Ask question

Ask question

All categories

3 years ago

13

Derek reads a rate of 25 words per min. How long would it take him to read a 225 work article if he read at a constant rate with

out stopping

1 answer:

atroni [7]3 years ago

6 0

Hello there! Your answer is (255/25=9) hope it helps and have a wonderful day!

You might be interested in

A uniform magnetic field of magnitude 0.23 T is directed perpendicular to the plane of a rectangular loop having dimensions 7.8

Arlecino [84]

Answer:

0.0025116weber/m²

Explanation:

Magnetic field density (B) is the ratio of the magnetic flux (¶) through the loop to its cross sectional area (A).

Mathematically;

B = ¶/A

¶ = BA

Given B = 0.23Tesla which is the magnitude of the magnetic field

Dimension of the rectangular loop = 7.8 cm by 14 cm

Area of the rectangular loop perpendicular to the field B = 7.8cm×14cm

= 109.2cm²

Converting this value to m²

Area of the loop = 109.2 × 10^-4

Area of the loop = 0.01092m²

Magneto flux = 0.23×0.01092

Magnetic flux = 0.0025116weber/m²

3 0

3 years ago

Read 2 more answers

In what direction is the weight vector always drawn?

Varvara68 [4.7K]

Yo sup??

The weight vector is usually drawn vertically downwards from the centre of the body.

It can be respectively resolved as well.

Hope this helps

3 0

3 years ago

***FIRST CORRECT ANSWER GETS A BRAINLIEST***

melomori [17]

Answer:

Explanation: It is halved

3 0

3 years ago

Read 2 more answers

Suppose that the hatch on the side of a Mars lander is built and tested on Earth so that the internal pressure just balances the

Kaylis [27]

Answer:

The value is $F_{net} = 4444 lb$

The force will be outward

Explanation:

From the question we are told that

The diameter of the disk is $d = 50.0 \ cm = \frac{50}{100} = 0.5 \ m$

The external pressure on Mars is $P = 650 \ N/m^2$

From the question we are told that

Internal pressure = External pressure

Generally the external Force on earth is

$F_E = P_{atm} * A$

Here $P_{atm}$ is the atmospheric pressure with value $P_{atm} = 1.013*10^{5}\ Pa$

So

$F_E = 1.013 *10^{5} * \pi * \frac{d^2}{4}$

=> $F_E = 1.013 *10^{5} *3.142 * \frac{0.50 ^2}{4}$

=> $F_E = 19893 \ N$

Generally the external Force on Mars is

$F= P * A$

$F = 650 * \pi * \frac{d^2}{4}$

=> $F = 650 *3.142 * \frac{0.5^2}{4}$

=> $F = 127.6 \ N$

Net force is mathematically represented as

$F_{net} = F_E -F$

=> $F_{net} = 19893 -127.6$

=> $F_{net} = 19765.6 \ N$ converting to pounds

$F_{net} = \frac{19765.6}{4.448}$

=> $F_{net} = 4444 lb$

Given that that the value is positive then the force will be outward

7 0

3 years ago

I’ll give brainliest!! please help and answer correctly! plsss answer quick

noname [10]

Answer:

i dont think anything will happen so maybe A: the wall is larger than you.

Explanation:

8 0

3 years ago