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tatiyna
3 years ago
8

the car starts from rest at s=0 and increases its speed at at=4m/s^2. Determine the time when the magnitude of acceleration beco

mes 20m/s^. At what point s does this occur
Physics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

<em>Time =  5 seconds</em>

<em>Distance = 50 meters</em>

Explanation:

<u>Constantly Accelerated Motion</u>

When the velocity of a moving object changes at a constant rate, called acceleration, the velocity changes in same amounts in the same times. The question has a mistake when asking when the acceleration is 20 m/s. If the acceleration is constant, the only variable that can change to that value is the velocity. The equation to calculate the speed is

v_f=v_o+a.t

And the distance s is

\displaystyle s=v_o.t+\frac{gt^2}{2}

Given the object starts from rest, vo=0 and vf=20 m/s at a=4\ m/s^2. We compute t

\displaystyle t=\frac{v_f-v_o}{a}=\frac{20-0}{4}

\boxed{t=5\ sec}

Now we compute s

\displaystyle s=0+\frac{4\times 5^2}{2}

\boxed{s=50\ m}

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3 0
3 years ago
Which objects in space formed from the huge disk cisce and debris beyond the outer planets? Select two option
Dennis_Churaev [7]

Answer:

Comets

Explanation:

The Kuiper Belt is a collection of trans-Neptunian objects that consist of comets and other dwarf planets, including Pluto.

4 0
4 years ago
A spring with spring constant 450 N/m is stretched by 12 cm. What distance is required to double the amount of potential energy
snow_lady [41]

Answer:

<em> The distance required = 16.97 cm</em>

Explanation:

Hook's Law

From Hook's law, the potential energy stored in a stretched spring

E = 1/2ke² ......................... Equation 1

making e the subject of the equation,

e = √(2E/k)........................ Equation 2

Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.

Given: k = 450 N/m, e = 12 cm = 0.12 m.

E = 1/2(450)(0.12)²

E = 225(0.12)²

E = 3.24 J.

When the potential energy is doubled,

I.e E = 2×3.24

E = 6.48 J.

Substituting into equation 2,

e = √(2×6.48/450)

e = √0.0288

e = 0.1697 m

<em>e = 16.97 cm</em>

<em>Thus the distance required = 16.97 cm</em>

6 0
3 years ago
The pressure of a gas changes from 120 kPa. The volume changes from 45 L to 40 L. If the initial temperature is 81 c, what is th
irina [24]

Explanation:

we are not given the pressure change, check yhe question please

7 0
3 years ago
Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki
irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

1)

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every  second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

a=1 m/s^2

which is equivalent to the gradient of the line in the velocity-time graph.

2)

In this part, we want to find after what time the body will stop its motion.

To do that, we can use the following suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

In this problem:

u = 10 m/s is the initial velocity of the body

a=-1 m/s^2 is the acceleration

v = 0 m/s, because we want to find the time T at which the body will stop

Re-arranging the equation, we find:

T=-\frac{u}{a}=-\frac{10}{-1}=10 s

3)

In order to find the total distance covered by the body during its accelerated motion, we have to use another suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

In this problem:

u = 10 m/s is the initial velocity

a=-1 m/s^2 is the acceleration

t = 10 s is the time it takes for the body to stop (found in part 2)

Solving for s, we find the distance covered:

s=(10)(10)+\frac{1}{2}(-1)(10)^2=50 m

7 0
4 years ago
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