The burning of coal is used to heat water in a coal-burning power plant. TRUE or FALSE?

The earth rotates about its pole once every 24 hrs. The distance from the pole to a location on the Earth is 35* north latitude

Oksana_A [137]

The angular velocity, ω=
2π/t; t = 24 hrs = 24 x 3600 seconds = 86400 s
ω = 7.27 x 10⁻⁵
v = ωr
= 7.27 x 10⁻⁵ x 3242.8 x 1.6 x 1000 (converting miles to meters)
= 377.2 m/s

4 0

2 years ago

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?

Eduardwww [97]

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

6 0

2 years ago

If the driver slammed on the brakes, what could happen to the crate?

stich3 [128]

The crate would slide forward

7 0

2 years ago

Read 2 more answers

The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of

PIT_PIT [208]

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

$\Phi_B=S\cdot B=SBcos\alpha$ (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

You solve the equation (1) for B, and replace the values of the other parameters:

$B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT$

The strength of the magntetic field is 4.1mT

4 0

2 years ago

A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00

ira [324]

Answer:

A. 51.42 m.

B. 17.14 s.

Explanation:

Using equations of motion:

vf^2 = vi^2 + 2 * aS

Where,

vf = final velocity

a = acceleration

S = distance to which swan traveled

vi = 0 m/s

6.00^2 = 2 * 0.350S

S = 36/0.7

= 51.42 m.

B.

vf = vi + at

6 = 0 + 0.35t

t = 6/0.35

= 17.14 s.

5 0

2 years ago