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meriva
3 years ago
13

Which statement is true about the Big Bang Theory?

Physics
1 answer:
zmey [24]3 years ago
4 0
I think it’s A.) it explains why the universe is made up of matter
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Sound is a ____ wave.
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Sound waves are longitudal waves meaning they go back and forth
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3 years ago
Which is measured by the power of a machine?
Alina [70]

the rate of work done or doing work is usually measured by the power of a machine .

7 0
3 years ago
What term describes the resistance that one encounters when moving over another
Anvisha [2.4K]
Friction? For example, like when a car's tires skid on rough concrete.
5 0
3 years ago
If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose
Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it  has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency  f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0
3 years ago
A particle’s position is ~r = (ct2 − 2dt3 )iˆ+ (2ct2 − dt2 )jˆ, where c and d are positive constants. Find the expressions for t
Mars2501 [29]

The velocity of the particle is given by the derivative of the position vector:

\vec v = \dfrac{\mathrm d\vec r}{\mathrm dt} = (2ct-6dt^2)\,\vec\imath + (4ct-2dt)\,\vec\jmath

(a) The particle is moving in the <em>x</em>-direction when the <em>y</em>-component of velocity is zero:

4ct-2dt = 2t (2c - d) = 0 \implies t=0

But we want <em>t</em> > 0, so this never happens, unless 2<em>c</em> = <em>d</em> is given, in which case the <em>y</em>-component is always zero.

(b) Similarly, the particle moves in the <em>y</em>-direction when the <em>x</em>-component vanishes:

2ct-6dt^2 = 2t (c - 3dt) = 0 \implies t=0 \text{ or }c-3dt = 0

We drop the zero solution, and we're left with

c-3dt = 0 \implies c=3dt \implies \boxed{t = \dfrac c{3d}}

In the case of 2<em>c</em> = d, this times reduces to <em>t</em> = <em>c</em>/(6<em>c</em>) = 1/6.

7 0
2 years ago
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