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dusya [7]
3 years ago
15

Which type of galaxy is too small for gravity to form it into an easily describable shape?

Physics
2 answers:
oksano4ka [1.4K]3 years ago
6 0
If you think about it, irregular galaxies dont really have a describable shape, as you can tell by the name. So B. 
ruslelena [56]3 years ago
4 0

The correct answer is B. irregular.

An irregular galaxy is a galaxy too small for gravity to form into a regular shape, or one shaped by collisions or the gravitational pull of other galaxies.

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What happens when a magenta light is shone on a green surface?
Sedbober [7]

Answer:

When Magenta light is shown on a green surface, it looks black.

Explanation:

It absorbs the Magenta light and also reflects none of the light.

5 0
3 years ago
Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil
Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

Q = ms\Delta T

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

Q_1 = (m_w s_w + m_ps_p)\delta T

Q_1 = (0.5(4186) + 0.750(s))\Delta T

Now similarly for other pan we have

Q_2 = (m_w s_w + m_ps_p)\delta T

Q_2 = (0.5(4186) + 1.50(s))\Delta T

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0
3 years ago
Physics Help Please!!!
liq [111]
The answer is the third graph
3 0
3 years ago
Read 2 more answers
James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a
Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0
3 years ago
Help ASAP plsssss:///
kenny6666 [7]

Answer:

c

Explanation:

8 0
3 years ago
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