The impact of the type of material of which the slope is made on acceleration?

2 answers:

6 0

If you are asking about the effect different materials used for the slope of a ramp have on acceleration, higher acceleration results with a smoother surface material, less friction resistance.

denpristay [2]3 years ago

3 0

The slope is big and small

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Hearing rattles from a snake, you make two rapid displacements of magnitude 1.8 m and 2.4m. Draw sketches, roughly to scale, to

seraphim [82]

Answer:

The answer is a 4.2m

Explanation:

Given data

Please see attached the rough drawing for your reference.

From the drawing, you ran 18m west and 2.4m south

The displacement is

= 1.8+2.4

=4.2m

8 0

3 years ago

An electric pump has 2kW power . How much water will it lift every minute if the height is 10 m ?

valkas [14]

Answer:

Given that,

Power = 2000 W
time = 60 seconds
distance= 10m

Power = work done ÷ time

Here, since the movement is vertical, w = mgh

So,

Power = mgh÷t

2000 = (m × 9.8 ×10) ÷ 60

m = (2000 ×60) ÷98

m = 1224.5kg

3 0

3 years ago

a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the

ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: $F=m*a$ with this we can get both accelerations; solving for acceleration $a=\frac{F}{m}$ . Now $a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2})$ and $a_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2})$ . Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, $x=\frac{1}{2}*a_{girl}*t^{2}$ and $15.0-x=\frac{1}{2}*a_{sled}*t^{2}$ , solving for the time we get: $t=\sqrt{\frac{2x}{a_{girl} } }$ and $t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } }$ with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get: $\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }$ . Finally we get: $\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} }$ and replacing the values we have got: $\frac{x}{0.14} =\frac{(15.0-x)}{0.73}$ so $5.33*x=15-x$ so x=2.37 (meters).