A gas has a volume of 1.5 L at 375 kPa.

A guitarist finds that the pitch of one of her strings is slightly flat—the frequency is a bit too low. Should she increase or d

Yuri [45]

Answer:

The guitarist should increase the tension of the string.

Explanation:

The frequency of a vibrating string is determined by fn=(n/(2L))√T/μ. So if the tension in the string increased, the rate at which it vibrates (the frequency) will also increase.

Therefore it is advisable that she increase the tension of the string.

I hope it helps, please give brainliest if it does

6 0

3 years ago

A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = $\frac{1}{2}mu^2$

Final kinetic energy = $\frac{1}{2}mv^2$

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2$

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2$

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0

3 years ago

A=(v – u)/t<br><br> i. V as subject<br> ii. U as subject<br> iii. T as subject

Norma-Jean [14]

Answer:

i.

$a = \frac{(v - u)}{t} \\ v - u = at \\ v = at + u$

ii.

$a = \frac{(v - u)}{t} \\ v - u = at \\ u = v - at$

iii.

$a = \frac{(v - u)}{t} \\ ta = v - u \\ t = \frac{(v - u)}{a}$

I hope I helped you^_^

8 0

3 years ago

When bouncing a ball, the bouncing motion results in the ball ____________.

yan [13]

Answer:

B) changing position

Explanation:

When a ball bounces to the ground it hits the ground with some energy. The amount of energy with which it hits the ground is kinetic energy. When it comes in the contact with the ground kinetic energy gets converted into potential energy. This potential energy again gets converted into kinetic energy and balls moves again from the ground and bounces multiple times. So, due to multiple bounce the position of the ball changes.

Thus, When bouncing a ball, the bouncing motion results in the ball changing position.

6 0

3 years ago

Read 2 more answers

What temperature is Sirius B?

Dmitry [639]

The answer would be 9,940 K is the temperature of Sirius B.

5 0

2 years ago