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inessss [21]
3 years ago
13

Jay bought a toy car. To make the car move, he must turn a key attached to a spring. Turning the key winds the spring tight. The

toy car moves when the key is released and the spring unwinds. What’s the main transformation of energy in this example?
Physics
2 answers:
lapo4ka [179]3 years ago
5 0
The chemical energy in Jay's body, to kinetic energy in the car
Brilliant_brown [7]3 years ago
4 0

Answer:

Elastic potential energy into kinetic energy

Explanation:

Initially, all the energy is stored in the spring, which is compressed. This energy is called elastic potential energy, and it is equal to

U=\frac{1}{2}kx^2

where k is the spring constant and x is the compression of the spring. When the key is turned, the spring is released, and all this energy is converted into kinetic energy, which is the energy due to the motion of the car. The equation that gives this energy is

K=\frac{1}{2}mv^2

where m is the mass of the car and v is the speed of the car.

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HELPPPPP !!!!
Nataly [62]

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

4 0
3 years ago
A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary
Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

  • <em>mass of the bullet, m = 23 g = 0.023 g</em>
  • <em>speed of the bullet, u = 230 m/s</em>
  • <em>mass of the wood, m = 2 kg</em>
  • <em>final speed of the bullet, v = 170 m/s</em>
  • <em>coefficient of friction, μ = 0.15</em>

The final velocity of the wood after the bullet hits is calculated as follows;

m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s

The acceleration of the wood is calculated as follows;

\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2

The distance traveled by the wood after the bullet emerges is calculated as follows;

v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

Learn more here:brainly.com/question/15244782

7 0
3 years ago
Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc
TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0
3 years ago
A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
Papessa [141]

Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

7 0
3 years ago
Read 2 more answers
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
3 years ago
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