Mickey had three substances. He hit each of them with a hammer. A red

A block is released to slide down a frictionless incline of 15∘ and then it encounters a frictional surface with a coefficient o

Elodia [21]

The block's potential energy at the top of the incline (at a height h from the horizontal surface) is equal to its kinetic energy at the bottom of the incline, so that

mgh = 1/2 mv²

where v is its speed at the bottom of the incline. It follows that

v = √(2gh)

If the incline is 20.4 m long, that means the block has a starting height of

sin(15°) = h/(20.4 m) ⇒ h = (20.4 m) sin(15°) ≈ 5.2799 m

and so the block attains a speed of

v = √(2gh) ≈ 10.1728 m/s

The block then slides to a rest over a distance d. Kinetic friction exerts a magnitude F over this distance and performs an amount of work equal to Fd. By the work-energy theorem, this quantity is equal to the block's change in kinetic energy, so that

Fd = 0 - 1/2 mv² ⇒ d = (-1293.58 J)/F

By Newton's second law, the net vertical force on the block as it slides is

∑ F [vertical] = n - mg = 0

where n is the magnitude of the normal force, so that

n = mg = (25 kg) g = 245 N

and thus the magnitude of friction is

F = -0.16 (245 N) = -39.2 N

(negative since it opposes the block's motion)

Then the block slides a distance of

d = (-1293.58 J) / (-39.2 N) ≈ 32.9994 m ≈ 33 m

5 0

2 years ago

A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T

Serga [27]

Answer:

a) $k_{e}$ = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m, e) v = 15.23 m / s

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

$k_{e}$ = ½ k x²

$k_{e}$ = ½ 2900 0.80²

$k_{e}$ = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

U = m h and

U = 8 9.8 (-0.80)

U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

K = ½ m v²

K = 0

d) write the energy at two points, maximum compression and maximum height

Em₀ = ke = ½ m x²

$E_{mf}$ = mg y

Emo = $E_{mf}$

½ k x² = m g y

y = ½ k x² / m g

y = ½ 2900 0.8² / (8 9.8)

y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

Y = y- 0.80

Y = 11.8367 -0.80

Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

Emo = ½ k x² = 928 J

$E_{mf}$ = ½ m v²

Emo = $E_{mf}$

Emo = ½ m v²

v =√ 2Emo / m

v = √ (2 928/8)

v = 15.23 m / s

8 0

3 years ago

It took 1500 Newton's of force to push a car 3 meters. How much work was done

denis23 [38]

Answer:

ow much work was done? W = F xD. IN X 2m = 2;. 2. A force of 15 newtons is ... 3. It took 50 joules to push a chair 5 meters across the floor. With what force was ... was done. How far was the rock lifted? W=FXD. D=1500 = 1.5m. Answer: :.5m ... A young man exerted a force of 9,000 newtons on a stalled car, but he was.

Explanation:

3 0

2 years ago

Which types of waves have the shortest wavelength?

Alla [95]

Radio waves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays are all types of electromagnetic radiation. Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.

8 0

2 years ago

Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject

Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

saturn orbit around the sun R_2 = 1.43 *10^9 km

earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

$t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}$

where

$\mu_{sun}$ is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2. $t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}$

t = 12,105.96 sec

6 0

3 years ago