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3 years ago

9

A biker pedals hard for 3 seconds. What is his initial velocity if he accelerated by 4m/s2 until he's going 20m/s. (Which equati

on should I use?)

1 answer:

____ [38]3 years ago

4 0

Answer:

answer is option 4

Explanation:

you have to use option 4 because u need to find out initial velocity (Vi)

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How often should muscle strengthening activities be performed?

LekaFEV [45]

Answer:

once a week

Explanation:

once a week so answer B

3 0

3 years ago

Read 2 more answers

A person is pushing a box.The net external force on the 60-kg box is stated to be 90 N. If the force of friction opposing the mo

Trava [24]

Answer:

a = 1 m/s²

Explanation:

given,

mass of the person = 60 Kg

Net External force exerted = 90 N

force of friction opposing the motion = 30 N

acceleration of the box = ?

Net force = External force applied - frictional force

Net force = 90 - 30

net force = 60 N

we know

F = mass x acceleration

60 = 60 x a

a = 1 m/s²

acceleration of the box is equal to a = 1 m/s²

7 0

3 years ago

What is 18 degrees celsius in fahrenheit ?

Butoxors [25]

18 degree is equal to 64.4 fahrenheit

4 0

3 years ago

As the temperature of an object increases, the wavelength of the brightest light emitted _______

alexira [117]

Answer:

option (B) decreases

Explanation:

According to the Wein's displacement law, the minimum wavelength of the radiated emission is inversely proportional to the absolute temperature of the body which emits radiation.

$\lambda_{m}\alpha \frac{1}{T}$

Where, T is the absolute temperature of the body and λm is the minimum wavelength of heat radiated.

Here, as the temperature increases, the wavelength decreases.

4 0

3 years ago

The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

2)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

3)

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

3 years ago