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Usimov [2.4K]
3 years ago
15

At what time were the racers at the same position? O 15 O 3s O 5s 0 4s

Physics
2 answers:
viktelen [127]3 years ago
7 0
The answer is B
Have a great day
Jlenok [28]3 years ago
5 0
The Answer is:

O 3s

Hope you got it right.
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A small car and a large heavier bus are traveling at the same speed. Which has more momentum?
lions [1.4K]

Answer:

small car since they weigh less than a bus

Explanation:

7 0
3 years ago
As the frequency of a wave increases the wavelength.
zaharov [31]

Answer:

decreases

Explanation:

Remeber:

There is always inverse relation between frequency and wavelength.

So if one of them increases, other decreases and vice-versa.

f ∝ 1 / λ

4 0
2 years ago
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What are conductors and insulators? Give at least five example of each ​
Leona [35]

Answer: conductors are substances that allow heat or electricity to pass through . It deals with only the flow of electrons.

Eg. water, copper wire, iron rod, some ceramic materials, metallic nail.

Insulators are materials that do not allow heat or electricity to pass through.

Eg. Book, plastic, rubber, glass, paper

Explanation:

8 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
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