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3 years ago

14

Please help with this problem ASAP!!!

1 answer:

Alika [10]3 years ago

3 0

Answer:

its point Y because the dot that represents Y is cose to s and can make a chan reaction

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Which law is associated with inertia

N76 [4]

Answer:

The focus of Lesson 1 is Newton's first law of motion - sometimes referred to as the law of inertia. An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

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3 years ago

What is the average speed of a boy who jogs 250 meters in 110 seconds

Ede4ka [16]

2.27 mps repeating.

This is the last question ill ever answer here. Thanks for being the last.

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3 years ago

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You lift a 2.0 kg book from a shelf that is 1 m from the floor. What is the potential energy with respect to floor?

sladkih [1.3K]

Potential energy = mgh
Potential energy = 2 x 9.8 x 1
Potential energy 19.6 J

8 0

3 years ago

Turning a corner at a typical large intersection in a city means driving your car through a circular arc with a radius of about

Naddik [55]

Answer:

9.89 m/s.

Explanation:

Given that,

The radius of the circular arc, r = 25 m

The acceleration of the vehicle is 0.40 times the free-fall acceleration i.e.,a = 0.4(9.8) = 3.92 m/s²

Let v is the maximum speed at which you should drive through this turn. It can be solved as follows :

$a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{3.92\times 25} \\\\=9.89 m/s$

So, the maximum speed of the car should be 9.89 m/s.

8 0

3 years ago

nata0808 [166]

Explanation:

Problem 1.

Initial speed of the runner, u = 0

Acceleration of the runner, $a=4.2\ m/s^2$

Time taken, t = 100 s

Let v is the speed of the runner now. Using the first equation of kinematics as :

$v=u+at$

$v=at$

$v=4.2\ m/s^2\times 100\ s$

v = 420 m/s

Problem 2.

Initial speed of the plane, u = 0

Distance covered, d = 300 m

Time taken, t = 25 s

Using the equation of kinematics as :

$d=ut+\dfrac{1}{2}at^2$

$d=\dfrac{1}{2}at^2$

$a=\dfrac{2d}{t^2}$

$a=\dfrac{2\times 300\ m}{(25\ s)^2}$

$a=0.96\ m/s^2$

Problem 3.

A ball free falls from the top of the roof for 5 seconds. Let it will fall at a distance of d. It is given by :

$d=ut+\dfrac{1}{2}gt^2$

$d=\dfrac{1}{2}\times 9.8\times (5)^2$

d = 122.5 meters

Let v is the final speed at the end of 5 seconds. Again using first equation of kinematics as :

$v=u+gt$

$v=9.8\ m/s^2\times 5\ s$

v = 49 m/s

Hence, this is the required solution.

3 0

3 years ago