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Kazeer [188]
3 years ago
13

Distant galaxies appear to be much larger than those nearby. true or false

Physics
1 answer:
jonny [76]3 years ago
3 0

Answer:True

Explanation: Distant galaxies appear to be larger than those close by because of cosmic expansion. light from nearer galaxies travel farther and makes the other to be billions of years away. The farther it takes to get to another galaxy, the bigger it appears

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POTENTIAL I KINETIC ENERGY
TEA [102]

7kinetic energy is decreasing in B

4 0
2 years ago
A sprinter accelerates from rest to 10.0 m/s in 1.28 s . Part A Part complete What is her acceleration in m/s2? a a = 7.81 m/s2
Mashutka [201]

Explanation:

It is given that,

Initial speed of sprinter, u = 0

Final speed of sprinter, v = 10 m/s

Time taken, t = 1.28 s

a. We need to find the acceleration of sprinter. It can be calculated using first equation of motion as :

a=\dfrac{v-u}{t}

a=\dfrac{10\ m/s}{1.28\ s}

a=7.81\ m/s^2

b. Final speed of the sprinter, v = 36 km/h

Time, t = 0.000355 h

Acceleration, a=\dfrac{36}{0.000355}

a=101408.45\ km/h^2

Hence, this is the required solution.

3 0
4 years ago
If you were capable of converting mass to energy with 100%, efficiency, how much mass would you need to produce 3.5x10^12 Joules
Alexeev081 [22]

Answer:

a) 3.9 x 10⁻⁵ kg

Explanation:

The amount of mass required to produce the energy can be given by Einstein's formula:

E = mc^2\\\\m = \frac{E}{c^2}

where,

m = mass required = ?

E = Energy produced = 3.5 x 10¹² J

c = speed of light = 3 x 10⁸ m/s

Therefore,

m = \frac{3.5\ x\ 10^{12}\ J}{(3\ x\ 10^8\ m/s)^2} \\\\m = 3.9\ x\ 10^{-5}\ kg

Hence, the correct option is:

<u>a) 3.9 x 10⁻⁵ kg</u>

7 0
3 years ago
A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra
Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

sin(\alpha) = \frac{Vyi}{Vi}

Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}

Vy in this problem will follow this equation =

Vy(t) = Vyi -g.t

where g is the gravity acceleration

Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t

This is equation (1)

For Y(t) :

Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}

We suppose yi = 0

Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s

So in t = 0.675 s  → Vy = 0. Now we calculate the y in which this happen using (2)

Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} }  .(0.675s)^{2} \\Y(0.675s) =2.236 m

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0
4 years ago
What is the gravitational field theory and how does it work? four mark answer. Don't give me random answers just for points or i
MrMuchimi
It is a theory on a show that people try to solve.
6 0
3 years ago
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