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Contact [7]
3 years ago
9

If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?

Physics
2 answers:
natulia [17]3 years ago
8 0
If a galaxy is located 200 million light years from Earth, you can conclude that t<span>he light will take 200 million years to reach Earth. </span>
Novay_Z [31]3 years ago
3 0

The answer is not D.  Just got it incorrect on my quiz.


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a sprinter with a mass of 80kg accelerates uniformly from 0 m/s to 9 m/s in 3 s. a.)what is the runners acceleration? b.) what i
Katyanochek1 [597]
Part A: 
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
             a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time

Substituting,
            a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>

Part B: 
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
               F = m x a
where F is force,
m is mass, and 
a is acceleration

Substituting,
              F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N

<em>ANSWER: 240 N </em>

Part C: 
The distance that the sprinter travel is calculated through the equation,
         d = V₀t + 0.5at²

Substituting,
            d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
             d = 13.5 m

<em>ANSWER: d = 13.5 m</em>

3 0
3 years ago
A boat is traveling upstream in the positive direction of the x axis at 10 km/h with respect to the water of a river. The water
sasho [114]

Answer:

The velocity of the boat with respect to the ground is 3 km/h

Explanation:

The speed of an object is different depending on the reference system you use. This is called relative speed.

A boat travels upstream, this means that it moves in the opposite direction to the river current.

A boat travels upstream, this means that it moves in the opposite direction to the river current. Then, if the boat moves in the positive direction of the x axis at 10 km / h with respect to the water of a river, the water flows in the negative direction of the x axis at 7 km / h with respect to the ground.

This causes the speed of the boat relative to the ground to be calculated as follows:  

<em>VbG = Vbw - VwG </em>

where VbG is the speed of the boat relative to the ground, Vbw is the speed of the boat relative to the water of the river and VwG is the speed of the water relative to the ground.

So: VbG=10 km/h – 7 km/h

<u><em>VbG= 3 km/h </em></u>

The direction of this velocity is in the positive x-direction.

4 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
4 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
4 years ago
Read 2 more answers
Name the type of reproduction process as shown in Fig 1 and Fig 2. State one point of difference between the two
likoan [24]

Figure 1= binary fission in amoeba

figure 2= budding in yeast

difference

1.Parent divides to form two daughter cells and itself gets disappeared in binary fission but in budding , a bud gets matured and detaches from the parent

6 0
3 years ago
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