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Lostsunrise [7]
3 years ago
6

I need help ASAP please............

Physics
1 answer:
aniked [119]3 years ago
6 0

Answer:

A satellite orbiting earth 1

Your shoulder 3

f=m x a 2

A soccer ball 2

A magician 3

When you paddle 3

A picture 1

Earth 1

A rocket 1

A speed skater 2

Explanation:

Law 1 states an object at rest , stay at rest and an object in motion stays in motion unless acted on by another force.

Law 2 states that the acceleration of an object depends on its mass and force applied.

Law 3 states every action creates an equal and opposite reaction.

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4 years ago
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A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
3 years ago
Alguien me ayuda con calculos estequiometricos?
ruslelena [56]

Answer:

SURE!!!...

But what to calculate!!!....

5 0
3 years ago
An object increases its velocity from 22 m/s to 36 m/s in 5.0 s. What is the average velocity of the object?
Luden [163]

Answer:

a=2.8\ m/s^2

Explanation:

Given that,

Initial velocity of an object, u = 22 m/s

Final velocity of an object, v = 36 m/s

Time, t = 5 s

It can be assumed to find the average acceleration of the object instead of average velocity.

The change in velocity per unit time is equal to average acceleration of an object. It can be given by :

a=\dfrac{v-u}{t}\\\\a=\dfrac{36\ m/s-22\ m/s}{5}\\\\\a=\dfrac{14}{5}\ m/s^2\\\\a=2.8\ m/s^2

So, the acceleration of the object is 2.8\ m/s^2.

7 0
3 years ago
when the cross product of vector A and B are 2i+3j+4k and the sum of vector A and B are 3i+8j+7k then what will be the two vecto
Kryger [21]

Explanation:

the two vectors are 5i+11j+11k

8 0
2 years ago
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