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pav-90 [236]
3 years ago
8

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of

the bike and the rider just before riding up the hill? if the rider just coasts up the hill, at what height will the bike come to a stop?(neglect friction)
Physics
1 answer:
Anestetic [448]3 years ago
3 0

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

KE = \frac{1}{2}m* v^{2}

It is given:

v = 8.5 m/s

m = 91 kg

So:

KE= \frac{1}{2}*91*8.5^{2} =3,287.4J


b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J

m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height


Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m

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Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So mg=Kx

K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

Now we know that \omega =\sqrt{\frac{K}{m}}

So \frac{2\pi }{T} =\sqrt{\frac{K}{m}}

\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

T=2.1371sec

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If light travels from oil (slower medium) to water (faster medium) at an angle, what happens to the direction of the light ray i
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A 10 ohms resistor is powered by a 5-V battery. The current flowing<br> through the source is:
mario62 [17]
  • Resistance=R=10ohm
  • Voltage=V=5V
  • Current=I

Applying ohm's law

\\ \sf\longmapsto \dfrac{V}{I}=R

\\ \sf\longmapsto I=\dfrac{V}{R}

\\ \sf\longmapsto I=\dfrac{5}{10}

\\ \sf\longmapsto I=0.5A

4 0
2 years ago
Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal
Step2247 [10]

Answer:

T = 1.2 s

T = 15.1 m = 15 m

Explanation:

This is a case of projectile motion:

TOTAL TIME OF FLIGHT:

The formula for total time of flight in projectile motion is:

T = 2 V₀ Sinθ/g

where,

T = Total Time of Flight = ?

V₀ = Launch Speed = 13.9 m/s

θ = Launch Angle = 25°

g = 9.8 m/s²

Therefore,

T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)

<u>T = 1.2 s</u>

<u></u>

RANGE OF BALL:

The formula for range in projectile motion is:

R = V₀² Sin2θ/g

where,

R = Horizontal Distance Covered by ball = ?

Therefore,

T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)

<u>T = 15.1 m = 15 m</u>

4 0
3 years ago
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