Ask question

Ask question

All categories

3 years ago

13

Fig. 3.1

1 answer:

goldfiish [28.3K]3 years ago

7 0

Weight is the force acting on a bode weight =mass times gravity

You might be interested in

Which equation can be used to solve for acceleration? <br><br>

antoniya [11.8K]

A = d/t
hope this helps x

3 0

4 years ago

Read 2 more answers

The achievement of lifting a rocket off the ground and into space can be explained by

lawyer [7]

Answer:

The achievement of lifting a rocket off the ground and into space can be explained by Newton's third law of motion. What is required for a rocket to lift off into space? Thrust is required for a rocket to lift off into space, ... An object that travels around another object in space is called a satellite.

Explanation:

4 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

aliya0001 [1]

Answer:

m=41 kg

v=0,02 ms

R=2,1 m

F-?

a=v²/R

a=(0,02) ²/2,1=0,0002

F=m*a

F=41*0,0002=0,0082 H

F=0,0082 H

3 0

3 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