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3 years ago

If I had a six pulley system and a resistance of 24,000 N, how much effort would I need to lift it?

1 answer:

6 0

Suppose you could rig up a pair of pulley systems and get a friend, who is as strong as you, to pull one of them. You'd still have trouble, but you might be able to do it because each pulley would lift half the weight, or 1,000 pounds, and a quarter of that weight is 250 pounds

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If the same experiment is repeated in different parts of the world by different scientists,

Misha Larkins [42]

The answer is D) The outcome of the experiment will be non observable

3 0

3 years ago

Read 2 more answers

QUESTION 7

ryzh [129]

Answer:

<em>The force required is 3,104 N</em>

Explanation:

<u>Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

$v_f=v_o+at$

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

We are given the initial speed as vo=20.4 m/s, the final speed as vf=0 (at rest), and the time taken to stop the car as t=7.4 s. The acceleration is:

$\displaystyle a=\frac{0-20.4}{7.4}$

$a=-2.757\ m/s^2$

The acceleration is negative because the car is braking (losing speed). Now compute the force exerted on the car of mass m=1,126 kg:

$F = 1,126\ kg * 2.757\ m/s^2$

F= 3,104 N

The force required is 3,104 N

6 0

2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

6 0

2 years ago

The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is

almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

$I = I_{0}e^{\frac{t}{\tau\\ } }$

$\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }$

Apply inverse logarithmic function:

$ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}$

The final form will be:

$\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}$

Inputing values for I, IO, and t:

$\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95$

3 0

3 years ago

A woman experiences an electrical shock. The electrons making the shock come from the A woman experiences an electrical shock. T

Zinaida [17]

The electrons making the shock come from the women's body.

<h3>What is Electric shock ?</h3>

When a high voltage current flows through the body, electrical shock results. When someone unintentionally touches an electrical source, this typically occurs. Treatment for both internal and exterior burns may be necessary as part of the aftercare.

The nervous system may be impacted by a shock.

The tissue that makes up nerves presents extremely minimal resistance to the flow of an electric charge. Electric shocks that impact nerves can cause pain, tingling, numbness, weakness, or trouble moving a limb. These effects might disappear with time or remain for good.

How to Prevent from Electrical Shocks –

Keep the Appliances Away from Moisture and Water.
Never Connect or Disconnect Under Load.
Be Careful with Capacitors.
Use Insulated Tools.
Turn Off the Power.
Check for Improper or Faulty Wiring.
Fix Extension Cord Problems.

to learn more about electric shock go to - brainly.com/question/8822505

#SPJ4

7 0

1 year ago