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PilotLPTM [1.2K]
3 years ago
9

A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which w

ater hits a wall at 8 m away from the hose (consider that acceleration due to gravity =9.8 m/s2)
Physics
1 answer:
irinina [24]3 years ago
5 0

Answer:

<em>The water hits the wall at a height of 5.38 m</em>

Explanation:

<u>Projectile Motion </u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

The object describes a parabolic path given by the equation:

{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}

Where:

y   = vertical displacement

x   = horizontal displacement

θ   = Elevation angle

vo = Initial speed

The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.

The height at which the water hits a wall located at x=8 m from the hose is:

{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}

Calculating:

y = 5.38 m

The water hits the wall at a height of 5.38 m

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(ii) Describe how the acceleration of the train at time t = 100 s differs from the acceleration
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Explanation:

Acceleration is the rate of change of velocity with time. When acceleration increases a body moves a faster velocity.

  • In the graph acceleration at time t= 100s is rapidly increasing.
  • At t = 20s, the acceleration of the body is getting started up.

A vehicle at time 100s will have a faster velocity compared to one at t = 20s

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3 years ago
A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . A
yaroslaw [1]

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

v_{f}^{2}=v_{0}^{2}+2gy\\v_{f}=\sqrt{0+2(9.8\frac{m}{s^{2}})(145m)}=53.31\frac{m}{s}

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

F=(3.8*10^{8})(1.602*10^{-19}C)(53.31\frac{m}{s})(0.205T)=6.65*10^{-10}N

B)

b.  south - north (by the rigth hand rule)

I hope this is usefull for you

regards

8 0
3 years ago
The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc
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The gravitational force between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is r=6940 km=6.94 \cdot 10^6 m, the gravitational force is F=9.21 \cdot 10^4 N and the mass of the Earth is m_1=5.98 \cdot 10^{24} kg, therefore we can rearrange the previous equation to find m2, the mass of the telescope:
m_2 =  \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg
6 0
3 years ago
Read 2 more answers
A 61.0-kg person jumps from rest off a 10.0-m-high tower straight down into the water. Neglect air resistance. She comes to rest
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Answer:

Explanation:

In this case, law of conservation of energy will be implemented. It states that "the energy of the system remains conserved until or unless some external force act on it. Energy of the system may went through the conversion process like kinetic energy into potential and potential into kinetic energy.But their total always remain the same in conserved systems."

Given data:

Height of tower = 10.0 m

Depth of the pool = 3.00 cm

Mass of person = 61.0 kg

Solution:

Initial energy = Final energy

U_{i} =  (K.E) + U_{f}

As the person was at height initially so it has the potential energy only.

mg(h_{1} +h_{2}) = K.E + mgh_{2}

K.E = mgh_{1}

K.E = (61.0)(9.8)(10)\\K.E = 5978 J

Lets find out the magnitude of the force that the water is exerting on the diver.

W =ΔK.E

F.h_{2} = 5978\\

F = \frac{5978}{3}

F = 1992.67 N

7 0
3 years ago
Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna
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Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

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