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Masteriza [31]
3 years ago
12

A constant magnetic field of 0.50 T is applied to a rectangular loop of area 3.0 × 10-3 m2. If the area of this loop changes fro

m its original value to a new value of 1.6 × 10-3 m2 in 1.6 s, what is the emf induced in the loop?
Physics
1 answer:
Nonamiya [84]3 years ago
8 0

Answer:

e= 0.440 mV(millivolt)

Explanation:

Given Magnetic field B= 0.50T

A_{1}=3\times10^{-3}

A_{2}=1.6\times10^{-3}\\

Δt= 1.6sec

induced emf e=\frac{\mathrm{d} \phi }{\mathrm{d} t}

⇒e=B\frac{\mathrm{d} A}{\mathrm{d} t}

e=0.50\times\frac{3\times10^{-3}-1.6\times10^{-3}}{1.6}

e=0.440\times10^{-3} volt

e= 0.440 mV(millivolt)

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When driving a Toyota avensis car along a straight road for 16.5km at
Vitek1552 [10]

The average velocity of the whole journey will be total distance covered divided by the total time. It will be approximately equal to 8 m/s. The right answer is option B

<h2>VELOCITY</h2>

Velocity is the distance travelled in a specific direction. While the average velocity of the whole journey will be total distance covered divided by the total time

When driving a Toyota avensis car along a straight road for 16.5km at

50km/h,

The velocity = 50 km/h

Distance = 16.5 km

Use the speed formula to calculate time.

Speed = distance / time

Time = distance / speed

Time = 16.5 / 50

Time = 0.33 s

If over the next 20min, you walked another 2.5km further along the road for a petrol station, Then,

average velocity = Total distance covered divided by total time taken.

Where

The time t = 20/60 = 0.333 h

Total time = 0.33 + 0.3333

Total time = 0.6633333

Total distance = 16.5 + 2.5

Total distance = 19 km

Average velocity = 19 / 0.66333

Average Velocity = 28.64 km/h

Now convert Km/h to m/s

(28.6432 x 1000) / 3600

286432 / 3600

7.956m/s

Therefore, the average velocity of the whole journey from beginning of the drive to the arrival at the filling station will be approximately 8 m/s

Learn more about velocity here: brainly.com/question/6504879

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3 years ago
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uranmaximum [27]

Answer:

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3 years ago
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A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge
mamaluj [8]

Answer:

The magnitude of the electric field is 8.6\times10^{2}\ N/C

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

a=\dfrac{v-u}{t}

Put the value into the formula

a=\dfrac{160-0}{2.10}

a=76.19\ m/s^2

We need to calculate the magnitude of the electric field

Using formula of electric field

E=\dfrac{F}{q}

E=\dfrac{ma}{q}

E=\dfrac{a+g}{\dfrac{q}{m}}

Put the value into the formula

E=\dfrac{76.19+9.8}{0.100}

E=8.6\times10^{2}\ N/C

The direction is upward.

Hence, The magnitude of the electric field is 8.6\times10^{2}\ N/C

4 0
4 years ago
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choli [55]

Answer:

B. Acceleration in the direction of motion speeds you up

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Acceleration is defined as when something gains speed. For example, when a car speeds up. Deceleration is when the car slows down, and looses speed. When defining these terms, think of a car going faster, then slowing down at a red light.

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A tennis ball with mass m travelling west,hits a wall with a velocity of v.It bounces back with a velocity 1 ,63 v an easterly d
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Answer:

163 v an easterly direction

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3 years ago
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