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Scrat [10]
3 years ago
6

A projectile lands at the same height from which it was launched. which initial velocity will result

Physics
1 answer:
Serhud [2]3 years ago
5 0

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

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Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle
padilas [110]

Answer:

Explanation:

If E₀ is the electric field outside the smaller sphere and r is the radius of larger sphere.

E₀ = kQ/r²

The radius of the larger sphere is 3r and the charge on both sphere is same then the electric field outside the larger sphere is given as

E = kQ/(3r)² = kQ/9r² = 1/9 (kQ/r²)= 1/9 x E₀

hence the correct option is e.

5 0
3 years ago
an astronaut on earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. if she were instead in
nadya68 [22]

The gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

<h3>What is gravitational force?</h3>

All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.

As we know, the gravitational force is given by:

\rm F = \dfrac{Gm_1m_2}{r^2}

Where G is the gravitational constant.

m1 and m2 are masses.

r is the distance between the masses.

Weight of ice cube = weight of soft drink displaced

mg = ρVg

m = ρV

g does not affect astronauts on earth in a lunar module.

Thus, the gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float:A) with more than 9/10 submerged. B) with 9/10 submerged.C) with 6/10 submerged.D) totally submerged.

Learn more about the gravitational force here:

brainly.com/question/12528243

#SPJ1

3 0
1 year ago
Acceleration is defined as the rate of change for which of the following
Minchanka [31]
_Award brainliest if helped!
Velocity


Note : Not speed as Acceleration is a vector!
6 0
3 years ago
Maxwell shoots a rubber band at his friend Jimmy. Which type of energy is converted into kinetic energy?
Leokris [45]

A stretched rubber band is storing <em>elastic potential energy. (A)</em>

3 0
3 years ago
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m,
Tems11 [23]

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

5 0
2 years ago
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