What is the name of the units that are normally used to measure energy?

I need help with number 8

Mashutka [201]

Answer:

cxvfbgfdnsdfisblkfuoasegfoucaegru

5 0

4 years ago

What s physics?

Alex777 [14]

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.

7 0

4 years ago

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For a charged solid metal sphere with total charge Q and radius R centered on the origin: Select "True" or "False" for each stat

mario62 [17]

Answer:

the answer is true

Explanation:

7 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in the linear speed

$\Delta t$ is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

$v=8322 m/s$

Therefore, its linear speed is not changing, so the change in linear speed is zero:

$\Delta v=0$

And therefore, the tangential acceleration is zero as well:

$a_t=\frac{0}{\Delta t}=0 m/s^2$

5 0

3 years ago

Two soccer players kick the same 2-kg ball at the same time in opposite directions one kicks with a force of 15 n the other kick

Illusion [34]

Because the direction of the kicks are opposite, the net force between the applied forces is their difference.
Fn = F₂ - F₁
Substituting,
Fn = 15 N - 5 N
Fn = 10 N

From Newton's second law of motion,
Fn = m x a
where m is mass and a is acceleration. Manipulating the equation so that we are able to calculate for a,
a = Fn / m

Substituting,
a = (10 N) / 2 kg
a = 5 m/s²

<em>ANSWER: 5 m/s²</em>

7 0

3 years ago

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