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2 years ago

14

A 60-kg skier starts at the top of a 10-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose

to friction?

1 answer:

Karo-lina-s [1.5K]2 years ago

4 0

The potential energy of the skier at the top is
60 kg (9.81 m/s2) (10 m) = 5886 J

Her kinetic energy at the bottom is
(1/2) (60 kg) (10 m/s)^2 = 3000 J

The energy lost to friction is
5886 J - 3000 J = 2886 J

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A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

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Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

Explanation:

Given that,

Electric field $E=-6.00\times10^{5}i\ N/C$

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

$F=F_{e}$

$ma=qE$

$a=\dfrac{Eq}{m}$

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

$a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}$

$a=-5.74\times10^{13}\ m/s62$

The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). We need to calculate the initial peed

Using equation of motion

$v^2-u^2=2as$

Where, s = distance

Put the value into the formula

$0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}$

$u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}$

$u=2834783.94$

$u=2.83\times10^{6}\ m/s$

The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

$t=\dfrac{u}{a}$

Where, u = initial velocity

a = acceleration

Put the value into the formula

$t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}$

$t=0.493\times10^{-7}\ sec$

Hence, (a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$

(b). The initial peed of the protion is $2.83\times10^{6}\ m/s$

(c). The time is $0.493\times10^{-7}\ sec$ .

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2 years ago

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

Semenov [28]

Answer:

A)Object 1 has the greater magnitude of its momentum.

B)The objects 2 have the greater kinetic energy.

Explanation:

For object 1 :

v₁ = v ,m₁ = 2 m

For object 2 :

$v_2=2\sqrt{v}$ ,m₂=m

We know that linear momentum given as

P = M V

M=Mass , V=Velocity

For object 1 :

P₁ =m₁ v₁

P₁ =2 m v

For object 2

$P_2=m_2v_2$

$P_2=2m\sqrt {v}$

We can say that object 1 have more momentum.

The kinetic energy

$KE_1=\dfrac{1}{2}m_1v_1^2$

$KE_1=\dfrac{1}{2}\times 2m\times v^2$

$KE_1=mv^2$

$KE_2=\dfrac{1}{2}m_2v_2$

$KE_2=\dfrac{1}{2}\times m\times 4v^2$

$KE_2=2mv^2$

Therefore both the object 2 have higher kinetic energy.

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