URGENT PHYSIC PLEASE.

Vsevolod [243]

The electrical symbols are very important especially when fixing electrical appliances because it tells you where the wire of neutral and live go.

8 0

4 years ago

What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight ou

Lubov Fominskaja [6]

Answer:

The torque about his shoulder is 34.3Nm.

The solution approach assumes that the weight of the boy's arm acts at the center of the boy's arm length 35cm from the shoulder.

Explanation:

The solution to the problem can be found in the attachment below.

6 0

3 years ago

If the strength of gravity is 9.8 newtons per kilogram, that means:

Igoryamba

Here, E<span>ach kilogram of mass weighs 9.8 newtons as it's unit value of it.

In short, Your Answer would be Option D

Hope this helps!</span>

4 0

4 years ago

Read 2 more answers

A volcanic eruption melts a large area of rock, and all gases are expelled. after cooling, 40ar (atomic number 18) accumulates f

chubhunter [2.5K]

Decay of Ar-40 produces 1.33 mmol of K-40. Remaining number of moles of Ar-40 is 1.50 mmol. Initial mmol of Ar-40 present will be sum of number of moles of K-40 and remaining number of moles of Ar-40.

$n_{Ar-40}=(1.33+1.50)m mol=2.83 mmol$

now, half life time of the reaction is $1.25\times 10^{9} year$

For first order reaction, rate constant and half-life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}$

Putting the value of $t_{1/2}$ ,

$k=\frac{0.6932}{1.25\times 10^{9} year}=5.54\times 10^{-10} year^{-1}$

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{[A_{0}]}{[A_{t}]}}$

Here, $[A_{0}]$ is initial concentration and $[A_{t}]$ is concentration at time t.

$t=\frac{2.303}{(5.54\times 10^{-10} year^{-1})}log\frac{(2.83 mmol)}{(1.50 mmol)}}=1.146\times 10^{9} year$

Therefore, time required to cool the rock is $1.146\times 10^{9} year$ .

4 0

3 years ago

A friction-less pulley in the shape of a solid cylinder of mass 2.90 kg and radius 30.0 cm is used to draw water from a well. A

WINSTONCH [101]

Answer:

$s=-38.332m$

Explanation:

From the question we are told that

Cylinder of mass $m=2.90 kg$

Radius $R=30.0 m$

Mass of bucket $M_b= 1.90 kg$

Fall time $T_f= 3.70 seconds$

Generally the tension the bucket is mathematically given by

$T-mg=ma$

For cylinder

$T=I*\alpha$

where

$I=1/2mR^2$

$a=R*\alpha$

Giving

$T=-1/2mR\alpha$

Therefore

$a=\frac{-mg}{m+1/2m}$

$a=\frac{-(1.9*9.81)}{1.9+0.5*(2.9)}$

$a=-5.563880597 \approx -5.6m/s^2$

Generally the Newton's equation for motion is mathematically represented as

$s=ut+1/2at^2$

$s=0(3)+1/2*-5.6*(3.7)^2$

$s=-38.332m$

8 0

3 years ago