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3 years ago

14

What is the kinetic energy of a baseball with a mass of 0.133 kg and moving at a speed of 28.7m/s

1 answer:

blsea [12.9K]3 years ago

3 0

Answer:

54.7 J

Explanation:

kinetic energy formula

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What kind of weather is signaled by cumulus clouds

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Cumulonimbus clouds are associated by severe weather. They are thunderstorm clouds that bring heavy rain, snow, hail, lightning and even tornadoes.

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18. The orbits of planets being elliptical was one the planetary laws developed by

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It was a man named <span>Johannes Kepler. </span>

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3 years ago

An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons

vlada-n [284]

Answer:

$v=2.02\frac{m}{s}$

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

$K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}$

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

$v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}$

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3 years ago

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km. What is his displacement from the

Readme [11.4K]

Explanation:

Given that,

A person walks 9.0 km directly east and then turns left and heads directly north for 12.0 km.

We need to find his displacement from the starting position.

We know that,

Displacement = shortest path covered

$D=\sqrt{9^2+12^2} \\\\D=15\ km$

For direction,

$\theta=\tan^{-1}(\dfrac{d_y}{d_x})\\\\\theta=\tan^{-1}(\dfrac{12}{9})\\\\= $$53.13^{\circ}$

Hence, this is the required solution.

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3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

4 years ago