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VMariaS [17]
3 years ago
14

A skier weighing 86.2 kg starts from rest and slides down a 32.0-m frictionless slope that is inclined at an angle of 15.0° with

the horizontal. Ignore air resistance.
a. calculate the work done by gravity on the skier.

b.calculate the work done by the normal force on the skier

Physics
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

A. = 26.11kJ

B = 0N

Explanation:

Mass = 86.2kg

Distance (s) = 32m

Angle of the slope = 15°

a. Work done due to gravity

W = force * distance * angle of displacement

W = F * S * cos θ

Force = mass * acceleration due to gravity

Force = 86.2 * 9.8

Force = 844.76N

Work = 844.76 * 32 * cos 15

Work = 26111.21J

Work = 26.11kJ

b. The workdone by the normal force is equal to zero since the normal force is perpendicular to the displacement.

Hence the normal force does not act on the body.

N = 0N

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A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph
kenny6666 [7]

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

4 0
3 years ago
The horizontal surface on which the block (mass 2.0 kg) slides is frictionless. The speed of the block before it touches the spr
ch4aika [34]

Answer:3.67 m/s

Explanation:

mass of block(m)=2 kg

Velocity of block=6 m/s

spring constant(k)=2 KN/m

Spring compression x=15 cm

Conserving Energy

energy lost by block =Gain in potential energy in spring

\frac{m(v_1^2-v_2^2)}{2}=\frac{kx^2}{2}

2\left [ 6^2-v_2^2\right ]=2\times 10^3\times \left [ 0.15\right ]^2

v_2=3.67 m/s

7 0
4 years ago
A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second, and the t
slega [8]

Answer:

The tension increases to four times its original value.

Explanation:

v = Velocity

r = Radius

m = Mass of stone

The centripetal force is

F_c=m\dfrac{v^2}{r}

The tension will balance the centripetal force

T=m\dfrac{v^2}{r}

\\\Rightarrow T\propto v^2

\dfrac{T_1}{T_2}=\dfrac{v_1^2}{v_2^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{v_1}{2^2v_1^2}\\\Rightarrow \dfrac{T_1}{T_2}=\dfrac{1}{4}\\\Rightarrow T_2=4T_1

The new tension will be 4 times the old tension

5 0
3 years ago
A load of 45N attached to a spring that is hanging vertically stretches the spring 0.14m. What is the spring constant?
Vesna [10]
6.3 That Would be the I answer I think but Check on Google For the formula
6 0
3 years ago
A 12-V battery is connected across a 100-Ω resistor. How many electrons flow through the wire in 1.0 min?
Vanyuwa [196]

Answer:

The quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

Explanation:

Given;

emf of the battery, V = 12 V

resistance of the resistor, R = 100-Ω

time of current flow, t = 1 min

charge of 1 electron = 1.602 x 10¹⁹ C

The current through this circuit is given by;

I = V / R

I = (12) / (100)

I = 0.12 A

The quantity of charge or electron flowing the wire in the given time is calculated as;

Q =It

where;

I is the current flowing through the wire

t is the time of current flow = 1 x 60s = 60 s

Q = 0.12 x 60

Q = 7.2 C

1.602 x 10⁻¹⁹ C --------------- 1 electron

7.2 C -----------------------------? electron

= \frac{7.2 }{1.602*10^{-19}} \\\\= 4.5*10^{19} \ electrons

Therefore, the quantity of charge or electron flowing the wire in the given time is 4.5 x 10¹⁹ electrons.

4 0
3 years ago
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