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MatroZZZ [7]
3 years ago
11

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. When the box reaches position

x2 (where x2>x1), how much work Wp has the person done on the box? Assume that the box reaches x2 after the person has accelerated it from rest to speed v1. Express the work in terms of m, v0, x1, x2, and v1.
Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

Answer:

w_p = \frac{1}{2}mv_1^2+\frac{1mv_0^2}{2x_1}(x_1-x_2)

Explanation:

First, the box goes from position x=0 to x=x1, so, the work made by the friction is calculated as:

-F_f(x_1) = \frac{1}{2}mV^2 -\frac{1}{2}mv_0^2

where F_f is the force of friction, x_1 the displacement, m the mass, V is the speed in rest and v_0 the initial speed.

V is equal to zero, so:

F_f(x_1)=\frac{1}{2}mv_0^2

it means that the force of friction F_f is equal to:

F_f = \frac{1mv_0^2}{2x_1}

On the other hand, when the box goes from position x=x1 to x=x2, the work-energy theorem said that:

-F_f(x_1-x_2)+w_p = \frac{1}{2}mv_1^2-\frac{1}{2}mV^2

Where w_p is the work done by the person.

Then, solving for w_p:

-F_f(x_1-x_2)+w_p = \frac{1}{2}mv_1^2

w_p = \frac{1}{2}mv_1^2+F_f(x_1-x_2)

Finally, replacing F_f by \frac{1mv_0^2}{2x_1}, we get:

w_p = \frac{1}{2}mv_1^2+\frac{1mv_0^2}{2x_1}(x_1-x_2)

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a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what
mamaluj [8]

0.04m²

Explanation:

Given parameters:

Pressure = 250000Pa

Weight = 40000N

Unknown:

Area of each foot = ?

Solution:

Pressure is the force exerted per unit area of a body

  Pressure = \frac{force}{area}

To find the area;

        Area = \frac{force }{pressure}

    Area = \frac{40000}{250000} = 0.16m²

The force exerted by all the four feet is 0.16m²

the area of each feet = \frac{0.16}{4} = 0.04m²

Learn more:

Pressure brainly.com/question/7139767

#learnwithBrainly

8 0
3 years ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, a=0.20 m/s^2

We have to find the final velocity at the end of the 100.0 m.

We know that

v^2-u^2=2as

Using the formula

v^2-(1.4)^2=2\times 0.20\times 100

v^2-1.96=40

v^2=40+1.96

v^2=41.96

v=\sqrt{41.96}

v=6.5 m/s

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0
2 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequ
Galina-37 [17]

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

5 0
2 years ago
) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular
DIA [1.3K]

Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

1 / 0,1 = 10 Hz

5 0
2 years ago
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