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3 years ago

The model of the universe that the earth is in the center is called

Physics

1 answer:

erastovalidia [21]3 years ago

3 0

It is the Geocentrism or the ptolemaic system
any of the 2 above

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HELP PLS and HURRY

tangare [24]

Answer:

C - 50,000 * 77 * 3

Explanation:

At the top of the hill the potential energy is E= mgh= (160 kg)(9.81 m s^-2)(30 m)= 47088

hope it helps ,

<u>help me by marking as brainliest....</u>

5 0

2 years ago

a particle moves along the x axis with an acceleration of a=18t, where a has units if m/s2. if the particle at time t=0 is at th

ludmilkaskok [199]

Answer:

Position at t= 4 seconds is 144 m

Explanation:

It is given that acceleration, a = 18 t, where t is the time.

We know that Velocity, $v = \int { a} \, dt$

Substituting value of a,

Velocity, $v = \int {18t} \, dt=\frac{18t^2}{2} +c=9t^2+c$

We know that at t = 0, v = -12 m/s

So, $9*0^2+c=-12\\ \\ c=-12m/s$

So velocity, $v = (9t^2-12)m/s$

We also know that displacement, $x = \int { v} \, dt$

Substituting value of v,

Displacement, $x=\int {(9t^2-12)} \, dt=\frac{9t^3}{3} -12t+c=3t^3-12t+c$

We know that at t = 0, particle is at origin, x =0.

So, $0=3*0^3-12*0+c\\ \\ c=0$

Displacement, $x = 3t^3-12t$

At t = 4 seconds

$x = 3*4^3-12*4=192--48=144m$

Position at t= 4 seconds is 144 m

4 0

3 years ago

What is a good definition of acceleration

lilavasa [31]

Rate of change of velocity is acceleration

8 0

3 years ago

The electric field between two parallel plates has a magnitude of 875 n/

Marizza181 [45]

The answer:
1.8 x 10^0 V

7 0

2 years ago

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

2 years ago