Answer:
The galaxy is moving away from us
Explanation:
Galaxy refers to a system of millions or billions of stars accompanied by gas and dust such that they are held together as a result of gravitational attraction.
When we observe a distant galaxy, we find that a spectral line of hydrogen that is shifted from its normal location in the visible part of the spectrum into the infrared part of the spectrum.
It means that the galaxy is moving away from us.
The specific heat of the substance will be 0.129 J/g°C.
<h3>What is specific heat capacity?</h3>
The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
The given data in the problem is;
Q is the amount of energy necessary to raise the temperature = 3,000.0 j
M is the mass= 0.465 kg.
Δt is the time it takes to raise the temperature.=50°c
s stands for specific heat capacity=?
Mathematically specific heat capacity is given by;

Hence the specific heat of the substance will be 0.129 J/g°C.
To learn more about the specific heat capacity refer to the link brainly.com/question/2530523
Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

as we know that




now from above equation we have



so the speed of combined system is 2 m/s
Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s

t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)