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3 years ago

13

If you change the magnetic field in a closed loop of wire You induce in the loop a(n) A. current. B. voltage. C. electric field.

D. All of these.

2 answers:

Luba_88 [7]3 years ago

8 0

d: all of these because more coil will increase the current,voltage,and the electrical field hope this helped

SashulF [63]3 years ago

8 0

Answer:

The correct answer is D.

Explanation:

Magnetic induction is the process by which magnetic fields generate electric fields in a conductor. When an electric field is generated in a conductive material, the load carriers are subjected to a force and an electric current is induced in the conductor. The magnetic field generates an electric current in a conductor, an induced fem appears, so that the charges of the conductor move generating an induced current.

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Explanation:

It is known that the relation between speed and distance is as follows.

velocity = $\frac{distance}{time}$

As it is given that velocity is 6 m/s and distance traveled by the bear is (d + 29). Therefore, time taken by the bear is calculated as follows.

$t_{bear} = \frac{(d + 29)}{6 m/s}$ ............. (1)

As the tourist is running in a car at a velocity of 4.2 m/s. Hence, time taken by the tourist is as follows.

$t_{tourist} = \frac{d}{4.2}$ ............. (2)

Now, equation both equations (1) and (2) equal to each other we will calculate the value of d as follows.

$t_{bear} = t_{tourist}$

$\frac{(d + 29)}{6 m/s}$ = $\frac{d}{4.2}$

4.2d + 121.8 = 6d

d = $\frac{121.8}{1.8}$

= 67.66

Thus, we can conclude that the maximum possible value for d is 67.66.

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A brick is dropped from a high scaffold. <br> a. How far does the brick fall during this time?

Alecsey [184]

Answer:

a: after 1 seconds it will have fallen 0.2452

after 2 seconds it will have fallen 0.981

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Explanation:

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g = G*M/R^2

earths gravitational constant is about 9.807

g = 9.807*M/R^2

The average weight of a brick is 5 pounds and I'm going to say it's 10 feet off the ground.

g = 9.807*5/10^2. g = 0.4905 so every second the brick will go 0.4905 fps faster. (fps means feet per second.)

after 1 seconds it will have fallen 0.2452

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after 3 seconds it will have fallen 2.2072

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4 years ago

Calculate the velocity displace = 3500km and time = 3 hours

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3 years ago

A ball is kicked at an angle . It is intended that the ball lands in 휃=44.5°the back of a moving truck which has a trunk of leng

evablogger [386]

Answer:

$V_{0min}=19.63m/s$

$V_{0max}=20.45m/s$

Explanation:

In order to solve this problem, we must start by drawing a diagram of the situation, this will help us understand the problem better so we can take the best possible approach. (See attached picture).

So next, we can see that the ball will have a parabolic trajectory, so we can split it into its x- and y-components. Let's start by analyzing the movement of the ball in the x-direction.

$x_{b}=x_{0b}+V_{0bx}t$

we know that the x-component of the velocity of the ball can be found by using the following equation:

$V_{0bx}=V_{0b}cos \theta$

So the equation can be rewritten as:

$x_{b}=x_{0b}+V_{0b}(cos \theta) t$

we can say that the ball will start its movement from the origin, so the initial displacement is zero. So the simplified equation is:

$x_{b}=V_{0b}(cos \theta) t$

So now we can analyze the vertical movement of the ball. We can describe it with the equation:

$y_{b}=y_{0b}+V_{0by}t+\frac{1}{2}at^{2}$

we can say that the ball will start its movement at the origin so we can say that the initial position of ball is equal to zero and the final position is zero as well, so the equation turns to:

$0=V_{0by}t+\frac{1}{2}at^{2}$

we can describe the vertical velocity of the ball with the following equation:

$V_{0by}=V_{0b}sin \theta$

and the acceleration as:

$a=-g=-9.81m/s^{2}$

so we can substitute them in the original equation:

$(V_{0b}sin \theta)t-\frac{1}{2}gt^{2}=0$

which can be solved for t by factoring, so we get:

$t(V_{0b}sin\theta - \frac{gt}{2})=0$

so we get:

t=0 which is the initial moment at which the ball starts moving

$V_{0b}sin\theta - \frac{gt}{2}=0$

which solves to:

$t=\frac{2V_{b0}}{g}sin \theta$

so we can substitute this into the x-movement equation:

$x_{b}=V_{0b}cos \theta(\frac{2V_{b0}}{g}sin \theta)$

which simplifies to:

$x_{b}=\frac{2V_{b0}^{2}cos\theta sin\theta}{g}$

By using trigonometric identities we can further simplify this to:

$x_{b}=\frac{V_{b0}^{2}sin(2\theta)}{g}$

We can leave that equation there by the time and start analyzing the movement of the truck. We can determine the distance between the back of the truck and the initial position of the ball with the following equation:

$x_{b}=x_{t}+d$

we know that:

$x_{t}=V_{t}t$

from the previos steps we know that the time t can be found by analyzing the y-movement of the ball, so we can use that same equation to substitute for t, so we get:

$x_{t}=V_{t}(\frac{2V_{b0}}{g}sin \theta)$

which can be substituted into the first equation:

$x_{b}=V_{t}(\frac{2V_{b0}}{g}sin \theta)+d$

and we can combine it with the equation we got for the x-movement of the ball.

$V_{t}(\frac{2V_{b0}}{g}sin \theta)+d=\frac{V_{b0}^{2}sin(2\theta)}{g}$

So we are dealing with a quadratic equation which can be rewritten like this:

$\frac{2V_{t}V_{0b}sin\theta}{g}+d=\frac{V_{0b}^{2}sin(2\theta)}{g}$

we can rewrite it in standard form so we get:

$\frac{V_{0b}^{2}sin(2\theta)}{g}-\frac{2V_{t}V_{0b}sin\theta}{g}-d=0$

So we can proceed and substitute the values we know, so we get:

$\frac{V_{0b}^{2}sin(2(44.5^{o})}{9.81}-\frac{2(11.4)V_{0b}sin(44.5^{o})}{9.81}-7.3=0$

which yields:

$0.10192V_{0b}^{2}-1.629V_{0b}-7.3=0$

which can be solved by using the quadratic formula:

$V_{b0}=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

so we get:

$V_{b0}=\frac{-(-1.629)\pm \sqrt{(-1.629)^{2}-4(0.10192)(-7.3)}}{2(0.10192)}$

which gives us two answers:

$V_{b0min}=19.63m/s$

or

$V_{b0min}=-3.65m/s$

Due to the nature of the problem we only take the positive answer, since

a negative answer would mean the ball is moving to the opposite side. So the answer is:

$V_{b0min}=19.63m/s$

In order to find the maximum velocity we follow the sme procedure, with the difference that in this case d=7.3m+2m=9.3m

The additional 2 meters come from the length of the trunk of the truck.

Turning the equation into:

$0.10192V_{0b}^{2}-1.629V_{0b}-9.3=0$

and changing the answers to:

$V_{b0}=\frac{-(-1.629)\pm \sqrt{(-1.629)^{2}-4(0.10192)(-9.3)}}{2(0.10192)}$

$V_{b0max}=20.45m/s$

or

$V_{b0min}=-4.46m/s$

so we take only the positive answer into account leaving us with the answer:

$V_{b0max}=20.45m/s$

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