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4 years ago

Big Bang theorist believe_____?

Physics

2 answers:

sashaice [31]4 years ago

8 0

<span>C. the universe started with an explosion and expansion of a very hot, dense, compact fireball</span>

harkovskaia [24]4 years ago

3 0

<span>C. the universe started with an explosion and expansion of a very hot, dense, compact fireball
</span>

You might be interested in

10. What is the acceleration of a 1000 kg car subject to a 500 N net force?

Simora [160]

Answer:

0.5m/s^2

Explanation:

We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.

F = ma

F/m = a

We know the net force and mass so substitute those values and simplify.

500/1000 = 0.5m/s^2

Best of Luck!

3 0

2 years ago

A 500 gram mass attached to a horizontal spring

Daniel [21]

Is that a question? .
..............................

8 0

4 years ago

A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between

Monica [59]

Answer:

a) $t=5.5\mu s$

b) $\Delta t'=12.5\mu s$

Explanation:

a) Let's use the constant velocity equation:

$v=\frac{\Delta x}{\Delta t}$

v is the speed of the muon. 0.9*c
c is the speed of light 3*10⁸ m/s

$t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}$

$t=5.5\mu s$

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

$\Delta t'=\gamma\Delta t$

$\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

$\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t$

$\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t$

$\Delta t'=\frac{1}{0.44}\Delta t$

No we use the value of Δt calculated in a)

$\Delta t'=2.27*5.5=12.5\mu s$

I hope it helps you!

8 0

4 years ago

A box is being pulled to the right. What is the direction of the gravitational force?

lapo4ka [179]

The correct answer to the question is vertically downward i.e towards the centre of earth.

EXPLANATION:

As per the question, the box is pulled to the right.

Hence, the direction of the applied force is towards right.

We are asked to determine the direction of the gravitational force that acts on the body.

Before answering this question, first we gave to understand the gravitational force of earth.

Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre. It is equivalent to the weight of the body.

The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.

Hence, the direction of the gravitational force which acts on the box is vertically downward.

7 0

4 years ago

Read 2 more answers

A pendulum consists of a 0.5 kg mass attached to the end of 1-meter-long rod of negligible mass. When the rod makes an angle of

Alenkinab [10]

Answer:

T=4.24 N.m

Explanation:

Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot

F=0.5 kg* 9.8 m/ $s^{2}$ = 4.9 N

and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus

so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N

and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m

anothe way to do it is,

T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result

7 0

3 years ago