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dmitriy555 [2]
4 years ago
6

found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it

. please explain​

Physics
1 answer:
evablogger [386]4 years ago
7 0

(h + .16) m g = 1/2 k x^2   total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2    PE released = PE of  spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

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in an automobile crash, a vehicle was stopped at a red light is rear-ended by another vehicle. The vehicles have the same mass.
Masja [62]

8 m/s

Explanation:

Using conservation of momentum :-

m1u1 + m2v1 = m1u2 + m2v2

Where:

m1 = Mass of first vehicle

m2 = Mass of second vehicle

u1 = initial speed of first vehicle

v1 = initial speed of second vehicle

u2 = Final speed of first vehicle

v1 = Final speed of second vehicle

From the received informations:

m1 = m2

v1 = 0

v2 = u2 = 4 \frac{m}{s}

So

m1u1 + 0 = 4m1 + 4m1

Now divide both sides by m1 :-

u1 = 4 + 4

u1 = 8m/s

Therefore, final answer is 8 m/s

4 0
3 years ago
In typical game play situation, (with no overtime), when is a game over?
Harman [31]
C) When the time runs out (usually in sports such as soccer, but i don’t know what sport you’re referring to)
3 0
4 years ago
The ball is displaced to the left and then oscillates backwards and forwards between the two plates. The ball touches a plate on
ELEN [110]

Answer:

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

Explanation:

The formula to be used here is

Q = It

where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C

I is current and it is measured in ampere (amps or A); unknown

t is time and it is measured in seconds (s); 0.05 s

Since, average current is what is unknown

I =Q/t

I = 0.000000028/0.05

I = 5.6 × 10⁻⁷ A

Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere

4 0
3 years ago
I need help with this Physics problem. I've been stuck forever:
snow_tiger [21]
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6 0
3 years ago
A 1.0-kilogram rubber ball traveling east at 4.0 meters per second hits a wall and bounces back toward the west at 2.0 meters pe
Mrrafil [7]

Answer:

8 J and 2 J

Explanation:

Given that,

Mass of the rubber ball, m = 1 kg

Initial speed of the rubber ball, u = 4 m/s (in east)

Final speed of the rubber ball, v = -2 m/s (in west)

We need to find the kinetic energy of the ball before it hits the wall, the kinetic energy of the ball after it bounces off the wall.

Initial kinetic energy,

K_i=\dfrac{1}{2}mv^2\\\\K_i=\dfrac{1}{2}\times 1\times (4)^2\\\\K_i=8\ J

Final kinetic energy,

K_f=\dfrac{1}{2}mv^2\\\\K_f=\dfrac{1}{2}\times 1\times (2)^2\\\\K_f=2\ J

So, the initial kinetic energy is 8 J and the final kinetic energy is 2 J.

8 0
3 years ago
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