PHYSICS COLLEGE

What is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s

Answers

Answer 1

Answer:

Answer:

8m/s

Explanation:

Vavg= 16-0/2=8m/s

Related Questions

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2dB) d/√2C) d/4D) 2dE) d/2
Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity of the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)
Why are continental rocks much older than oceanic crust?A. Oceanic crust is continually recycled through convection in the earth's mantleB. Oceanic crust is made out of much less dense material than continental crustC. Continental crust is continually renewed through convection in the earth's mantleD. Continental crust eats oceanic crust for breakfast
You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle
Define reflection of sound?

An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge −Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. Show that the electric field inside this atom is : Ein=Ze4πϵ0(1r^2−rR^3). b. What is the electric field at the surface of the atom? Is this the expected value? Explain.c. A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field at r = R/2?

Answers

Answer:

Part a)

$E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))$

Part b)

$E = 0$

Yes it is the expected value of electric field at the surface of an atom

Part c)

$E = 4.64 * 10^(13) N/C$

Explanation:

Since negative charge of electrons in uniformly distributed in the atom while positive charge is concentrated at the nucleus

So the electric field due to positive charge of the nucleus is given as

$E = (kq)/(r^2)$

$E_1 = (Ze)/(4\pi \epsilon_0 r^2)$

now charge due to electrons inside a radius "r" is given as

$q = (-Ze r^3)/(R^3)$

now we will have electric field given as

$E_2 = ((-Zer^3)/(R^3))}{4\pi\epsilon_0 r^2}$

now net electric field is given as

$E = E_1 + E_2$

$E = (Ze)/(4\pi \epsilon_0 r^2) - (Zer)/(4\pi \epsilon_0 R^3)$

$E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))$

Part b)

At the surface of an atom

$r = R$

$E = 0$

Yes it is the expected value of electric field at the surface of an atom

Part c)

If Z = 92

R = 0.10 nm

$r = (R)/(2)$

so we will have

$E = 92(1.6 * 10^(-19)) * (9 * 10^9)((4)/(R^2) - (1)/(2R^2))$

$E = (4.64 * 10^(-7))/((0.10 * 10^(-9))^2)$

$E = 4.64 * 10^(13) N/C$

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of 5.81 T, and the current in the solenoid is 3.79 × 102 A. What is the number of turns per meter of length of the solenoid?

Answers

Answer:

n = 1.22 10⁴ turns/m

Explanation:

The magnetic field in a solenoid is proportional to the intensity of the current, the number of turns per unit length (n) and the magnetic permeability (myo), is described by the equation

B = μ₀ n I

Let's clear the density of turns

n = B / (μ₀ I)

Let's replace and calculate

n = 5.81 / (4pi 10-7 3.79 102)

n = 5.81 105 / 47.63

n = 1.22 10⁴ turns / m

A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have to be transferred between a carpet and a shoe for there to be a 5.8 kV potential distance d = 2.8 mm ? Approximate the area of a shoe as 30 cm x 8 cm. Express your answer using two significant figures.

Answers

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$(Q)/(V)=(A\epsilon_0)/(d)\n\Rightarrow \Q=V(A\epsilon_0)/(d)\n\Rightarrow Q=5.8* 10^3(0.024* 8.85* 10^(-12))/(0.0028)\n\Rightarrow Q=4.4* 10^(-7)\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$

$(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h$

$h\approx 1.6420\,m$

above the normal hanging position.

Each mass in the figure is 3 kg. Find the magnitude and direction of the net gravitational force on mass A due to the other masses.A. 2.45 × 10–7 N toward B
B. 3.75 × 10–7 N toward C
C. 2.00 × 10–7 N toward D
D. 1.15 × 10–7 N toward D

Answers

The magnitude and direction of the net gravitationalforce on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this case, all of the masses are equal to 3 kg, and the distance between mass A and mass D is 3 m.

The gravitational force between mass A and mass D is therefore:

F = G * m_A * m_D / r²

= 6.674 × 10⁻¹¹ N m² / kg² * 3 kg * 3 kg / 3 m²

= 1.15 × 10⁻⁷ N

The direction of the gravitational force is towards mass D.

Therefore, the net gravitational force on mass A due to the other masses is 1.15 × 10⁻⁷ N toward D.

To learn more about gravitational force, here

brainly.com/question/32609171

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Answer:

THE ANSER IS B

Explanation:

Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.