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3 years ago

What would be the mass of an object that is moving at 5 m/s, with a total momentum of 1500 kg*m/s southeast?

Physics

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

The mass of the object is 300 kg

Explanation:

The given parameters are;

The velocity of the object = 5 m/s southeast

The total momentum of the object = 1500 kg·m/s

The equation for linear momentum, is given as follows;

Momentum = Mass × Velocity

Therefore, the mass of the object can be found as follows;

$The \ mass \ of \ the \ object= \dfrac{The \ momentum \ of \ the \ object}{The \ velocity \ of \ the \ object}$

Therefore, we have;

$The \ mass \ of \ the \ object= \dfrac{1500 \ kg \cdot m/s}{5 \ m/s} = 300 \ kg$

The mass of the object = 300 kg.

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When we advance our technology we increase the things we can learn and scientists can increase the amount of scientific knowledge we have.

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3 years ago

When is the force on a current-carrying wire in a magnetic field at its strongest?

Hitman42 [59]

The forces on a current-carrying wire in a magnetic field are at their strongest when the current is at a 90-degree angle to the field. Option D is correct.

<h3>What is a magnetic field?</h3>

It is the type of field where the magnetic force is obtained. The magnetic force is obtained by the field felt around a moving electric charge.

The complete question is;

"When is the force on a current-carrying wire in a magnetic field at its strongest?

-when the current is at a 0-degree angle to the field

-when the current is at a 30-degree angle to the field

-when the current is at a 45-degree angle to the field

-when the current is at a 90-degree angle to the field"

The magnetic force is found as;

F=BILSINΘ

Where,

Magnetic Field, B

Length of the wire, L

The angle between field and current, Θ

When Θ=90°

The value of the magnetic force is;

F=BIL

When the current is flowing at a 90-degree angle to the magnetic field, the forces acting on a wire carrying a current are the strongest.

Hence, option D is correct.

To learn more about the magnetic field, refer to the link;

brainly.com/question/19542022

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1 year ago

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leav

nata0808 [166]

Answer:

F = -307.4 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 60 m/s

Final speed of the baseball, $v=65\ cos(30)=56.29\ m/s$

Time of contact, $t=1.75\ ms=1.75\times 10^{-3}\ s$

(a) It is assumed to find the horizontal component of average force. It is given by :

$F=m\dfrac{v-u}{t}$

$F=0.145\dfrac{56.29-60}{1.75\times 10^{-3}}$

F = -307.4 N

So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.

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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

3 years ago

A body is oscillating up and down at the end of a spring. Let’s consider when the body is at the top of its up-and-down motion.

Klio2033 [76]

The velocity of the body is zero; option A

<h3>What is the motion of an oscillating body?</h3>

The motion of an oscillating body is known as simple harmonic motion.

Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.

For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.

In conclusion, oscillating bodies undergo simple harmonic motion.

Learn more about simple harmonic motion at: brainly.com/question/24646514

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3 0

1 year ago