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3 years ago

A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What

Physics

1 answer:

enot [183]3 years ago

4 0

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

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What is work please give ans

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Work= force x displacement :)

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2 years ago

Compare relative age with absolute age.

oee [108]

Answer & Explanation:

Relative age is the age of a rock layer (or the fossils it contains) compared to other layers. It can be determined by looking at the position of rock layers. Absolute age is the numeric age of a layer of rocks or fossils.

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3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

4 0

2 years ago

What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?

skad [1K]

Answer:

100,048

Explanation:

K.E = 1/2 m (v)^2

K.E = 1^/2 * 74 * (52)^2

K.E = 100,048J =100.048kJ

8 0

3 years ago

When light is reflected, the incident rays are bent and change direction.<br> True<br> False

Helga [31]

Answer: True

Explanation: When light is reflected off lets say a mirror it is bent and changes direction to bounce off of another wall or object. For example if you take a flash light and shine it into a mirror the light reflects into a different direction your welcome

8 0

3 years ago

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