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2 years ago

Calculate the kinetic energy of a .30 kg ball thrown at a velocity of 96.6 m/hr in J.

Physics

1 answer:

spin [16.1K]2 years ago

4 0

Answer:

1.08 x 10^-4 J

Explanation:

mass of ball, m = 0.3 kg

Velocity of ball, v = 96.6 m/hr

First convert m/hr into m/s.

As we know that 1 hour = 3600 second

So, 96.6 m / hr = 96.6 / 3600 m/s = 0.0268 m/s

The formula for the kinetic energy is given by

Kinetic energy = 1/2 mv^2

Kinetic energy = 0.5 x 0.3 x 0.0268 x 0.0268 = 1.08 x 10^-4 J

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Convert 13.1 miles to feet. Using one step conversion

Marat540 [252]

Answer:

69,168 ft

Explanation:

6 0

3 years ago

A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi

Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0

2 years ago

What average force is needed to accelerate a 7.00-gram pellet from rest to 155 m/s over a distance of 0.600 m along the barrel o

leonid [27]

Answer: The force needed is 140.22 Newtons.

Explanation:

The key assumption in this problem is that the acceleration is constant along the path of the barrel bringing the pellet from velocity 0 to 155 m/s. This means the velocity is linearly increasing in time.

The force exerted on the pellet is

F = m a

In order to calculate the acceleration, given the displacement d,

$d = \frac{1}{2}at^2\implies a=\frac{2d}{t^2}$

we will need to determine the time t it took for the pellet to make the distance through the barrel of 0.6m. That time can be determined using the average velocity of the pellet while traveling through the barrel. Since the velocity is a linear function of time, as mentioned above, the average is easy to calculate as:

$\overline{v}=\frac{1}{2}(v_{end}-v_{start})=\frac{1}{2}(155-0)\frac{m}{s}=77.5\frac{m}{s}$