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1 year ago

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Suppose you run into a wall at 4.5 meters per second (about 10 mph). Let's say the wall brings you to a complete stop in 0.5 sec

ond. Find your deceleration and estimate the force (in newtons) that the wall exerted on you during the stopping?

1 answer:

STatiana [176]1 year ago

3 0

Answer with Explanation:

We are given that

Initial velocity,u=4.5 m/s

Time=t =0.5 s

Final velocity=v=0m/s

We have to find the deceleration and estimate the force exerted by wall on you.

We know that

Acceleration= $\frac{v-u}{t}$

Using the formula

Acceleration= $a=\frac{0-4.5}{0.5}$

deceleration=a= $-9m/s^2$

We know that

Force =ma

Using the formula and suppose mass of my body=m=40 kg

The force exerted by wall on you

Force= $40\times (9)=360N$

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Two billiard balls, assumed to have identical mass, collide in a perfectly elastic collision. Ball A is heading East at 12 m/s.

meriva

Answer:

Ball A will move towards West with speed 8 m/s

Ball B will move towards East with speed 12 m/s

Explanation:

As we know that during collision there is no external force on the system

So here we can use momentum conservation for finding the final velocities of two balls

so we have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

since two balls are identical so we have

$m_1 = m_2$

$m(12) - m(8) = mv_1 + mv_2$

$v_1 + v_2 = 4$

also we know that for elastic collision we have

$v_{2f} - v_{1f} = v_{1i} - v_{2i}$

so we have

$v_2 - v_1 = 12 - (-8)$

$v_2 - v_1 = 20$

so we have

$v_2 = 12 m/s$ Towards East

$v_1 = 8 m/s$ Towards West

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1 year ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

Tema [17]

Answer:

The value of developed electric force is $3.516\times 10^{- 7} N$

Solution:

As per the question:

Mass of the droplet = 1.8 mg = $1.8\times 10^{- 6} kg$

Charge on droplet, Q = $25 pC = 25\times 10^{- 12} C$

Distance between the 2 droplets, D = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

$F_{E} = \frac{1}{4\pi epsilon_{o}}.\frac{Q^{2}}{D^{2}}$

$\frac{1}{4\pi epsilon_{o}} = 9\times 10^{9} m/F$

$F_{E} = (9\times 10^{9}).\frac{(25\times 10^{- 12})^{2}}{0.004^{2}}$

$F_{E} = 3.516\times 10^{- 7} N$

The magnitude of force is too low to be noticed.

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1 year ago

Read 2 more answers

Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.

AleksandrR [38]

Answer:

$q=390.285x10^{-12}C$

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

$7.0cm*\frac{1m}{100cm}=70x10^{-3}m$

$A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62$

Capacitance

$C=E_o*\frac{A}{d}$

$E_o=8.85x10^{-12}\frac{C^2}{N*m^2}$

$C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}$

$C=43.365x10^{-12}F$

$C=43.365pF$

The charge is find by the equation

$q=C*V$

$q=43.365pF*9V$

$q=390.285x10^{-12}C$

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2 years ago

A straight wire 20 cm long, carrying a current of 4 A, is in a uniform magnetic field of 0.6 T. What is the force on the wire wh

zheka24 [161]

Answer:

Magnetic force, F = 0.24 N

Explanation:

It is given that,

Current flowing in the wire, I = 4 A

Length of the wire, L = 20 cm = 0.2 m

Magnetic field, B = 0.6 T

Angle between force and the magnetic field, θ = 30°. The magnetic force is given by :

$F=ILB\ sin\theta$

$F=4\ A\times 0.2\ m\times 0.6\ T\ sin(30)$

F = 0.24 N

So, the force on the wire at an angle of 30° with respect to the field is 0.24 N. Hence, this is the required solution.

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2 years ago

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In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r

scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W = 2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

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2 years ago