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Anna11 [10]
3 years ago
9

A student stays at her initial position for a bit of time, then walks slowly in a straight line for a while, then stops to rest

awhile and finally runs quickly back to her initial position along a straight line.
Required:
Draw displacement versus time plots best represents the student’s trip?
Physics
1 answer:
True [87]3 years ago
6 0

Answer:

The first interval is walked slowly, this is a straight line with a small slope

Second interval stops, which gives a horizontal line, indicating the same position

Third interval, walk back, straight downhill

Explanation:

In this problem we have a uniform movement, this means that the acceleration in each intervals

              x = v t

 

The first interval is walked slowly, this is a straight line with a small slope

Second interval stops, which gives a horizontal line, indicating the same position

Third interval, walk back, straight downhill

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
Compare and contrast repetition and replication in a science investigation
Bas_tet [7]
Repetition occurs when multiple sets of measurements are made during one scientific investigation. Replication occurs when a scientific investigation is <span>reproduced by another person.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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3 0
3 years ago
A constant 20 N force is applied to a 7 kg box to push it along the ground. How
Elis [28]

Answer:

40 joules

Explanation:

Work Done=Force*Distance

6 0
3 years ago
How does the input distance of a third-class lever compare to the output distance​
Alexandra [31]

Answer:

A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force

4 0
3 years ago
Arocket launches at an angle of 33.6 degrees from the horizontal at a
babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, \theta=33.6

Initial velocity of the rocket is, u=58.5

A vector at an angle \theta with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle \theta with the horizontal, then the horizontal and vertical components are given as:

A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

u_y=u\sin \theta

Plug in the given values and solve for u_y. This gives,

u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})

Therefore, the Y component of initial velocity is 32.37.

4 0
3 years ago
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