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2 years ago

Suppose you and your bicycle have a total mass of 70 kg. You are moving at 10 m/s. How much kinetic do you and the bicycle have?

Physics

1 answer:

Margarita [4]2 years ago

8 0

Hi,

we Know that,

Kinetic energy of an object which is moving with velocity v and having mass m $Kinetic energy=mxv^2/2.........................(1)here m=70kg\\and v=10m/s\\Put values of m and v in equation 1\\kinetic energy= 70x10^2/2\\kinetic energy=35x100\\=3500joule$

therefore kinetic energy of bicycle and you=3500Joule

Hope this answer will help you

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How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.

-BARSIC- [3]

Answer:

4.2s

Explanation:

Given parameters:

Power = 2190W

Mass of box = 1.47 x 10⁴g

distance = 6.34 x 10⁴mm

Unknown:

Time = ?

Solution:

Power is the rate at which work is done;

Mathematically;

Power = $\frac{work done}{time}$

Time = $\frac{work done}{power}$

Work done = weight x height

convert mass to kg;

100g = 1kg;

1.47 x 10⁴g = 14.7kg

convert the height to m;

1000mm = 1m

6.34 x 10⁴mm gives 63.4m

Work done = 14.7 x 9.8 x 63.4 = 9133.4J

Time taken = $\frac{9133.4}{2190}$ = 4.2s

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2 years ago

A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w

nirvana33 [79]

Answer:

$|D_{depth} |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\ -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m$

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3 years ago

Consider two waves X and Y traveling in the same medium. The two carry the same amount of energy per unit time, but X has one-se

RideAnS [48]

Answer:

7 / 1

Explanation:

The ratio of their amplitude = one-seventh and the ratio of their amplitude = the ratio of their wavelength

Ax / Ay = λx / λy = 1 / 7

λy / λx = 7 / 1

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2 years ago

Circuit has a single pathway for the electrons of flow.<br> series<br> parallel

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Series circuit!!!!!!!!!!

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3 years ago

Read 2 more answers

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

7 0

3 years ago