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Andrei [34K]
3 years ago
9

Two 0.006 Kg bullets are fired with speeds of 20.0 m/s and 50.0 m/s respectively. What are their kinetic energies? Which bullet

has more kinetic energy? What is the difference of their kinetic energies?
Physics
1 answer:
SashulF [63]3 years ago
5 0

Answer:

a) Kinetic energies

K₁ = 1.2 J

K₂ = 7.5 J

b) The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s , K₂ = 7.5 J

c) K₂ -K₁  = 6.3 J

Explanation:

The kinetic energy (K) is that due to the movement of a body and is calculated as follows:

K = (1/2) m*v²  (J)

Where :

m : the mass of the body ( kg)

v is the speed of the body (m/s)

Data

m₁ = m₂ = 0.006 Kg

v₁ = 20 m/s

v₂ = 50 m/s

a)Calculation of the kinetic energy

K₁ = (1/2) (m₁)*(v₁)²

K₁ = (1/2) (0.006)*(20)²

K₁ = 1.2 J

K₂= (1/2) (m₂)*(v₂)²

K₂ = (1/2) (0.006)*(50)²

K₂ = 7.5 J

b) K₂ ˃ K₁

The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s, K₂ = 7.5 J

c) Difference of their kinetic energies (K₂ -K₁)

K₂ -K₁  = 7.5 J - 1.2 J = 6,3 J

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Given that,

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F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N

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Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

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4 years ago
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eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

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But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

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