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Studentka2010 [4]
2 years ago
12

How could you increase the sliding friction between the sled and track? How could you decrease it?​

Physics
2 answers:
Oksanka [162]2 years ago
7 0

Explanation:

<h2>If the normal force is the weight of objects you are sliding across the floor, adding or subtracting weight will increase or decrease the sliding friction. But also, if you are pulling an object with a rope, pulling at an upward angle will reduce the normal force.</h2>
Rudik [331]2 years ago
6 0

Answer:

You could put sheet metal lining the bottom or something similar to that

Explanation:

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PLEASE HELP!!!!!!! MY HOMEWORK IS DUE TODAY !!!!!!!!
bogdanovich [222]

Answer:

Distance = 25000000 miles

Time = 50 hours

Explanation:

Venus is the closest planet to Earth. It is about 25 million miles away from Earth. Its precise distance depends on where both Venus and Earth are in their respective orbits

Given that

Speed V = 500000 mph

Distance d = 25 000,000 miles

Speed = distance/ time

Time = distance/speed

Time = 25000000/500000

Time = 50 hours

It will therefore take 50 hours to get to venus at that speed.

5 0
3 years ago
As a wave travels through a medium, it displaces particles in a direction parallel to the motion of the wave. We can conclude th
Gre4nikov [31]
Transvere wave because the direction which the particles are being displaced
5 0
3 years ago
A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum h
Nikitich [7]

Hope this helps a little

initial distance up = 2

initial velocity component up = 9 sin 60 = 7.79

v = 9 sin 60 - 9.8 t

when v = 0, we are there

9.8 t = 7.79

t = .795 seconds to top

h = 2 + 7.79(.795) - 4.9(.795^2)

7 0
4 years ago
Read 2 more answers
A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f
hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

h=ut+\frac{1}{2}gt^2

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2    .... (2)

By equation the equation (1) and (2), we get

41.7=1.12 u +4.9 \times 1.12^{2}

u = 31.75 m/s

5 0
3 years ago
Marcia flew her ultralight plane to a nearby town against a head wind of 15 km/h in 2h 20 min. the return trip under the same wi
insens350 [35]

Let the distance between the towns be d and the speed of the air be s.

distance = speed * time

convert the minutes time into hours.

When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:

d

s−15

=

7

3

return trip is then :

d

s+15

=

7

5

Cross-multiplying both we get the two-variable system:

3d=7∗(s−15)5d=7∗(s+15)

3d=7s−1055d=7s+105

subtract first equation from second equation we get

2d=210d=105km

Substitute the value of d in the above equations for s.

5∗105=7s+1057s=420s=60km/hr

8 0
3 years ago
Read 2 more answers
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