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larisa86 [58]
3 years ago
13

How far would a jet going 155 m/s travel in 9 s?

Physics
1 answer:
docker41 [41]3 years ago
5 0

Answer:

1,395 m

Explanation:

155×9

multiply m/s by 9s

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What work is the person doing doing to the box if the box is just being held and it is not moving?
kobusy [5.1K]
Allowing for a platform in which the box holds potential energy
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One horsepower is equal to how many watts
Keith_Richards [23]
One horsepower is about 745.7 watts
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When we add or remove energy from a substance, what kind of changes can we observe? Can they happen at the same time ?
goldenfox [79]

Answer:

Removing Energy: Removing energy will cause the particles in a liquid to begin locking into place. A. Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas.

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5 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
3 years ago
What is the energy of a rock with a mass of 10.2 kg on a cliff that is 300 m height?
Anuta_ua [19.1K]

The potential energy of the rock is 30,000 J

Explanation:

The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):

E=PE+KE

where

PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field

KE is the kinetic energy, which is the energy possessed by the object due to its motion

In this problem, the rock is at rest, so its kinetic energy is zero:

KE = 0

Therefore, the energy of the rock is just equal to its potential energy, which is:

E=PE=mgh

where

m = 10.2 kg is the mass of the rock

g=9.8 m/s^2 is the acceleration of gravity

h = 300 m is the height of the rock above the ground

Substituting and solving, we find

PE=(10.2)(9.8)(300)=30,000 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

4 0
3 years ago
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