3 years ago

A race car makes one lap around a track of radius 50m in 90sec. what is the average velocity?

Physics

1 answer:

drek231 [11]3 years ago

7 0

Average SPEED = 3.491 m/s.

Average VELOCITY = zero.

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34.6 cL= (blank) hL convert

shepuryov [24]

Answer

0.00346 hL

Explanation

cL means Centilitre while hL means Hectolitre.

10,000 cL = 1 hL

∴ 34.6 cL = 34.6/10,000 hL

= 0.00346 hL

3 0

4 years ago

If you run at 12 m/s fr 15 minutes, how far will you go

notsponge [240]

Answer:

180m

Explanation:

We can use the formula [ d = st ].

12 * 15 = 180m

Best of Luck!

4 0

3 years ago

Can somone pls help me??!! i’m very stuck

Alchen [17]

Answer:

its the third one

Explanation:

3 0

3 years ago

A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

ac = 3.92 m/s²

5 0

3 years ago

if you have a mass of 55 kg and you are standing 3 meters away from your car, which has a mass of 1234 kg, how strong is the for

bagirrra123 [75]

Gravitational force between two masses is given by formula

$F = \frac{Gm_1m_2}{r^2}$

here we know that

$m_1 = 55 kg$

$m_2 = 1234 kg$

$r = 3 m$

$G = 6.67 \times 10^{-11} Nm^2/kg^2$

now from the above equation we will have

$F = \frac{(6.67 \times 10^{-11})(55)(1234)}{3^2}$

$F = 5.03 \times 10^{-7}N$

so above is the gravitational force between car and the person

5 0

3 years ago