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3 years ago

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A car is traveling at an angular acceleration of 0.04 rad/s2 on a circular part of a road. If the radius of the circular part is

200m, find the net acceleration of the car when the angular velocity of the car is 0.2rad/s.

1 answer:

natta225 [31]3 years ago

4 0

Answer:

8 m/s²

Explanation:

α = 0.04 rad/s²

w = 0.2 rad/s

t = ?

α = w/t

===> t = w/α = 0.2/0.04 = 5 sec

v = wr = 0.2 × 200 = 40 m/s

the net acceleration a = v/t = 40/5 = 8 m/s²

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Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

8)

The flux is calculated as

$\Phi = BA sin \theta$

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

$\theta=18^{\circ}$ is the angle between the direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

$A=(0.15 m)(0.08 m)=0.012 m^2$

So the flux is

$\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb$

See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

5 0

3 years ago

5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m

iris [78.8K]

Answer:

<h2>470.59 kg</h2>

Explanation:

The the mass of the car can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{400}{0.85} \\ = 470.58823...$

We have the final answer as

<h3>470.59 kg</h3>

Hope this helps you

7 0

3 years ago

What is the main difference between an internal cumbustion engine and a external combination engine

aleksandrvk [35]

Internal and external combustion engines are two types of heat engines: they convert thermal energy into mechanical energy. The main difference between internal and external combustion engine is that in internal combustion engines, the working fluid burns inside the cylinder, whereas in external combustion engines, combustion takes place outside the cylinder and heat is then transferred to the working fluid.

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3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

leonid [27]

Answer:

$KE_A=33\ J$

$KE_B=99\ J$

Explanation:

Given:

Let mass of the particle B be, $m_B=m$

then the mass of particle A, $m_A=3m$

Energy stored in the compressed spring, $E=132\ J$

Now when the compression of the particles with the spring is released, the spring potential energy must get converted into the kinetic energy of the particles and their momentum must be conserved.

Kinetic energy:

$\frac{1}{2}m_A.v_A^2+\frac{1}{2}m_B.v_B^2=132$

$3m.v_A^2+m.v_B^2=264$ .............................(1)

<u>Using the conservation of linear momentum:</u>

$m_A.v_A+m_B.v_B=0$

$3m.v_A+m.v_B=0$ .............................(2)

Put the value of $v_A$ from eq. (2) into eq. (1)

$3m\times (\frac{-v_B}{3})^2+m.v_B^2=264$

$v_B^2=\frac{198}{m}$ ...........................(3)

<u>Now the kinetic energy of particle B:</u>

$KE_B=\frac{1}{2}\times m_B\times v_B^2$

$KE_B=\frac{1}{2}\times m\times \frac{198}{m}$

$KE_B=99\ J$

Put the value of $v_B^2$ form eq. (3) into eq. (1):

$v_A^2=\frac{22}{m}$

<u>Now the kinetic energy of particle A:</u>

<u /> $KE_A=\frac{1}{2}m_A.v_A^2$ <u />

<u /> $KE_B=\frac{1}{2}\times 3m\times \frac{22}{m}$ <u />

$KE_A=33\ J$

6 0

3 years ago

You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work

hammer [34]

Answer:

1) F = 71.6 N , 2) F = 120.2 N , 3) P_m= 68.600 Pa, 4) V = 2.4210-5 m³

Explanation:

This is a problem of fluid mechanics, to find the force we must use its definition

P = F / A

F = P A

The area of the circular pipe is

A = π r² = π d²/4

The pressure is given by the expression

P = P_atm + ρ g h

1) the force on the outer side is

P = P_atm

we substitute in the expression of force

F = P_atm π d² / 4

let's calculate

F = 1,013 10⁵ π 0.03²/4

F = 7.16 10¹ N

F = 71.6 N

2) the force on the inner side

the pressure

P = P_atm + ρ g h

P = 1,015 10⁵ + 1,000 9.8 7

P = 1,701 10⁵ Pa

F = 1,701 10⁵ π 0.03² / 4

F = 1,202 10²

F = 120.2 N

3) manometric pressure

Pm = ρ g h

P m = 1000 9.8 7

P_m= 68.600 Pa

4) In this part they ask for the volume that comes out in time t= 3 h

to calculate this volume we can use the flow ratio

Q = A v

V t = A v

V = A v / t

sent is the velocity of the water that comes out, to calculate it we use the Bernoulli equation

we will use index 1 for the lake surface and ionice 2 apa the position of the plug

P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

As the lake has much more capacity than the pipeline, the velocity of the surface of the lake is peeling, in this case we approximate it steel

(P₁-P₂) + ρ G (y₁ -y₂) = ρ g v₂²

1000 9.8 v₂² = ρ g h + 1000 9.8 (7-0)

9800 v₂² = 1000 9.8 7 + 68600

v₂ = √ (137200)

v₂ = 370.4 m / s

t = 3 h (3600s / h) = 10800 s

we substitute in the volume equation

V = π d²/4 370.4 / 10800

V = 2.4210-5 m³

4 0

3 years ago