Ask question

Ask question

All categories

2 years ago

7

A 100kg car starts from rest at the top of a hill with a height of 50m. which of the following is this cars kinetic energy at th

e bottom of the hill?
98,000 j
5000 j
49,000 j
150 j

1 answer:

Anarel [89]2 years ago

8 0

Explanation:

KE =1/2 MV^2

HOPE YOU COULD SOLVE FROM NOW..

You might be interested in

A mirror with a flat surface is a

USPshnik [31]

A- plane mirror because its surface is plane

7 0

3 years ago

Read 2 more answers

Describe the conditions necessary for sublimation to occur please help

katovenus [111]

Answer:

Sublimation occurs more readily when certain weather conditions are present, such as low relative humidity and dry winds. Sublimation also occurs more at higher altitudes, where the air pressure is less than at lower altitudes. Energy, such as strong sunlight, is also needed.

6 0

2 years ago

You dip your finger into a pan of water twice each second, producing waves with crests that are separated by 0.15 m. Determine t

PIT_PIT [208]

Frequency = rate of sploosh = 2 per second = 2 Hz.

Period = ( 1/frequency ) = 1/2 second

Speed = (wavelength) x (frequency) = (0.15m) x ( 2/sec) = 0.075 m/s .

5 0

3 years ago

Which describes Einstein’s second postulate about the special theory of relativity? The speed of light in a vacuum is constant,

crimeas [40]

Answer:A

Explanation:

5 0

2 years ago

Read 2 more answers

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

(c) From any of the equations, solve for Length of the string (L);

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0

2 years ago