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2 years ago

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A student throws a softball horizontally from a dorm window 15.0 m above the ground. Another student standing 10.0 m away catche

s the ball at a height of 1.50 m above the ground. What is the initial speed of the ball?

1 answer:

Nonamiya [84]2 years ago

5 0

Answer:

Speed of ball equals 6.024 m/s.

Explanation:

Let the student throw the ball with a velocity of 'v' m/s horizontally.

Now the time in which the ball travels 10.0 meter horizontally shall be equal to the time in which it travels (15.0-1.50) meters vertically

Hence the time taken to cover a vertical distance of 13.50 meters is obatined using 2 equation of kinematics as

$s=\frac{1}{2}gt^{2}\\\\t=\sqrt{\frac{2s}{g}}\\\\t=\sqrt{\frac{2\times 13.5}{9.81}}\\\\\therefore t=1.66 seconds$

Since there is no acceleration in horizantal direction we infer that in time of 1.66 seconds the ball travels a distance of 10 meters

Hence the spped of throw is obatines as

$Speed=\frac{Distance}{Time}\\\\v=\frac{10}{1.66}=6.024m/s$

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Answer:

$x=2t^2+15t+5$

Explanation:

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Two identical carts travel at the same speed toward each other, and then a collision occurs. The graphs show the momentum of eac

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Explanation :

The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.

In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.

After the collision, both the object sticks over each other and moves with one velocity.

Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.

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A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva

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Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

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2 years ago

A heat engine (Power Cycle) with a thermal efficiency of 35 percent efficiency produces 750 kJ of work. Heat transfer to the eng

frosja888 [35]

Answer:

a) The schematic illustrating is attached

b) The heat transfer to the heat engine is 2142.86 kJ, the heat transfer from the heat engine is 1392.86 kJ

c) The heat transfer to the heat engine is 1648.35 kJ, the heat transfer from the heat engine is 898.35 kJ

Explanation:

b) The heat transfer to the engine and the heat transfer from the engine to the air is:

$Q_{1} =\frac{W}{n}$

Where

W = 750 kJ

n = 35% = 0.25

Replacing:

$Q_{1} =\frac{750}{0.35} =2142.86kJ$

$Q_{2} =Q_{1} -W=2142.86-750=1392.86kJ$

c) The efficiency of Carnot engine is:

$n=1-\frac{300K}{550K} =0.455$

The heat transfer to the heat engine is:

$Q_{1c} =\frac{750}{0.455} =1648.35kJ$

The heat transfer from the heat engine is:

$Q_{2c} =1648.35-750=898.35kJ$

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2 years ago

A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between

Oduvanchick [21]

Answer:

Net electric field, $E_{net}=91406.24\ N/C$

Explanation:

Given that,

Charge 1, $q_1=7.5\ nC=7.5\times 10^{-9}\ C$

Charge 2, $q_2=-2.9\ nC=-2.9\times 10^{-9}\ C$

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

$E_1=\dfrac{kq_1}{r^2}$

$E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}$

$E_1=65917.96\ N/C$

Electric field due to charge 2 is given by :

$E_2=\dfrac{kq_2}{r^2}$

$E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}$

$E_2=25488.28\ N/C$

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

$E_{net}=E_1+E_2$

$E_{net}=65917.96+25488.28$

$E_{net}=91406.24\ N/C$

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

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2 years ago