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4 years ago

What is the force being applied to an object with a mass of 12 kg and an acceleration of 4 m/s2?

Physics

1 answer:

Jet001 [13]4 years ago

7 0

Answer:

48 Newtons

Explanation:

Newtons Second law Force = Mass* Acceleration

12*4 = 48 Newtons Of force

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A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo

Alex

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

Speed = $\frac{distance}{time taken}$

Since time is unknown, we make it the subject of the expression;

time = $\frac{distance }{speed}$ = $\frac{400}{600}$

time = 0.67s

Learn more:

Speed brainly.com/question/10048445

#learnwithBrainly

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3 years ago

The luminous star Alnilam in the Orion belt is 1,340 light-years away from Earth. Use the conversion factor 1 parsec = 3.262 lig

svlad2 [7]

The answer is 410.8 pc.

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4 years ago

Read 2 more answers

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

3 years ago

How to atoms behave in non-magnetic items?

Anastaziya [24]

Answer:

By altering the quantum interactions of the electrons in the atoms of a metal's atoms, scientists from the University of Leeds have generated magnetism in metals that aren’t normally magnetic.

Explanation:

5 0

3 years ago

A student is building a simple circuit with a battery, light bulb, and copper wires. When she connects the wires to the battery

denis23 [38]

It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open

5 0

3 years ago

Read 2 more answers