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3 years ago

How does a bicycle rider's energy of motion changes as the gain speed while riding down a hill?

1 answer:

8 0

It is b. It gains momentum when going down. Think of it like a sled. When you go up the hill, it goes slower, but when you go down the hill, you gain speed(momentum) and it increases

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PLEASE HELP!!<br> What is osteoporosis? What are the symptoms and treatments?

Korvikt [17]

Explanation:

Osteopetrosis is a condition in which bones become weak and brittle. It causes the body to constantly absorb and replace bone tissue. With osteoporosis, new bone creation doesn't keep up with old bone removal.

While some don't experience symptoms, common symptoms can include bone fracture and a decrease in height.

Most treatments include supplements and diet modifications.

3 0

2 years ago

Read 2 more answers

A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow

shusha [124]

Answer:

150000000

$\dfrac{F_e}{F_g}=0.00000203873598369$

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

$1\ C=6.25\times 10^{18}\ electrons$

Number electrons is

$n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000$

The number of electrons added or removed was 150000000

Force is given by

$F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N$

The ratio is

$\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369$

The ratio is $\dfrac{F_e}{F_g}=0.00000203873598369$

Balancing the forces we get

$Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C$

The electric field required is 49050000 N/C

4 0

3 years ago

An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Exp

Artemon [7]

Explanation:

It is given that,

Mass of the football player, m = 92 kg

Velocity of player, v = 5 m/s

Time taken, t = 10 s

(1) We need to find the original kinetic energy of the player. It is given by :

$k=\dfrac{1}{2}mv^2$

$k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2$

k = 1150 J

In two significant figure, $k=1.2\times 10^3\ J$

(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0

i.e. $P=\dfrac{W}{t}=\dfrac{\Delta K}{t}$