What is it called when dolphins communicate underwater through sound waves?

A family uses an electric frying pan with a power rating of 1.2 X 10^3 W. Although the pan is thermostatically controlled, its e

kirill115 [55]

Answer:

378 KWh

Explanation:

We'll begin by converting 1.2×10³ W to KW. This can be obtained as follow:

10³ W = 1 KW

Therefore,

1.2×10³ W = 1.2×10³ W × 1 KW / 10³ W

1.2×10³ W = 1.2 KW

Next, we shall convert 6.3×10² mins to hours (h). This can be obtained as follow:

60 mins = 1 h

Therefore,

6.3×10² mins = 6.3×10² mins × 1 h / 60 mins

6.3×10² mins = 10.5 h

Finally, we shall determine the electrical energy in KWh used for 1 month (i.e 30 days). This can be obtained as follow:

Power (P) = 1.2 KW

Time (t) for 1 month (30 days) = 10.5 h × 30

= 315 h

Energy (E) =?

E = Pt

E = 1.2 × 315

E = 378 KWh

Thus, the electrical energy used for 1 month (i.e 30 days) is 378 KWh.

8 0

3 years ago

Explain why during a cold winter, the air temperatures are generally higher on snowy days than on clear days.

Aleks04 [339]

I don’t know that is weird

4 0

4 years ago

A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at

zhuklara [117]

Answer:

$B=1.21*10^{-10}T$

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

$B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)$

$\mu_0$ is the permeability of free space, $I$ is the flowing current through the ring, $R$ is the ring's radius and $r$ is the distance to the center of the ring.

The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period $T=\frac{2\pi}{\omega}$ . So, we have:

$I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A$

Now, replacing in (1):

$B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T$

6 0

4 years ago

The force measured to move a sled was 10 N and the sled was moved 10 meters,

sergey [27]

Answer:

work done would be I don't know

3 0

3 years ago

A bear scratches his claws on a big pine tree. Does the bear do work on the tree? why or why not?

Ganezh [65]

Answer:

See the explanation below.

Explanation:

Work in physics is defined as the product of force by the distance that the body travels in the direction of the force.

It can be represented by means of the following equation.

$W=F*d$

where:

W = work [J]

F = force [N]

d = distance [m]

In the given example the work is zero since the tree does not move, therefore the bear exerts a force on the tree. But there is no talk of movement of the tree, therefore the work is zero.

8 0

4 years ago