Explain why sound energy is not a type of potential energy.

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

2 years ago

What is the distance a bullet will travel when launched at 166 meters per second for 5.75seconds?

vodka [1.7K]

166x5.75= 954.5 meters in 5.75 seconds.

3 0

3 years ago

A 97 kg man lying on a surface of negligible friction shoves a 62 g stone away from himself, giving it a speed of 2.6 m/s. What

tangare [24]

Answer:

man will move in opposite direction with speed

$v_1 = 1.66 \times 10^{-3} m/s$

Explanation:

As we know that man is lying on the friction-less surface

so here net force along the surface is zero

so if we take man + stone as a system then net change in momentum of this system will become zero

so here we have

$P_i = P_f$

$0 = m_1v_1 + m_2v_2$

here we have

$0 = (97)v_1 + 0.062(2.6)$

$v_1 = -\frac{0.1612}{97}$

$v_1 = -1.66 \times 10^{-3} m/s$

3 0

2 years ago

A car slows from 27 m/s to 5 m/s with a constant acceleration for 6.87 s. What is the car’s acceleration?

Kobotan [32]

In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

a = (Vf-Vo)/ t

a: acceleration (m/s2)

Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s<span>^2</span>

5 0

2 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

6 0

3 years ago

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