8 cups water a day is that measuring cups?

If the weight of a person is 500 newton what is his mass on the earth ?

miskamm [114]

Answer:

The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?

f= Force (of gravity)=500N

g=acceleration of gravity=1.6m/s^2

m=mass=312kg

m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg

his mass is 312kg

8 0

2 years ago

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a

Burka [1]

Solution :

a). B at the center :

$=\frac{u\times I}{2R}$

Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.

Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE

b). Also, the sum of the fields must be zero.

Therefore,

$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$

So,

$\frac{I_1}{d_1}= \frac{I_2}{d_2}$

$=\frac{16}{21}=\frac{I_2}{32}$

$I_2=24.38$ A

Therefore, the current in the outer wire is 24.38 ampere.

3 0

2 years ago

Read 2 more answers

A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below

siniylev [52]

Answer:

125.83672 seconds

Explanation:

P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)

m = Mass of professor = 103 kg

g = Acceleration due to gravity = 9.8 m/s²

h = Height of professor = 93 m

Work done would be equal to the potential energy

$W=mgh\\\Rightarrow W=103\times 9.8\times 93\\\Rightarrow W=93874.2\ J$

Power is given by

$P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds$

The time taken by the horse to pull the professor is 125.83672 seconds

6 0

3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$