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Savatey [412]
3 years ago
12

A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with th

e horizontal. If an average friction force of 4,000 N impedes the motion of the car, find the speed of the car at the bottom of the driveway.
Physics
1 answer:
ehidna [41]3 years ago
5 0

Answer:

The speed at the bottom of the driveway is3.67m/s.

Explanation:

Height,h= 5sin20°= 1.71m

Potential energy PE=mgh= 2000×9.8×1.71

PE= 33516J

KE= PE- Fk ×d

0.5mv^2= 33516 - (4000×5)

0.5×2000v^2= 33516 - 20000

1000v^2= 13516

v^2= 13516/1000

v =sqrt 13.516

v =3.67m/s

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A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in met
Tamiku [17]

Answer:

4 tonne/m³

Explanation:

ρ = m / V

ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))

ρ = 0.0041 g/mm³

Converting to tonnes/m³:

ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³

ρ = 4.1 tonne/m³

Rounding to one significant figure, the density is 4 tonne/m³.

6 0
3 years ago
Why is the Earth generally warmer at the equator than at the poles?
Elenna [48]
Because of the earths curve and temperature. The temperature in an area depends on the amount of the Sun's energy reaching the surface in that area. The equator tilts closer to the sun than the poles do. I hope this helps!
5 0
2 years ago
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A car passes you at 18 m/s to the north and increases its speed at a rate of 3.0 m/s. Determine the car's velocity when it has a
monitta

122

...................................................................,.......,..................................

8 0
2 years ago
A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the
mash [69]

Answer:

\Delta x = v_0 t + \frac{1}{2}at^2

Explanation:

To solve this problem, we can use the following suvat equation:

\Delta x = v_0 t + \frac{1}{2}at^2

where

\Delta x is the vertical displacement of the frog

v_0 is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from v_0, all the other quantities are known. In fact:

\Delta x =0.1 m is the vertical displacement

t = 2 s is the total time of flight

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

Therefore, solving for v_0, we find the initial velocity of the frog:

v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s

4 0
3 years ago
If a 0.50-kg block initially at rest on a frictionless, horizontal surface is acted upon by a force of 3.7 N for a distance of 6
mel-nik [20]

The block's velocity is determined as 10.03 m/s.

<u>Explanation:</u>

According to work energy theorem, the work done on an object is equal to the change in kinetic energy of the object.

So, work done = Kinetic energy

\text {Force } \times \text { displacement }=\frac{1}{2} \times m \times v^{2}

Thus, the velocity can be determined as

\text {velocity}=\sqrt{\frac{2 \times \text {Force} \times \text {displacement}}{m}}

\text {velocity}=\sqrt{\frac{2 \times 3.7 \times 6.8}{0.50}}=\sqrt{\frac{50.32}{0.50}}=\sqrt{100.64}

Velocity = 10.03 m/s.

So the block's velocity is determined as 10.03 m/s.

5 0
3 years ago
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