The "Big Dig" was the nickname of the civil engineering project that redesigned the highway Infrastructure for the city of

Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,

Tems11 [23]

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

F

=

(

20

×

10

−

9

C

)

(

10

m/s

)

(

5

×

10

−

5

T

)

=

1

×

10

−

11

(

C

⋅

m/s

)

(

N

C

⋅

m/s

)

=

1

×

10

−

11

N

6 0

3 years ago

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What was a campaign belief in the 1980 presidential election? Carter called for a stronger national defense. Carter promised to

andreyandreev [35.5K]

Answer:

C

Explanation:

On edge 2021

6 0

3 years ago

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A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago

List two ways you can make an informal survey

solniwko [45]

The two ways you can use to make an informal survey are:

make field observations
interview people using informal unstructured techniques

<h3>What are informal surveys?</h3>

In informal surveys can be regarded as a type of survey that can be made by the researcher by going to the field themselves and this can be done by using different methods or ways.

For instance, the researcher can go out to interview people that can give the data that is needed about the research such as informally asking them questions, unstructured techniques can also be used to solve critical issues.

learn more about survey at: brainly.com/question/6947486

#SPJ9

8 0

1 year ago

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Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.

-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

r = 0.5*θ

θ = 0.5*t^2 ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

dr/dt = 0.5*dθ/dt

d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

dθ/dt = t

d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

tan Ψ = r / (dr/dθ) = 1 / 0.5

Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

a_r = d^2r / dt^2 - r * dθ/dt

a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

N_c = 3.03 N

F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

F = 1.81 N

4 0

3 years ago