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N76 [4]
3 years ago
15

A student creates a model by placing raisins in bread dough and allowing the dough to rise for several hours. The student sketch

es the model before and after the dough rises.
The student claims that the model accurately describes the motion of the galaxies in the universe. What evidence from the model supports the student’s claim?


As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.
As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.
As the dough rises, the size of the raisins increases, resembling the expansion of the galaxies.

As the dough rises, the distance between the raisins decreases, indicating that the galaxies will collide.
As the dough rises, the distance between the raisins decreases, indicating that the galaxies will collide.

As the dough rises, the size of the raisins decreases, resembling the emission of light from stars.
As the dough rises, the size of the raisins decreases, resembling the emission of light from stars.

As the dough rises, the distance between the raisins increases, indicating that the galaxies are moving apart.
Physics
1 answer:
lbvjy [14]3 years ago
3 0

As the dough rises, the distance between the raisins increases, indicating that the galaxies are moving apart. D

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7.
Igoryamba

D. Free fall

Explanation:

An object is said to be in free fall when there is only one force acting on the body, which is the force of gravity.

Near the Earth's surface, the force of gravity acting on a body is given by

F = mg

where

m is the mass of the body

g is the acceleration of gravity (its value is 9.8 m/s^2)

The direction of this force is downward (towards the Earth's centre).

If we apply Newton's second law on an object in free-fall, we can find its acceleration. In fact, we have:

a=\frac{F}{m}

And substituting F,

a=\frac{mg}{m}=g=9.8 m/s^2

So, every object in free-fall accelerates at 9.8 m/s^2 towards the ground.

Learn more about free fall here:

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brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

5 0
3 years ago
Read 2 more answers
A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera
Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m=2.1\times 10^3 kg

Initial velocity,u=0

Distance,s=5.9 m

\theta=19^{\circ}

Average friction force,f=4.0\times 10^3 N

We have to find the speed of the car at the bottom of the driveway.

Net force,F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3

Where g=9.8 m/s^2

Acceleration,a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}

v=\sqrt{2as}

v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}

v=3.9 m/s

7 0
3 years ago
A parallel-plate air capacitor is made from two plates 0.190 m square, spaced 0.770 cm apart. It is connected to a 120 V battery
Naddik [55]

Answer:

aaksj

Explanation:

a) the capacitance is given of a plate capacitor is given by:

C = \epsilon_0*(A/d)

Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:

The plates are squares so their area is given by:

A = L^2 = 0.19^2 = 0.0361 m^2

C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F

b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:

Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C

c) The electric field on a capacitor is given by:

E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]

E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m

d) The energy stored on the capacitor is given by:

W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J

6 0
3 years ago
Read 2 more answers
A driver in a car traveling at a speed of 21.8m/s sees a cat 101 m away on the road. How long will it take for the car to accele
kobusy [5.1K]
The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:

v^2 = v_0^2 + 2ad\ \to\ a = \frac{-v_0^2}{2d} = -\frac{21.8^2\ m^2/s^2}{2\cdot 99\ m} = -2.4\frac{m}{s^2}

We can clear time in the speed equation:

v = v_0 + at\ \to\ t = \frac{-v_0}{a} = \frac{-21.8\ m/s}{-2.4\ m/s^2} = \bf 9.08\ s

If you find some mistake in my English, please tell me know.
3 0
3 years ago
Read 2 more answers
In an eslastic ,the momentum is_______and the mechanical energy is​
DIA [1.3K]

Answer:

In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.

Explanation:

There are two types of collisions:

- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.

- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)

3 0
3 years ago
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