Which is an example of transfroming potential energy to kinetic

PLEASE PROVIDE AN EXPLANATION<br><br> THANK YOU!

Rus_ich [418]

Answer:

(a) 0.993 s

(b) 14.0 N/m

(c) -3.02 m/s

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m). The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

(c) Energy is conserved:

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

6 0

3 years ago

Read 2 more answers

I was reading an old thermodynamics textbook and came across this equation describing change in internal energy. What does the (

RoseWind [281]

Explanation:

I remember that notation! The expression

$dQ = dU = (\dfrac{\partial U}{\partial T})_{V} dT+ (\dfrac{\partial U}{\partial V})_{T}dV$

is the 1st law of thermodynamics and it refers to the heat supplied to the system dQ which is also a change in its internal energy dU. The first term is the <u>partial</u> derivative of the internal energy U with respect to temperature T while the volume V is kept constant, as denoted by the subscript V. The 2nd term is similar but this time, temperature is kept constant while its volume partial derivative is being taken.

Ah, memories!

8 0

2 years ago

Pls help it's due today

mario62 [17]

Answer:

Breh seriously. Ugh fine.

1.B

2.D

3.C

4.C

5.D,A and B

6.A,C and D

3 0

1 year ago

What is the kinetic energy of a vehicle that has a mass of 3,500 kg and is moving at 40 m/s

stiv31 [10]

Answer:2800000j

Explanation:

For us to know the kinetic energy of the vehicle,

Where m is the mass

And v is the velocity

Then, K.E=1/2mv^2

While, K.E=1/2×3500×40^2

Therefore, our answer will now be

K.E=2800000j

5 0

2 years ago

A car travels 50 km in 25 km /h and then travels 60km with a velocity 20 km/h and then travels 60km with a velocity 20 km/h in t

Butoxors [25]

Answer:

v = 21.25 km/h

The average velocity is 21.25km/h

Explanation:

Average velocity = total displacement/time taken

v = d/t

Given;

A car travels 50 km in 25 km /h

d1 = 50km

v1 = 25km/h

time taken = distance/velocity

t1 = d1/v1

t1 = 50/25 = 2 hours

and then travels 60km with a velocity 20 km/h

d2 = 60km

v2 = 20km/h

t2 = d2/v2 = 60/20

t2 = 3 hours

and then travels 60km with a velocity 20 km/h in the same direction

d3 = 60km

v3 = 20km/h

t3 = d3/v3 = 60/20

t3 = 3 hours

Average velocity = total displacement/total time taken

v = (d1+d2+d3)/(t1+t2+t3)

v = (50+60+60)/(2+3+3)

v = 170/8

v = 21.25 km/h

The average velocity is 21.25km/h

3 0

2 years ago

Which is an example of transfroming potential energy to kinetic​

Which is an example of transfroming potential energy to kinetic