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3 years ago

PLEASE HELP!! ILL MARK BRAINLYEST!!!

Physics

2 answers:

zhannawk [14.2K]3 years ago

8 0

Answer:

it's true...........

podryga [215]3 years ago

6 0

It is true, 100 percent true in just typing for more words

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A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .

Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

6 0

3 years ago

What would the distance be if a bicyclist traveled 4.5 hours at a speed of 20.0km/h?

Mice21 [21]

5.5 is the answer i think

3 0

3 years ago

What would happen if a group of force of 45 N compete against a group of 50 N

larisa [96]

The 50N group of force would be greater
So whichever object is being pulled will be pulled towards the 50N force

7 0

3 years ago

A car goes from point A to point B, five miles away and then returns to point A. The car is going 15 mph.

Alex73 [517]

Answer: B) 0 mph

Explanation:

Velocity is a vector quantity. It can be measured as rate of change of displacement.

$velocity\frac{Displacement}{time}$

Displacement is the straight path length between final and initial position of the body in motion.

It is given that the car goes from point A to point B and the returns to Point A. Hence, the net displacement of the car is 0 as the car comes back to its initial position.

Therefore, the velocity of the car is 0 mph.

Option B is correct.

3 0

3 years ago