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garik1379 [7]
4 years ago
15

HELP!! PICTURE ATTACHED!

Physics
2 answers:
tamaranim1 [39]4 years ago
8 0
Question 17:

At point A, the snowboarder is on the point of moving and the potential energy would be at its maximum (the particle has to work against the force of gravity). The kinetic energy is zero since the snowboarder is not yet moving (has no velocity). 

Question 18:

At point C, the kinetic energy will be zero and the potential energy will be minimum. As the snowboarder moves from point C to B, there will be a transfer between the kinetic energy to the potential energy. At point B, the potential energy will be back to maximum.
yulyashka [42]4 years ago
4 0

17. At point A, the snowboarder has potential energy

Explanation:

There are two types of energy involved here:

- Gravitational potential energy: this is the energy related to the height of the snowboarder, and it is given by U=mgh, where m is the mass of the snowboarder, g is the gravitational acceleration and h is the height of the snowboarder relative to the ground

- Kinetic energy: this is the energy related to the motion of the snowboarder, and it is given by K=\frac{1}{2}mv^2, where v is the speed of the snowboarder

We see that at point A the snowboarder is located at a higher point, so h is larger, therefore he has gravitational potential energy.


18. From point C to point B, kinetic energy is converted into potential energy

Explanation:

The law of conservation of energy states that the mechanical energy (sum of potential and kinetic energy: E=U+K) is constant. At point C, the snowboarder is located at height h=0, so he only has kinetic energy (in fact, its speed is maximum at point C). As he moves towards point B, he increases his height: therefore, its potential energy increases, while his kinetic energy decreases (this implies that his speed decreases as well). Therefore, kinetic energy is converted into potential energy.

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kaheart [24]

Answer:

its sort of like how Pounds are measured in lbs

Explanation:

Pounds are measured in lbs when they dont sound anything similar, same thing applies to gun technology.

5 0
3 years ago
Be sure to show all 4 steps for each of the problems below.
Kisachek [45]

Answer:

1=920

2=2300

Explanation:

from the question,

M¹=23

H¹=4

g=10(constant)

using the formula

P. E,= mgh

P. E= 23×10×4=920

P. E=23×10×10=2300

7 0
4 years ago
Write an expression for the magnitude of the normal force, fn, acting on the block, in terms of f2 and the other variables of th
mario62 [17]
<span>If the force is fn, then it acting on the block, in the term of f2 and the other variables of the problems. then we assume the surface is rest on it's rigid, then the magnitude of the force is, fn =f2 sin(?) +m g</span>
7 0
4 years ago
Newton's law of cooling states that the rate of change of temperature of an object in a surrounding medium is proportional to th
Nastasia [14]

Answer:

T =  40 +  e^{3.68t} e^{2.99}

Explanation:

The differential equation for given  is given as

\frac{dT}{dt} = - k(T-T_s)

integrating above equation we have

ln(T-T_s) = -kt + C

At t = 0 , T(0) =  60

ln(60- 40) = -k\times 0 + C

2.99 = C

 At t =1 , T(1) = 40.49887

ln(40.49787 - 40) = -k\times 1 +  2.99

- k = -3.687

So we have

T- 40  = e^{3.68t + 2.99}

T =  40 +  e^{3.68t} e^{2.99}

4 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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