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denis23 [38]
3 years ago
12

Please helpList 5 best career

Physics
1 answer:
lukranit [14]3 years ago
4 0

1. Software Developer

2. Doctor

3. Neurosurgeon

4. Anesthesiologist

5. Financial Analyst

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please help. Sound travels at 330 m/s. If a lightning bolt strikes the ground 5 m away from you, how long will it take for the s
Olenka [21]

Answer:

0.015 seconds

Explanation:

5/330=0.015

3 0
3 years ago
A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop
liq [111]

Answer:

692.31 N

Explanation:

Applying,

F = ma............... Equation 1

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 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
What is magnification
il63 [147K]
The action or process of magnifying something or being magnified, especially visually. Hope this helped
5 0
4 years ago
Identify the forces acting on the cars and explain why the cars do not slide down the hill
Dvinal [7]
<span>Examples of outside forces acting on a car is gravity, wind, and other cars. Cars do not slide down hills because their weight, combined with the friction of their tires against the road, hold them in place. </span>
5 0
4 years ago
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