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Vlad1618 [11]
4 years ago
6

A racecar driver on a flat racetrack steps on the gas, changing her velocity from 10 m/sec to 30 m/sec in 5 seconds. What is the

acceleration of the racecar during that time frame?
Physics
2 answers:
erastova [34]4 years ago
8 0

Answer:

4m/s2

Explanation:

The following data were obtained from the question:

U (initial velocity) = 10m/s

V (final velocity) = 30m/s

t (time) = 5secs

a (acceleration) =?

Acceleration is the rate of change of velocity with time. It is represented mathematically as:

a = (V - U)/t

Now, with this equation i.e

a = (V - U)/t, we can calculate the acceleration of the race car as follow:

a = (V - U)/t

a = (30 - 10)/5

a = 20/5

a = 4m/s2

Therefore, the acceleration of the race car is 4m/s2

VLD [36.1K]4 years ago
8 0

Answer:

a = 4\,\frac{m}{s^{2}}

Explanation:

The acceleration of the racecar is:

a = \frac{v-v_{o}}{\Delta t}

a = \frac{30\,\frac{m}{s} - 10\,\frac{m}{s} }{5\,s}

a = 4\,\frac{m}{s^{2}}

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Choose the word that BEST completes the following sentence. Saturn’s rings have been assigned a letter, based on their location.
puteri [66]

Answer:

The correct answer would be Saturn's Cassini Division.

Explanation:

Read about it here.

https://caps.gsfc.nasa.gov/simpson/kingswood/rings/

Hope this helps! :)

5 0
3 years ago
The intensity of light from a star varies inversely as the square of the distance. If you lived on a planet ten times farther aw
frosja888 [35]

Answer:

the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.

That is,

Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100

Explanation:

Let the intensity of light be represented by I

Let the distance of the star be d

I ∝ (1/d²)

I = k/d²

For the earth,

Iₑ = k/dₑ²

k = Iₑdₑ²

For the other planet, let intensity be Iₒ and distance be dₒ

Iₒ = k/dₒ²

But dₒ = 10dₑ

Iₒ = k/(10dₑ)²

Iₒ = k/100dₑ²

But k = Iₑdₑ²

Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100

Iₒ = Iₑ/100

Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.

3 0
4 years ago
When the weight of the object increase block what is the force of friction applied? Explanation?
erik [133]

Answer:

There is absolutely No relationship between the weight of an object (which is constant) and the frictional force. If a block is sliding on a surface, that surface will be exerting a force on the block. That force can be resolved into a component parallel to the surface (which we call the frictional component), and a component perpendicular to the surface (called the normal component). For many situations, we find experimentally that the frictional component is approximately proportional to the normal component. The frictional component divided by the normal component is defined to be a quantity called the coefficient of kinetic or sliding friction. The coefficient of kinetic friction obviously depends on the nature of the surfaces involved. The normal component on an object can be decreased if you pull in the direction of the normal component (the weight does not change). However pulling this way on the object not only decreases the normal component, but it also decreases the frictional component since they are proportional. This is why it is easier to slide something if you pull up on it while you push it. If you push down, the normal and frictional components increase so it is harder to slide the object. The weight of an object is the downward force exerted by Earth’s gravity on that object, and it does not change no matter how you push or pull on the object.

8 0
3 years ago
Which scenario will drain the battery in the circuit diagram fastest: three 20 W bulbs in positions one, two, and three or 80 W
kakasveta [241]

Answer:

See the explanation below.

Explanation:

If you connect the three bulbs 20 W each will have a total power of 60W.

Now we need to understand to assign the meaning of the word gap, that is, if the circuit is open at that point or if there is no bulb connected at that point.

If the circuit is open at Point 2, there will be no current in the circuit, so the battery will drain faster with the three bulbs 20W.

In the second event, where gap means that there are no bulbs connected at that point, it means that you have two bulbs connected in series of 80W each.

In this case the bulbs will consume 160W thus drain the battery faster than the three 20W bulbs connected in series.

3 0
3 years ago
A 1180kg car is moving at a speed of 85.5 km/h. Find the force needed to bring the car to rest in a distance of 300m
yarga [219]

Answer:

1110 N

Explanation:

First, find the acceleration.

Given:

Δx = 300 m

v₀ = 85.5 km/h = 23.75 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (23.75 m/s)² + 2a (300 m)

a = -0.94 m/s²

Find the force:

F = ma

F = (1180 kg) (-0.94 m/s²)

F = -1110 N

The magnitude of the force is 1110 N.

3 0
3 years ago
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