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inna [77]
2 years ago
7

_____ of rough surfaces reduces friction

Physics
1 answer:
Radda [10]2 years ago
3 0

Answer:

Hey shaikaadil700 !

<u> </u><u>Lubricating</u><u> </u> of rough surfaces reduces friction.

Explanation:

• Lubricating is the smoothening or polishing of the surfaces

.

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Two planes leave wichita at noon. one plane flies east 30 mi/h faster than the other plane, which is flying west. at what time w
scoundrel [369]
You have to take note of the individual directions of the plane. Since one is heading east, and the other is heading west, the planes are heading at opposite directions. So, it means that their distance between each other would be equal to 1,200 miles which accounts for the sum of their individual distances. The equation is as follows:

Total Distance = Distance of slower plane + Distance of faster plane
1,200 miles = st + (30+s)(t)
where
s is the speed of the slower plane and t is the time. Since both are not given, the final answer would just be in terms of s.
1,200 = t(s + 30 + s)
t = 1200/(30+2s)
t = 600/(15+s)
4 0
3 years ago
A car has a force of 2000N and a mass of 1000kg. What is the acceleration
Anastaziya [24]

Answer:

2m/s/s

Explanation:

The formula goes- F=MA

F-Force M-Mass & A-Acceleration

We need to rearrange this formula to find the acceleration-

A=F/M

All we need to do now is substitute the values in

A=2000N/1000kg

A=2m/s^2

In the given option the last option (2m/s/s) would be the ans, as it's the same as 2m/s^2

So ya, I guess that's all

6 0
2 years ago
A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas
Radda [10]

Answer:

change in momentum, \Delta p=7.65 \,kg.m.s^{-1}

  • \Delta p_x= 6.6251 \,kg.m.s^{-1}
  • \Delta p_y= 3.825 \,kg.m.s^{-1}

Average Force, F=144.3396\,N

  • F_x=125.0018\,N
  • F_y=72.1698\,N

Explanation:

Given:

angle of kicking from the horizon, \theta= 30^{\circ}

velocity of the ball after being kicked, v=18 m.s^{-1}

mass of the ball, m=0.425\, kg

time of application of force, t=5.3\times 10^{-2}\,s

We know, since body is starting from the rest

\Delta p=m.v.....................(1)

\Delta p=0.425\times 18

\Delta p=7.65 \,kg.m.s^{-1}

Now the components:

\Delta p_x= 7.65\times cos 30^{\circ}

\Delta p_x= 6.6251 \,kg.m.s^{-1}

similarly

\Delta p_y= 7.65\times sin 30^{\circ}

\Delta p_y= 3.825 \,kg.m.s^{-1}

also, impulse

I=F\times t.........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

F\times t=m.v

F\times 5.3\times 10^{-2}= 7.65

F=144.3396\,N

Now, the components

F_x=144.3396\times cos 30^{\circ}

F_x=125.0018\,N

&

F_y=144.3396\times sin 30^{\circ}

F_y=72.1698\,N

6 0
3 years ago
The closer together a magnets field lines are,
elena55 [62]
The stronger they will be
4 0
3 years ago
A 58 kg skier is going down a 35 degree slope. The areaof each
maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

N = mgcos\theta

Since the skier is standing on two skis, his weight will be divide by two

N' = \frac{mgcos\theta}{2}

Pressure is given as the force applied in a given area, that is

P = \frac{F}{A}

Replacing F with N'

P = \frac{N'}{A}

P = \frac{\frac{mgcos\theta}{2}}{A}

Our values are given as,

m = 58kg

g = 9.8m/s^2

\theta = 35\°

A = 0.3m^2

Replacing we have that

P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}

P = 776.01Pa

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0
3 years ago
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