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olga nikolaevna [1]
3 years ago
13

If a roller coaster has 50,000 J of potential energy at the top of the first hill, how much kinetic energy does it have at the l

owest point?
Physics
1 answer:
Fed [463]3 years ago
4 0

Answer:

50,000 J because of the transformation of energy

Explanation:

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True or False:
bulgar [2K]

Answer:

False

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8 0
3 years ago
If a ball with an original velocity of zero is dropped from a tall structure and it takes 7 seconds to hit the ground, what velo
Colt1911 [192]
The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:

<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s

The negative sign denotes the direction of the ball.
5 0
3 years ago
Read 2 more answers
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc
julia-pushkina [17]

Answer:

E = 2.84 * 10^5 N/C

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

v^2 = u^2 + 2as

(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C

8 0
3 years ago
A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag
Kobotan [32]
Do you still need help with this question??
5 0
3 years ago
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g/m^{2} = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = \frac{1}{2L}\sqrt{\frac{T}{M} }

⇒      T = 4L^{2}f^{2}M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × 2^{2} ×65^{2} × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

4 0
3 years ago
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