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prisoha [69]
2 years ago
13

A roller coaster has a mass of 450 kgIt sits at the top of a hill with height 49 m. If it drops from this hill, how fast is it g

oing when it reaches the bottom? Assume there is no air resistance or friction.)
Physics
1 answer:
fgiga [73]2 years ago
6 0

The speed of the roller coater at the bottom of the hill is 31 m/s.

<h3>Speed of the roller coater at the bottom of the hill</h3>

Apply the principle of conservation of mechanical energy as follows;

K.E(bottom) = P.E(top)

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

  • v is the speed of the coater at bottom hill
  • h is the height of the hill
  • g is acceleration due to gravity

v = √(2 x 9.8 x 49)

v = 31 m/s

Thus, the speed of the roller coater at the bottom of the hill is 31 m/s.

Learn more about speed here: brainly.com/question/6504879

#SPJ1

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Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

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2 years ago
Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t
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Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}

Where K, U, W are linear kinetic energy, gravitational potential energy and work, respectively.

8 0
3 years ago
Sort the statements about heat transfer into the correct columns.
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A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl
kondaur [170]

Answer:

Part a)

y_m = 0.157 mm

part b)

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

Explanation:

As we know that the speed of wave in string is given by

v = \sqrt{\frac{T}{m/L}}

so we have

T = 17.5 N

m/L = 5.4 g/cm = 0.54 kg/m

now we have

v = \sqrt{\frac{17.5}{0.54}}

v = 5.69 m/s

now we have

Part a)

y_m = amplitude of wave

y_m = 0.157 mm

part b)

k = \frac{\omega}{v}

here we know that

\omega = 2\pi f

\omega = 2\pi(92.2) = 579.3 rad/s

so we  have

k = \frac{579.3}{5.69}

k = 101.8 rad/m

Part c)

\omega = 579.3 rad/s

Part d)

here since wave is moving in negative direction so the sign of \omega must be positive

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3 years ago
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