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Sati [7]
3 years ago
11

Two ice skaters stand together as illustrated in the attached figure. They "push off" and travel directly away from each other,

the boy with a velocity of v = +0.50 m/s. If the boy weighs 710 N and the girl 480 N, what is the girl's velocity in m/s after they push off? (Consider the ice to be frictionless.)
Physics
1 answer:
Tema [17]3 years ago
6 0
By definition, the momentum is given by:
 p = m * v
 Where,
 m = mass
 v = speed.
 On the other hand,
 F = m * a
 Where,
 m = mass
 a = acceleration:
 For the boy we have:
 p1 = m * v
 p1 = (F / a) * v
 p1 = ((710) / (9.81)) * (0.50)
 p1 = 36.19 Kg * (m / s)
 For the girl we have:
 p2 = m * v
 p2 = (F / a) * v
 p2 = ((480) / (9.81)) * (v)
 p2 = 48.93 * v Kg * (m / s)
 Then, we have:
 p1 + p2 = 0
 36.19 + 48.93 * v = 0
 Clearing v:
 v = - (36.19) / (48.93)
 v = -0.74 m / s (negative because the velocity is in the opposite direction of the boy's)
 Answer:
 the girl's velocity in m / s after they push off is -0.74 m / s
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6 0
3 years ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
Which has the larger kinetic energy, a 10 g bullet fired at 400 m/s or a 80 kg bowling ball rolled at 6.5 m/s ?
mixer [17]
Formulae for Kinetic energy is:
Kinetic Energy= 1/2xmassx(velocity)^2

For comparison we need to have same units,thus we convert 10g into Kg.
10g/1000=0.01Kg

Input the value of bullet in the formulae;
Kinetic Energy= 1/2x0.01kgx(400)^2
K.E=800J

Input value of the ball:
Kinetic Energy=1/2x80kgx(6.5)^2
K.E=1690J

Which means that th Energy of the ball is more than the bullet.
7 0
3 years ago
An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the
Lerok [7]
Answer:

Explanation:
Kinematics equation for first Object:

but:
The initial velocity is zero

it reach the water at in instant, t1, y(t)=0:


Kinematics equation for the second Object:
The initial velocity is zero

but:

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s


The velocity is negative, because the object is thrown downwards
7 0
3 years ago
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