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Helen [10]
3 years ago
15

A 25.0 kg projectile is fired by accelerating it with an electromagnetic rail gun on the earth's surface. the rail makes a 30.0

degree angle with the horizontal, and the gun applies a 1250 n force on the projectile for a distance of 7.50 m along the rail. (a) ignoring air resistance and friction, what is the net work done on the projectile, by all the forces acting on it, as it moves 7.50 m along the rail? (b) assuming it started at rest, what is its speed after it has moved the 7.50 m?
Physics
1 answer:
stich3 [128]3 years ago
8 0

(a)

The work done on the projectile is 9375 joule.

The work on the projectile is calculated as

W=F×d

=1250×7.5

=9375 joule

(b)

The speed of the projectile after 7.5 m is 27.38 m/s

First we need to find out the acceleration of the projectile

F=m×a

1250=25×a

a=50 m/s^{2}

Now the velocity of the projectile after 7.5 m is calculated as

v^2=u^2+2a×s

v^2=0+2×50*7.5

v=27.38 m/s

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Answer:

the answer is D

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Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg
nata0808 [166]

Answer:

a) U_{e} = 3 \times 10^{10}\,J, b) v \approx 7745.967\,\frac{m}{s}

Explanation:

a) The potential energy is:

U_{e} = Q \cdot \Delta V

U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)

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b) Maximum final speed:

U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }

The final speed is:

v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }

v \approx 7745.967\,\frac{m}{s}

3 0
3 years ago
) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

f= 466 Hz

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s

The angular frequency instead is given by

\omega = 2\pi f

And substituting

f = 466 Hz

We find

\omega = 2\pi (466 Hz)=2926.5 rad/s

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz

While the angular frequency is given by:

\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s

(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

7 0
3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
2 years ago
Pls answer B And C Will Mark Brainlist. And 30 Points. :)
Vika [28.1K]

Answer:

use the formula for option B ( d/t = s )

and

look at the graph representation to explain

5 0
3 years ago
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