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Gennadij [26K]
3 years ago
5

Determine which of the following moving objects obey the equations of projectile motion developed in this topic. A ball is throw

n in an arbitrary direction. A jet airplane crosses the sky with its engines thrusting the plane forward. A rocket leaves the launch pad. A rocket moves through the sky after its engines have failed. A stone is thrown under water.
Physics
1 answer:
Morgarella [4.7K]3 years ago
5 0

Answer:

A ball is thrown in an arbitrary direction and a rocket moves through the sky after its engines have failed.

Explanation:

The concept applied is that of projectile motion which is the free fall for both then horizontal and vertical component of a constant velocity. it is also the motion of an object thrown under the influence of acceleration due to gravity.

The correct option are as follows ; A ball is thrown in an arbitrary direction and a rocket moves through the sky after its engines have failed.

For the first case ; A ball is thrown in an arbitrary direction ; after the ball must have been thrown, the projectile further moves in a curved path.

for the second case ; A rocket moves through the sky after its engines have failed ; the rocket also moves in a curved path under the influence of the acceleration due to gravity.

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Does gravity increase or decrease with greater distance?
PolarNik [594]
Gravitational force decreases with increasing distance. So it decreases!
6 0
3 years ago
A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object
Tamiku [17]

Answer:

So the mass of the second object M will be 1.951 kg    

Explanation:

We have given mass of the first object m_1=3.5kg and its velocity v_1=8.1m/sec

Mass of the second object m_2=M  it is at rest so its velocity v_2=0

From conservation of momentum we know that

Initial momentum = final momentum

So m_1v_1+m_2v_2=(m_1+m_2)v

3.5\times 8.1+M\times 0=(3.5+M)5.2

28.35=18.2+5.2M

M = 1.951 kg

8 0
3 years ago
A pipe open at both ends has a fundamental frequency of 220 Hz when the temperature is 0°C. (a) What is the length of the pipe?
Allushta [10]

Explanation:

It is given that,

Fundamental frequency, f = 220 Hz

(a) We know that at 0 degrees, the speed of sound in air is 331 m/s.

For open pipe, \lambda=2l

l is the length of pipe

Also,

v=f\lambda

l=\dfrac{v}{2f}=\dfrac{331}{2\times 220}=0.75\ m

(b) Let f' is the fundamental frequency of the pipe at 30 degrees and v' is its speed.

v'=331\sqrt{1+\dfrac{T}{273}}

v'=331\sqrt{1+\dfrac{30}{273}}

v' = 348.71 m/s

So, f'=\dfrac{v'}{\lambda}

f'=\dfrac{348.71}{2\times 0.75}

f' = 232.4 Hz

Hence, this is the required solution.

8 0
3 years ago
Read 2 more answers
Can anyone answer this fast pls
nekit [7.7K]
I believe the answer would be 4.5. because it wouldnt be c or d. and 2 seems too small.
3 0
3 years ago
b. The stream of water flowing through a hole at depth h = 10 cm in a tank holding water to height H = 40 cm. . 3 At what distan
topjm [15]

Answer:

34.64 cm

Explanation:

Given that:

The depth of the hole h = 10 cm

height of the water holding in the tank H = 40 cm

For a stream of flowing water, the distance (x) at which the stream strikes the floor can be  computed by using the formula;

x = 2 \sqrt{h(H-h)}

x = 2 \sqrt{10(40-10)}

x = 2 \sqrt{10(30)}

x = 2 \sqrt{300}

x = 2 \times 17.32

x = 34.64 cm

5 0
3 years ago
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