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Phoenix [80]
3 years ago
5

What is a situation in which friction is helpful?

Physics
1 answer:
bagirrra123 [75]3 years ago
8 0
When driving a car on the road
You might be interested in
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
A racquet ball with mass m = 0.221 kg is moving toward the wall at v = 13.9 m/s and at an angle of θ = 25° with respect to the h
Salsk061 [2.6K]

Answer:

1) 3.07kgm/s

2) 5.56kgm/s

3) 76.16N

4) 4.33kgm/s

5) 0.57s

6) -8.66J

Explanation:

Given

m = 0.221kg

v = 13.9m/s

θ = 25°

t = 0.073s

1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,

P(i) = mv(i)

P(i) = 0.221 * 13.9

P(i) = 3.07kgm/s

2) Magnitude of the change in momentum of the ball,

P(i,x) = P(i) cos θ

P(i,x) = 3.07 * cos25

P(i,x) = 3.07 * 0.9063

P(i,x) = 2.78

ΔP = 2P(i,x)

ΔP = 2 * 2.78 = 5.56kgm/s

3) magnitude of the average force exerted by the wall,

F(ave) = ΔP/Δt

F(ave) = 5.56/0.073

F(ave) = 76.16N

4) ΔP(z) = mv(f) - mv(i)

ΔP(z) = 0.221*-7.8 - 0.221*11.8

ΔP(z) = -1.72 - 2.61

ΔP(z) = 4.33kgm/s

5) F(ave) = ΔP/Δt

Δt = ΔP/F(ave)

Δt = 4.33 / 76.16

Δt = 0.57s

6) KE(i) = 0.5mv(i)²

KE(f) = 0.5mv(f)²

ΔKE = 0.5m[v(f)² - v(i)²]

ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]

ΔKE = 0.1105 ( 60.84 - 139.24 )

ΔKE = 0.1105 * -78.4

ΔKE = -8.66J

6 0
3 years ago
The total amount of energy and mass in the universe is _____.
bezimeni [28]
The answer is it stays constant.
5 0
3 years ago
Read 2 more answers
A satellite moves in a stable circular orbit with speed Vo at a distance R from the center of a planet. For this satellite to mo
slava [35]

Answer:

The new speed must be \frac{V_0}{\sqrt{2}}

Explanation:

In order for the satellite to be on a stable orbit around the planet, the gravitational attraction must be equal to the centripetal force that keeps the satellite in circular motion:

G \frac{Mm}{R^2}= m\frac{V_0^2}{R}

where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, V0 the speed of the satellite at distance R from the center of the planet.

We can re-write V0, the initial satellite speed, by re-arranging the equation:

V_0 = \sqrt{\frac{GM}{R}}

Now, if we want the satellite to orbit at a distance of 2R, the new tangential speed must be:

V' = \sqrt{\frac{GM}{2R}}=\frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}}= \frac{V_0}{\sqrt{2}}

3 0
3 years ago
A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumpe
AURORKA [14]

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

\triangle m = \frac{2d}{\lambda} (n - 1)

where

d = 3.95 \times 10^{-2}m

\lambda = 400 \times 10^{-9}m

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

=  \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

8 0
3 years ago
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