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AleksAgata [21]
3 years ago
13

A roller coaster car with a mass of 17,690 kg is moving at a speed of 5 m/s over a hill that is 87.4 m above the

Physics
1 answer:
Sati [7]3 years ago
3 0

Answer:

E=1.53\times 10^7\ J

Explanation:

Given that,

The mass of the car, m = 17,690 kg

The speed of the car over the hill, v = 5 m/s

The car is 87.4 m above the  ground.

We need to find the total mechanical energy of the car-Earth system. Mechanical energy is equal to the sum of kinetic energy and potential energy. So,

E=mgh+\dfrac{1}{2}mv^2\\\\E=17690\times 9.8\times 87.4+\dfrac{1}{2}\times 17690 \times (5)^2\\\\E=1.53\times 10^7\ J

So, the total mechanical energy of the car-Earth system is 1.53\times 10^7\ J.

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The angular momentum of an electron in each of its orbits must be some multiple of:
Nastasia [14]
The Bohr's proposal for the angular momentum of an electron in Bohr's model of the hydrogen atom is:

L=(n*h)/(2π), where n is the number of the energy level and h is the Planck's constant. This equation shows us the quantization of angular momentum of the electron. So the correct answer is the second one: Planck's constant. 
7 0
3 years ago
The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie
pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

\bf \longrightarrow  {v}^{2}  =  {u}^{2}  + 2as \\  \\ \rm \longrightarrow  {0}^{2}  =  {15}^{2}  + 2 \times a \times 63 \\  \\ \rm \longrightarrow 0 = 225 + 126a \\  \\ \rm \longrightarrow 126a =  - 225 \\  \\ \rm \longrightarrow a =  -  \dfrac{225}{126}  \\  \\ \rm \longrightarrow a =  - 1.78 \: m {s}^{ - 2}

\therefore Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0
3 years ago
Neritic Sediments are deposited on the ocean floor in a sorted manor. In which order are the sediments ordered moving from the s
BaLLatris [955]
B <span> of Earth’s surface is covered by water. Very little or no light penetrates beyond a few hundred feet in water</span>
6 0
4 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
If a tank contains water 4 m deep, what is the pressure at the bottom of the tank? (Neglect the atmospheric pressure.)
makkiz [27]

Given:

Water 4 m deep

Required:

Pressure at the bottom of the tank

Solution:

p2 – p1 = gh

p2 – p1 = p = gh

p = gh = 1000kg/m3 (9.8m/s2)(4m)

<span>p = 39200 Pa</span>

6 0
4 years ago
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