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Bess [88]
4 years ago
8

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

itially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 130 ms. What is the average induced emf around the border of the circular region?
Physics
1 answer:
Daniel [21]4 years ago
3 0

Answer:81.57\mu V

Explanation:

Given

radius of circular region r=1.50 mm

A=\pi r^2=7.069\times 10^{-6}\ m^2

Magnetic Field B=1.50\ T

time t=130 ms

Flux is given by

\phi =B\cdot A

change in Flux d\phi =(B_f-B_i)A

Emf induced is e=\frac{\mathrm{d} \phi}{\mathrm{d} t}

e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}

e=81.57 \mu V

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When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver
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Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times m\times(32.3)^2

We need to calculate the escape speed of the planet

Using formula of kinetic energy

\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}

\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2

\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2

v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}

v=41.29\ m/s

Hence, The escape speed of the planet is 41.29 m/s.

4 0
3 years ago
Which answer accurately describes the difference between speed and velocity? Question 8 options:
Snowcat [4.5K]
A., Because speed is constant unless you stop it, or slow down <span />
7 0
4 years ago
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After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 48.0 cm . The explorer finds that
Ber [7]

Answer:

g = 12.22 m/s²

Explanation:

The time period of this pendulum is given as follows:

T = \frac{time\ taken}{no. of cycles}\\\\T = \frac{144\ s}{93}\\\\T = 1.55\ s

but the formula for the time period of a simple pendulum is as follows:

T=2\pi \sqrt{\frac{l}{g}}\\

where,

L = length of pednulum = 48 cm = 0.48 m

g = magnitude of th gravitational acceleration on this planet = ?

Therefore,

1.55\ s=2\pi \sqrt{\frac{0.48\ m}{g}}\\\\\sqrt{g} = 2\pi \sqrt{\frac{0.48\ m}{1.55\ s}}\\\\g = 3.49^{2}\\

<u>g = 12.22 m/s²</u>

6 0
3 years ago
Help please
NARA [144]

Explanation:

help please

A lamp is marked 1.8w in normal brightness it carries a

3 0
3 years ago
A car increases its speed from 20km/hr to 50km/hr in seconds. Its acceleration is _____.
mash [69]

Answer:

Option D

0.83 m/s2

Explanation:

Time is assumed as 10 seconds

First, convert the speeds from km/h to m/s

20 km/h\times \frac {1000 m}{3600 s}=5.555555556 m/s \approx 5.56 m/s

50 km/h\times \frac {1000 m}{3600 s}=13.88888889  m/s \approx 13.89 m/s

Acceleration, a=\frac {v-u}{t} where u and v are the initial and final velocities respectively, t is the time taken to accelerate.

Substituting 13.89 m/s for v, 5.56 m/s for u and 10 s for t then

a=\frac {13.89-5.56}{10}=0.8333333 m/s^{2}\approx 0.83 m/s^{2}

7 0
3 years ago
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