All categories

2 years ago

As Luke rides his bike down a hill, potential energy is converted to kinetic energy. What is this an example of?

Physics

1 answer:

Musya8 [376]2 years ago

5 0

Answer:

If I am correct, the answer is D. Law of conservation of energy

Explanation:

the potential energy "converts" to kinetic once Luke is in motion.

You might be interested in

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone'

Helen [10]

We can use 3rd equation of motion to find the velocity of the stone just before it strikes the ground.

Height = S = 318 meters
Acceleration = a= 9.8 m/s²
Initial velocity = u = 0
Final velocity = v = ?

According to the 3rd equation of motion:

2aS = v² - u²

2(9.8)(318)=v²

v²=6232.8

⇒

v = 78.95 m/s

So, the velocity rounded of to nearest integer will be 79 meters per second.

Thus, C option is the correct answer

5 0

2 years ago

Moon A has a mass of 3M and a radius of 2R. Moon B has a mass of 4M and a radius of R. What is the ratio of the force of gravita

zubka84 [21]

Answer:

Explanation:

5.3" (and any subsequent words) was ignored because we limit queries to 32 words.

4 0

2 years ago

Answer key says b but i think it is false please help me

klasskru [66]

Energy in a spring:

E = 0.5 * k * x²

k spring constant = 800 n/m
x stretch of the spring = 5 cm = 0.05 m

E = 0.5 * 800 * 0.05² = 1

5 0

2 years ago

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

7 0

2 years ago

Is it possible to produce a continuous and oriented aramid fiber–epoxy matrix composite having longitudinalandtransverse moduli

Masja [62]

Answer:

Not possible

Explanation:

$E_{cl}$ = longitudinal modulus of elasticity = 35 Gpa

$E_{ct}$ = transverse modulus of elasticity = 5.17 Gpa

$E_m$ = Epoxy modulus of elasticity = 3.4 Gpa

$V_{\rho l}$ = Volume fraction of fibre (longitudinal)

$V_{\rho t}$ = Volume fraction of fibre (transvers)

$E_f$ = Modulus of elasticity of aramid fibers = 131 Gpa

Longitudinal modulus of elasticity is given by

$E_{cl}=E_m(1-V_{\rho l})+E_fV_{\rho l}\\\Rightarrow 35=3.4(1-V_{\rho l})+131V_{\rho l}\\\Rightarrow 35=3.4-3.4V_{\rho l}+131V_{\rho l}\\\Rightarrow V_{\rho l}=\frac{35-3.4}{131-3.4}\\\Rightarrow V_{\rho l}=0.24764$

Transverse modulus of elasticity is given by

$E_{ct}=\frac{E_mE_f}{(1-V_{\rho t})E_f+V_{\rho t}E_m}\\\Rightarrow 5.17=\frac{3.4\times 131}{(1-V_{\rho t})131+V_{\rho t}3.4}\\\Rightarrow \frac{3.4\times 131}{5.17}-131=-127.6V_{\rho t}\\\Rightarrow V_{\rho t}=\frac{\frac{3.4\times 131}{5.17}-131}{-127.6}\\\Rightarrow V_{\rho t}=0.35148$

$V_{\rho l}\neq V_{\rho t}$

Hence, it is not possible to produce a continuous and oriented aramid fiber.

5 0

3 years ago