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RoseWind [281]
3 years ago
13

A 1.2-kg mass is projected from ground level with a velocity of 30 m/s at some unknown angle above the horizontal. A short time

after being projected, the mass barely clears a 16-m tall fence. Disregard air resistance and assume the ground is level. What is the kinetic energy of the mass as it clears the fence
Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0

Answer:

K_f = 351.84 J

Explanation:

Using the conservation of energy K:

E_i = E_f

so:

\frac{1}{2}mv^2 = K_f + mgh

where m is the mass, v the initial velocity, K_f is the kinetic energy of the mass as it clears the fence, g the gravity and h the altitude.

Then, replacing values, we get:

\frac{1}{2}(1.2kg)(30m/s)^2 = K_f + (1.2kg)(9.8m/s^2)(16m)

solving for K_f:

K_f = 351.84 J

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Power is the rate at which energy is transformed. A person is limited in the total work he or she can do only by the total energ
Sergio039 [100]

Answer:

Power is the rate at which energy is transformed.

The SI unit of power is the watt.

Power is the rate at which work is done.

Explanation:

In this question we have to choose the statements related to Power that are True.

In this sense, Power P is the speed with which work W is done. Its unit is Watts (W), being 1 W=\frac{1 Joule}{1 s}.  

Power is mathematically expressed as:

P=\frac{W}{t}

Where t is the time during which work W is performed.

On the other hand, the Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path. It is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter (1J=(1N)(1m)=Nm ).

So, according to this explanation, the true statements are:

Power is the rate at which energy is transformed.

The SI unit of power is the watt.

Power is the rate at which work is done.

4 0
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
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