All categories

3 years ago

In each diagram I'm to label the reaction forces I'm a bit confused with the question

Physics

1 answer:

Furkat [3]3 years ago

5 0

If what I see is correct the middle picture is the answer

If you throw a bowling ball at the pins, the pins will fall down (that is if you don't miss). The force of the ball will knock down the pins, therefore the pins are reacting to the force of the bowling ball which is reacting to how the person threw it. Does that make sense?

You might be interested in

Light rays in a material with index of refraction 1.31 can undergo total internal reflection when they strike the interface with

3241004551 [841]

Answer:

The refractive index of the material is 1.28.

Explanation:

It is given that,

Refractive index of medium 1, n₁ = 1.31

Critical angle, $\theta_c=78.3$

At critical angle rays will reflects at 90 degrees. Using Snell's law as:

$n_1sin\theta_1=n_2sin\theta_2$

At critical angle, $n_1sin\ \theta_c=n_2\ sin90$

$n_1sin\ \theta_c=n_2$

$n_2=1.31\times sin(78.3)$

$n_2=1.28$

So, the refractive index of the material is 1.28. Hence, this is the required solution.

3 0

2 years ago

Photons are also known as beta particles. (true or false)

Llana [10]

No. Beta particles are electrons.

5 0

3 years ago

Near the center of our galaxy, hydrogen gas is moving directly away from us in its orbit about a black hole. The electromagnetic

klemol [59]

To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.

By definition the wavelength is given defined by,

$\lambda_{obs} = \lambda_s \sqrt{\frac{1+u/c}{1-u/c}}$

Where

$\lambda_{obs}$ = Observed wavelength

$\lambda_s$ = Wavelength of the source

c = Speed of light in vacuum

u = Relative velocity of the source to the observer

According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is $\lambda_{obs}=1945nm$

Therefore replacing in the previous equation we have,

$1945 = 1875 \sqrt{\frac{1+\frac{u}{c} }{1-\frac{u}{c} }}$

$\sqrt{\frac{1+u/c}{1-u/c}} = 1.03733$

$\frac{1+\frac{u}{c} }{1-\frac{u}{c} }=1.03733^2$

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

Solving for u,

$1+\frac{u}{c} =1.03733^2*(1-\frac{u}{c} )$

$1+\frac{u}{c} =1.03733^2-1.03733^2(\frac{u}{c} )$

$\frac{u}{c} +1.03733^2\frac{u}{c} =1.03733^2-1$

$2.88595\frac{u}{c}=1.03733^2-1$

$\frac{u}{c} = \frac{1.03733^2-1}{2.88595}$

$u = \frac{1.03733^2-1}{2.88595}*c$

$u = 0.02635c$

Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.

6 0

3 years ago

A current-carrying wire of length 52.0 cm is positioned perpendicular to a uniform magnetic field. If the current is 15.0 A and

nata0808 [166]

Answer:

The magnetic field strength is 0.29 T.

Explanation:

Given that,

Length of current-carrying wire, L = 52 cm = 0.52 m

Current, I = 15 A

Magnetic force, F = 2.3 N

We need to find the magnetic field strength. We know that the magnetic force is given by :

$F=ILB$

B is magnetic field strength

$B=\dfrac{F}{IL}\\\\B=\dfrac{2.3}{0.52\times 15}\\\\B=0.29\ T$

So, the magnetic field strength is 0.29 T.

6 0

3 years ago

The velocity of a passenger relative to a boat is -vpb. The velocity of the boat relative to the river it is moving on is vbr. T

RideAnS [48]

Answer:

vps = vbr + vrs - vpb

Explanation:

If the passenger were at rest, his speed relative to the shore will be identical to the boat's, as follows:
vps = vbr + vrs
As he is moving in a direction opposite to the boat's, his velocity relative to the shore must be less than if he were at rest, in the same quantity that he was moving opposite to the boat, as follows:
vps = vbr+ vrs -vpb

5 0

2 years ago