Which area of the solar system is the largest

An object moving in a constant velocity will always have a

jeka94

Answer:

constant velocity unless acted on my an opposite force

7 0

3 years ago

Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22

natka813 [3]

Answer:

2420 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 22.9 N

Angle (θ) = 35°

Distance (d) = 129 m

Workdone (Wd) =?

The work done can be obtained by using the following formula:

Wd = Fd × Cos θ

Wd = 22.9 × 129 × Cos 35

Wd = 22.9 × 129 × 0.8192

Wd ≈ 2420 J

Thus, the workdone is 2420 J.

3 0

3 years ago

A circular loop of wire with a radius of 15.0cm and oriented in the horizontal xy-plane is located in a region of uniform magnet

Drupady [299]

Answer:

See answer

Explanation:

The area of the circular loop is given by:

$A = \pi r^2$

The magnetic flux is given by:

$\phi = \int \vec{B} \cdot d\vec{A}$

$d\vec{A}$ is parallel to $\vec{B}$ and $\vec{B}$ is constant in magnitude and direction therefore:

$\phi = \int \vec{B} \cdot d\vec{A}= \int BdAcos(0)= B\int dA= B*(\pi r^2)= \pi Br^2$

Part A)

initially the flux is $\phi =\pi B r^2$

after the interval $\Delta t= 2.4 [m/s]$

the flux is

$\phi = 0$

now, the EMF is defined as:

$\epsilon =- \frac{d \phi}{dt}$ ,

if we consider $\Delta t= 2.4 [m/s]$ very small then we can re-write it as:

$\epsilon =- \frac{\Delta \phi}{\Delta t}$

$\Delta \phi = 0 - \pi B r^2=-\pi (1.7) (0.15)^2=-0.12$

then:

$\epsilon =- \frac{-0.12}{0.0024} = 50 [V]$

Part B)

When looked down from above, the current flows counter clockwise, according to the right hand rule, if you place your thumb upwards (the direction of the magnetic field) and close your fingers, then the current will flow in the direction of your fingers.

3 0

3 years ago

1. Explain who is doing more work and why: a bricklayer carrying bricks and placing them on the wallof a building being

Basile [38]

Both are doing because they have chorus

7 0

2 years ago

What must be the acceleration of a train in order for it to stop 12 m/s in a distance if 541 m

earnstyle [38]

Answer:

The acceleration of the train must be - 0.133 m/s²

Explanation:

Lets explain how to solve the problem

A train in order for it to stop 12 m/s in a distance if 541 m

That means the initial velocity of the train is 12 m/s

Its final velocity is zero (stop)

The distance it covers is 541 m

We want to find its acceleration

The acceleration will be negative quantity because the train reduced its

velocity from 12 m/s to zero

We need rule contains velocity, acceleration and distance

So we will use ⇒<em> v² = u² + 2as</em>, where v is the final velocity, u is the

initial velocity, a is the acceleration and s is the distance

v = 0, u = 12 m/s, s = 541 m

Substitute these values in the rule

(0)² = (12)² + 2(a)(541)

0 = 144 + 1082 a

Subtract 144 from both sides

-144 = 1082 a

Divide both sides by 1082

- 0.133 = a

<em>The acceleration of the train must be - 0.133 m/s²</em>

3 0

3 years ago