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sattari [20]
2 years ago
10

A ball is projected horizontally from the top of a hill with a velocity of 30m/s if it reaches the ground 5sec later the height

of the hill is???
Physics
1 answer:
Vikki [24]2 years ago
7 0

Answer:

The hill is a total of 150 meters tall

Explanation:

30×5=150

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Which best describes the electric field created by a positive charge?
Vlad [161]
Its rays point away from the charge
3 0
3 years ago
Read 2 more answers
Which nucleus completes the following equation?<br> A. 299 Np<br> B. 20Pa<br> C. 2 Pa<br> D. - Np
blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

Collect like terms

92 + 1 = x

93 = x

x = 93

239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0
3 years ago
A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio
docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

Explanation:

- write the equation F(r) = -Kr^{4} with angular momentum <em>L</em>

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = \sqrt{\frac{7K}{m} } x [\frac{L^{2}}{mK}]^{3/14}

3 0
3 years ago
Which example describes a nonrenewable resource?
ioda

Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.

8 0
2 years ago
A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water a
Kaylis [27]

Answer:

Cp= 0.44 J/g.C

This is heat capacity of metal.

Explanation:

From energy conservation

Heat lost by metal = Heat gain by water +Heat gain by  calorimeter

Because here temperature of metal is high that is why it loose the heat.The temperature of water and  calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

Q= m Cp ΔT

m=Mass

Cp=Specific heat capacity

ΔT=Temperature difference

By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

Cp ( 99 .5- 30.5) = 30.65

Cp= 0.44 J/g.C

This is heat capacity of metal.

4 0
3 years ago
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