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Novay_Z [31]
3 years ago
10

A 40kg-skier starts at the top of a 12-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose t

o friction
A.She doesn't lose any because mechanical energy is always conserved

B.4,704 J

C.2,000 J

D.2,704 J
Physics
2 answers:
dexar [7]3 years ago
6 0
I took this quiz the answer was D.2,704 J. Just trying to help.
KIM [24]3 years ago
4 0
Energy at top = U = mgh = 40 * 9.8 * 12 = 4704 J

Energy at bottom = 1/2 mv² = 1/2 * 40 * 10² = 4000 / 2 = 2000 J

Energy Lost = Final - Initial = 4704 - 2000 = 2704 J

In short, Your Answer would be Option D

Hope this helps!
You might be interested in
A lamp is connected to the power supply.
Vika [28.1K]

Answer:

There are 45 turns in the secondary coil.

Explanation:

Given that,

Input potential of the lamp, V_{in}=5\ V

The output potential of the lamp, V_{out}=1.5\ V

Number of turns in primary coil, N_P=150

We need to find the number of turns needed on the secondary coil. We know that the ratio for a transformer is as follows :

\dfrac{V_{out}}{V_{in}}=\dfrac{N_s}{N_P}\\\\N_s=\dfrac{V_{out}N_P}{V_{in}}\\\\N_s=\dfrac{1.5\times 150}{5}\\\\N_s=45\

So, there are 45 turns in the secondary coil.

5 0
2 years ago
At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor
Irina18 [472]

Answer:

The charge flows in coulombs is

dq=1.843x10^{-5}C

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

I=N*\frac{dF}{Rdt}

\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}

dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}

N=1300, \beta=0.703, A=\pi*r^2=\pi*0.10^2=0.01\pi m^2, R=99.4+202=301.4Ω

\alpha_f=14.6,\alpha_i=165.4

Replacing :

dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}

dq=1.843x10^{-5}C

5 0
3 years ago
If the mass of a material is 99 grams and the volume of the material is 22 cm3, what would the density of the material be?
Kazeer [188]
You are asked to give the answer in <span>g/cm3. So without knowing any single formulae you can just divide grams by cm3. 

</span>\frac{99 grams}{22 cm^{3} }

 = 4.5 g/cm3
7 0
3 years ago
A Porsche challenges a Honda to a 400 m race. Because the Porsche's acceleration of 3.4 m/s2 is larger than the Honda's 3.0 m/s2
padilas [110]

Answer:

Winner wins by 0.969 s

Explanation:

For the Porche:

Given:

Displacement of Porsche s = 400 m

Acceleration of Porsche a = 3.4 m/s^2

From Newton's second equation of motion,

s = ut + (1/2) a t^2 (u = 0 as the car was initially at rest)

Substituting the values into the equation, we have

t^2 = (2 * 400) / 3.4

= 235.29 / 3.4

t = 15.33 s

For the Honda:

Displacement of Honda = 310 m

Acceleration of Honda = 3 m/s^2

Applying Newton's second equation of motion

s = ut + (1/2) a t^2 (u = 0 for same reason)

Substituting the values into the equation, we obtain

t^2 = (2 * 310) / 3

= 620 / 3

t = 14.37 s

Hence

The winner (honda) wins by a time interval of = 15.33 - 14.37    

=0.969 s

8 0
2 years ago
MATHPHYS HELP
rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0
3 years ago
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