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Ray Of Light [21]
3 years ago
10

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. what is the ball's horizont

al acceleration at the top of its trajectory?
Physics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

the horizontal acceleration of the ball at the top of the trajectory is zero.

in case of projectile motion,only the force of gravity acts on the object along the vertical. There are no horizontal forces acting on the object.

now force = m a

m= mass

a= horizontal acceleration

f= horizontal force=0

so a= F/m

a=0/m=0

so the horizontal acceleration at all the points including the top of the trajectory is zero.

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The correct answer is option b, 63.6 kPa

Given,

The diameter of the heel, d=7 cm=0.07 m

The mass of the child, m=25 kg

The pressure is given by the ratio of the force to the area through which the force is applied.

The force applied on the floor by the child is equal to its weight.

Thus the pressure applied on the floor by the child is given by,

\begin{gathered} P=\frac{F}{A} \\ =\frac{mg}{\pi(\frac{d}{2})^2} \end{gathered}

Where A is the area of the cross-section of the heel.

On substituting the known values,

\begin{gathered} P=\frac{25\times9.8}{\pi(\frac{0.07}{2})^2} \\ =63.6\times10^3\text{ Pa} \\ =63.6\text{ kPa} \end{gathered}

Thus the pressure applied on the floor by the heel is 63.6 kPa

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11 months ago
When two systems in contact are not at the same temperature, _____ occurs.expansionheat flowfrictionevaporation
Murljashka [212]

Heat flow occurs when two systems in contact are not at the same temperature.

What is heat flow?

  • when two bodies of different temperatures are in contact, heat flow takes place.
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Phyllis is calculating the path of a space probe that will travel to Pluto. On the way to Pluto, the probe will pass Mars, Satur
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Explanation:

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2 years ago
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The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

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2) circuit Q

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R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

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