = Vp - Va

Vp is velocity of plane while Va is velocity of air

and since distance/time = velocity ;

Time = velocity/distance and thus;

Time during this west period ;Tw = X/(Vp - Va)

Now during the east trip,

Ve = Vp + Va

And Te = X/(Vp + Va)

From the question, the plane can fly 4.9 hours on a full load of fuel. Let's convert this to seconds because velocity is in m/s

Thus, 4.9 hours = 4.9 x 60 x 60 = 17640 seconds

So, this time will be equal to the sum of that in the west and east directions.

Thus; T = Tw + Te

From above we know Tw and Te.

Let's substitute them into the equation;

T = [X/(Vp - Va)] + [X/(Vp + Va)]

T = X[(Vp + Va + Vp - Va)/((Vp)² — (Va)²)

T = X[(2Vp)/((Vp)² — (Va)²)

Making X the subject to obtain;

X = [T((Vp)² — (Va)²)]/(2Vp)

X = [17640((188)² — (78.2)²)]/(2 x 188)

X = 515595326.4/376 = 1371264.17m = 1371.26 Km

6 0

3 years ago

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0

3 years ago

A toy train is pushed forward and released at 2.0m with a speed of 1.0m/s. it rolls at a steady speed for 2.0s, then one wheel b

Mila [183]

Recall that

${v_f}^2-{v_0}^2=2a\Delta x$

where

- $v_f$ is the final velocity of the train; 0 in this case, because the train eventually stops

- $v_0$ is the initial velocity of the train; 1.0 m/s in this case, because this initial velocity refers to velocity it starts with after one of the wheels gets stuck, and the train is initially traveling at a constant speed of 1.0 m/s

- $a$ is the acceleration we want to find

- $\Delta x$ is the change in the train's position; 5.0 m in this case

So we have

$-\left(1.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2a(5.0\,\mathrm m)$

$\implies a=-0.10\,\dfrac{\mathrm m}{\mathrm s^2}$

4 0

4 years ago