1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
aksik [14]
3 years ago
8

Later that day, Martin does the same thing with Josh, who is 2 times heavier than Martin. If the collision is totally inelastic,

what height does Josh reach?
a. h/25
b. h/16
c. h/8
d. h/4
e. h/2
f. h
g. None of the above.
Physics
1 answer:
charle [14.2K]3 years ago
6 0

Answer:

Josh reaches at  \dfrac{h}{9}  height.

(g) is correct option

Explanation:

Given that,

Mass of martin = m

Mass of josh m'= 2m

Martin and Paul are skateboarding in a large semicircular halfpipe. Martin starts out from rest at a height h and collides with Paul standing at the bottom. Martin and Paul have about the same mass. After the collision,

We need to calculate the speed before the collision

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

v=\dfrac{2gh}

We need to calculate the final speed after collision

Using conservation of momentum

mv_{i}=(m+2m)v_{f}

Put the value into the formula

v_{f}=\dfrac{\sqrt{2gh}}{3}

We need to calculate the height

Using conservation of energy

\dfrac{1}{2}\times m\times v^2=mgH

Put the value into the formula

\dfrac{1}{2}\times 3m\times(\dfrac{2gh}{9})=3mgH

H=\dfrac{h}{9}

Hence, Josh reaches at \dfrac{h}{9} height.

You might be interested in
Kepler’s first law states that the orbits of planets are ellipses with the Sun at one ____.
oee [108]
The answer is focus.
4 0
3 years ago
Read 2 more answers
Calculate the amount of heat needed to raise 1.0 kg of ice at -20 degrees Celsius to steam at 120 degree Celsius
CaHeK987 [17]

Answer:

801.1 kJ

Explanation:

The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.

The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m =  mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20  °C = 4216 J

The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m =  mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100  °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m =  mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20  °C = 39920 J

The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J

+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ

4 0
3 years ago
A ball is thrown horizontally at a speed of 16 m/s from the top of a cliff. If the ball hits the ground 6.0 s later, approximate
lina2011 [118]

Answer:

Y = 176.4 m

Explanation:

For the height of cliff we will analyze the vertical motion. We will apply the 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = ?

V₀y = Initial Vertical Velocity = 0 m/s (since, ball is thrown horizontally)

t = time = 6 s

g = 9.8 m/s²

Therefore,

Y = (0 m/s)(6 s) + (0.5)(9.8 m/s²)(6 s)²

<u>Y = 176.4 m</u>

3 0
3 years ago
PLEASE HELP ME!!!
denis-greek [22]
The reason for that is that P-waves (primary waves) travel faster than S-waves (secondary waves).

If we call v_p the speed of the primary waves and v_s the speed of the secondary waves, and we call S the distance of the seismogram from the epicenter, we can write the time the two waves take to reach the seismogram as
t_P =  \frac{S}{v_P}
t_S= \frac{S}{v_S}

So the lag time between the arrival of the P-waves and of the S-waves is
\Delta t = t_S-t_P= \frac{S}{v_S}- \frac{S}{v_P}= S(\frac{1}{v_S}- \frac{1}{v_P})
We see that this lag time is proportional to the distance S, therefore the larger the distance, the greater the lag time.
6 0
2 years ago
The law of repulsion by Coulomb agrees with:
r-ruslan [8.4K]
<span>In Coulomb's law, however, the magnitude and sign of the electric force are determined by the electric charge, rather than the mass, of an object. ... Thus, two negative charges repel one another, while a positive charge attracts a negative charge. The attraction or repulsion acts along the line between the two charges</span>
8 0
3 years ago
Read 2 more answers
Other questions:
  • Which of the following objects is usually the smallest?
    10·2 answers
  • Two small plastic spheres between them has magnitude 0.22 N. What is the charge on each sphere is one the other? Explain whether
    12·1 answer
  • Which of the following is a likely outcome of a failed experiment?
    12·1 answer
  • A ball is thrown upward in the air with an initial velocity of 40 m/s. How long does it take to
    14·1 answer
  • When a compass needle settles down in a magnetic field, _______. the needle aligns itself with the field, the south end of the c
    8·1 answer
  • Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
    7·1 answer
  • an object is thrown straight up. which of the following is true abought the sign of workdone by the force of gravity on the obje
    6·1 answer
  • Equation for distance
    5·1 answer
  • In addition to ozone, what four other greenhouse gases or groups of greenhouse gases are included in nearly all the climate mode
    8·1 answer
  • What is the change in entropy of 0.130 kg of helium gas at the normal boiling point of helium when it all condenses isothermally
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!