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sdas [7]
2 years ago
8

Peter throws a snowball at his car parked in the driveway. The snowball disintegrates as it hits the car. By Newton’s third law,

you know that both the car and the snowball exert a force on each other. What can you say about the magnitude of the forces exerted by the snowball and the car?
Physics
1 answer:
elena55 [62]2 years ago
3 0
Well, they are equal and opposite, but the force is high enough to break up the snowball.  That is where the energy is dissipated.
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If an object is neutrally buoyant (does not sink or float) in fresh water, the same object placed into salt water wouldA sink.B
Grace [21]

Answer:

D float.

Explanation:

Here, neutrally buoyant in fresh water

Now,

since the specific gravity of salt water is higher than the specific gravity of the fresh water therefore, the salt water will apply more buoyant force on the object.

And as the object is neutral in fresh water, more buoyant force will make the object float in the salt water.

4 0
3 years ago
A light spring with a force coefficient 11.85 N/m is compressed by 14 cm as it is held between a 0.27 kg block on the left and a
lilavasa [31]

Answer:

A) Left = 6.14 m/s2 Right=2.765 m/s2   B) Left = 4.59 m/s2 Right= 1.215 m/s2

Explanation:

In the question we are given the spring constant which is 11.85 N/m and the compression of the spring is 14 cm. There are two blocks in front of the springs which are 0.27 kg and 0.6 kg respectively.

First, we need to calculate how much force the spring are going to exert on the masses when they are released. Since this force is not going to change for both cases, we only need to do it once.

The formula for calculation the force is F = k.x where k is the spring constant or force coefficient 11.85 N/m and the x is the compression which is 14 cm or 0.14 meters. If we put them in the equation we can find that F = 1.659 N

A) In the first case scenario, where the friction is equal to 0, we can use the formula F=m.a where F is the force applied by the release of the spring, m is the mass of the block and a is the acceleration.

For the first block, when we put 1.659 N and 0.27kg in the equation, a is calculated to be 6.14 m/s2.

For the second block, the same force of 1.659 N and 0.6 kg, a is calculated to be 2.765 m/s2.

B) In the second case scenario, where the friction is equal to 0.158, we first need to calculate its effect on each block. We need to use the formula      Fk = μ.N where μ is the friction constant and N is the normal force of the block which is m.g (where g = 9,81 m/s2).

The friction force for the first block is calculated to be 0.4184 N and the friction force for the second block is calculated to be 0.9299 N.

The total force for the first block is 1.659 N - 0.4184 N = 1.2405 N.

The total force for the second block is 1.659 N - 0.9299 N = 0.7290 N.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the first block as 1.2405 = 0.27 x a  and a = 4.59 m/s2.

When we insert these numbers into the same equation we used to find the acceleration in the first case scenario, we can calculate the acceleration of the second block as 0.7290 = 0.6 x a  and a = 1.215 m/s2.

4 0
3 years ago
The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to t
kompoz [17]

Answer:

3.6 kHz

Explanation:

The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.

The first harmonic will be as seen in the figure attached.

The length of the first harmonic will be λ/4.

λ/4=2.4 cm

λ=2.4 * 4=9.6 cm 0.096 m

Speed of Sound- 344 m/s(in air)

velocity(v) * Time Period(T) = Wavelength (λ)

Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}

\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency

Plugging in the values into the equation,

Frequency = \frac{344}{0.096} Hz

                  = 3583.3 Hz≈3600 Hz= 3.6 kHz

Frequency= 3.6 kHz

7 0
2 years ago
Read 2 more answers
During both the vernal equinox and _____________, there is an equal amount of night and day.
choli [55]
During both the Vernal Equinox and Autumnal equinox, there is an equal amount of night and day
Vernal Equinox marks the beginning of Spring and Autumnal Equinox marks the beginning of Fall

hope this helps
5 0
3 years ago
Read 2 more answers
A body weighing 250 grams was dropped from a helicopter flying at an altitude of 100 meters. Determine the potential energy of t
kogti [31]

Answer:

the potential energy of this body is 245 J.

Explanation:

Given;

mass of the body, m = 250 g = 0.25 kg

height from which the body was dropped, h = 100 m

acceleration due to gravity, g = 9.8 m/s²

The potential energy of this body is calculated as;

P.E = mgh

substitute the given values and solve for the potential energy of this body;

P.E = 0.25 x 9.8 x 100

P.E = 245 J.

Therefore, the potential energy of this body is 245 J.

7 0
2 years ago
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