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lesya692 [45]
4 years ago
8

What does the sky look like by a black hole? astronomy

Physics
1 answer:
Komok [63]4 years ago
4 0

By a black hole, the sky looks normal then the stars closer to the black holes look like they are spinning around the black hole.

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Determining the pH of substances such as purple grape juice and catsup using test strips can be difficult. Why?
Charra [1.4K]

Answer:

Because they would naturally dye the test strips in the colors violet and red, regardless of their pH values

(would really appreciate the brainliest)

4 0
4 years ago
Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a
FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

d_1 = 1* t

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

d_2 = 2*(t - 120)

now it is given that Blaise wins by 10 m distance

d_1 - d_2 = 10

1* t - 2*(t - 120) = 10

t - 2t + 240 = 10

t = 230 s

now the distance moved by Blaise is given by

d_1 = 1*230 = 230 m

6 0
3 years ago
Q1) An action exerted on an object which may change the object's
Sunny_sXe [5.5K]

Answer: force...

Explanation:

4 0
3 years ago
"a 10 kg rock is pushed off the edge of a bridge 50 meters above the ground. what was the kinetic energy of the rock at the midw
valentinak56 [21]
Let's call h the initial height of the rock (h=50 m, the height of the bridge).

Initially, the rock has only gravitational potential energy, which is given by
U_i=mgh
where m=10 kg is the mass of the rock while g is the gravitational acceleration. So, the total mechanical energy of the rock at this point is
E_i = U_i = mgh=(10 kg)(9.81 m/s^2)(50 m)=4905 J

At midway point of its fall, its height is \frac{h}{2}, so its potential energy is
U_f = mg \frac{h}{2} = (10 kg)(9.81 m/s^2)(25 m)=2452.5 J
But now the rock is also moving by speed v, so it also has kinetic energy:
K_f =  \frac{1}{2}mv^2
So the total energy at the midway point of the fall is
E_f = U_f + K_f (1)

The mechanical energy must be conserved, so E_i = E_f, so we can rewrite (1) and solve it to find the kinetic energy of the rock at midway point of its fall:
E_i = U_f + K_f
K_f = E_i - U_f = 4905 J - 2452.5 J=2452.5 J
8 0
3 years ago
A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one
Fittoniya [83]
I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)
8 0
3 years ago
Read 2 more answers
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