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2 years ago

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A 15kg hoop with a radius of 3 m is rolling at on a level surface when it reaches a 25 degree incline. If it reaches a height of

4 m, what was the hoop's initial velocity?

1 answer:

dimulka [17.4K]2 years ago

4 0

The initial velocity of the hoop is determined as 8.854 m/s.

<h3>Conservation of energy</h3>

The initial velocity of the hoop can be determined from the principle of conservation of energy.

Final potential energy = Initial kinetic energy

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

2gh = v²

√2gh = v

√(2 x 9.8 x 4) = v

8.854 m/s = v

Thus, the initial velocity of the hoop is determined as 8.854 m/s.

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The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa

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Explanation:

Formula to represent thrust is as follows.

F = $\frac{dP}{dt}$

= $\frac{2p}{dt}$

or, p = $\frac{E}{c}$

$\frac{p}{dt} = \frac{W}{c}$

F = $\frac{2IA}{c}$

= $\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}$

= 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

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3 years ago

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

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To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

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3 years ago

Spaceship 1 and spaceship 2 have equal masses of 300 kg. spaceship 1 has a speed of 0 m/s, and spaceship 2 has a speed of 6 m/s.

Kipish [7]

The final speed is 3 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces, the total momentum of the two spaceships must be conserved. So we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

where

$m_1 = m_2 = 300 kg$ is the mass of each spaceship

$u_1 = 0 m/s$ is the initial velocity of spaceship 1

$u_2 = 6 m/s$ is the initial velocity of spaceship 2

v is the combined velocity of the two spaceships after the collision

Solving for v, we find

$v=\frac{m_2 u_2}{m_1+m_2}=\frac{(300)(6)}{300+300}=3 m/s$

So, their final speed is 3 m/s.

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3 years ago

Radio waves travel at the speed of light, which is 3.00 u 108 m/s. How many kilometers will radio messages travel in exactly one

egoroff_w [7]

Answer:

Distance traveled will be $9.462\times 10^{12}km$

Explanation:

We have given speed of the light $v=3\times 10^8m/sec$

We have to find the distance traveled by light in 1 year

We know that 1 year = 8760 hour

And 1 hour = 60×60 = 3600 sec

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We know that distance = speed × time

So distance traveled by light in one year $=3.154\times 10^7\times 3\times 10^8=9.462\times 10^{15}m$

We have to fond the distance in km

As we know that 1 km = 1000 m

So $9.462\times 10^{15}m=\frac{9.462\times 10^{15}}{1000}=9.462\times 10^{12}km$

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2 years ago

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Contact area is not a factor of the amount of friction force acting between two bodies in contact, preventing rolling or sliding

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The coefficient of friction is higher in softer tires than hard tires which are required to be wider to reduce the pressure that can cause deformation withstood by hard tires.

The wide tires in drag racing cars can be thought of being made from the soft compound tire material having a higher coefficient of friction that increases the friction force.

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2 years ago