All categories

3 years ago

Most people would consider a comfortable room temperature to be

Physics

2 answers:

Alla [95]3 years ago

4 0

A is extremly hot

and c is -330.07 degrees farenheit

Tanzania [10]3 years ago

3 0

Answer:

Most people would consider a comfortable room temperature to be

B. 72 degrees F

Explanation:

You might be interested in

A fire helicopter carries a 700-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

rodikova [14]

Answer:

F = 50636.873 N

Explanation:

given,

bucket of water = 700-kg

length of cable = 20 m

Speed = 40 m/s

angle of the cable = 38.0°

let air resistance be = F

tension in rope be = T

T cos 38° = m×g..................(1)

$T sin 38^0= \dfrac{mv^2}{l} + F$ ..........(2)

equation (1)/(2)

$tan 38^0 =\dfrac{\dfrac{mv^2}{l} + F}{mg}$

$0.781=\dfrac{\dfrac{700\times 40^2}{20} + F}{700\times 9.8}$

F = 50636.873 N

Hence the force exerted on the bucket is equal to F = 50636.873 N

5 0

3 years ago

The Hubble Telescope The Hubble Telescope is a large telescope that orbits Earth. It was named after Edwin Hubble. Hubble was an

Marrrta [24]

Quoting from the article itself:

"Since it is above Earth's atmosphere, it gives us clearer pictures of space than telescopes on Earth can."

3 0

3 years ago

Calculate the speed for a car that went a distance of 125 meters in 2 seconds time.

stiv31 [10]

Answer:

Explanation:

The speed of the car can be found by using the formula

$s = \frac{d}{t} \\$

d is the distance

t is the time

From the question we have

$s = \frac{125}{2} = 62.5 \\$

We have the final answer as

Hope this helps you

5 0

2 years ago

The magnetic field at the center of a 1.50-cm-diameter loop is 2.70 mT . Part A. What is the current in the loop?

elixir [45]

Explanation:

It is given that,

Diameter of the circular loop, d = 1.5 cm

Radius of the circular loop, r = 0.0075 m

Magnetic field, $B=2.7\ mT=2.7\times 10^{-3}\ T$

(A) We need to find the current in the loop. The magnetic field in a circular loop is given by :

$B=\dfrac{\mu_o I}{2r}$

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.7\times 10^{-3}\times 0.0075}{4\pi \times 10^{-7}}$

I = 32.22 A

(b) The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 32.22}{2\pi \times 2.7\times 10^{-3}}$

r = 0.00238 m

$r=2.38\times 10^{-3}\ m$

Hence, this is the required solution.

8 0

3 years ago

Which of the following best describes the location of the

marshall27 [118]

Answer:

The mantle exists above the crust of the earth

8 0

3 years ago