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vladimir1956 [14]
3 years ago
13

A horizontal black line at left labeled A has a blue arrow pointing down to it and another blue arrow pointing down from it. A h

orizontal black line at right labeled B has a blue arrow pointing down to it and a blue wave running down from its bottom. Which diagram (A or B) shows transmission?
Physics
2 answers:
never [62]3 years ago
7 0

Answer:

The Correct answer is B

Sergio [31]3 years ago
4 0

Answer:

A is the answer it shows transmission

Explanation:

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A 33 kg gun is standing on a frictionless surface. The gun fires a 57.7g bullet with a muzzle velocity of 325m/s. The positive d
Stella [2.4K]
Kinetic Energy =  (1/2) mv^2.

m = 57.7 g = 57.7/1000 = 0.00577 kg.
v = 325 m/s.

E = 0.5 * 0.00577 * 325^ 2.    Use your calculator.

E = 304.728125  J.

That's the kinetic energy.
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3 years ago
What do you need to measure speed?
Furkat [3]

Answer:

The equation for speed is : distance divided by time.

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3 years ago
Let w(x)=3x-7.If w(x)=14, find x
dlinn [17]

Answer:

7

Explanation:

We are given:

    w(x) = 3x   -   7

     w(x)  = 14

The problem here entails us to solve for x;

To solve for x; equate the two expressions:

         

          3x  - 7  = 14

           3x  = 14 + 7

           3x  = 21  

             x = 7

So the value of x  = 7

6 0
3 years ago
In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of
nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

F_g = F_c

\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}

now we can say that kinetic energy of satellite is  given as

KE = \frac{1}{2}M_s v^2

KE = \frac{GM_sM_p}{2r}

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

U = -\frac{GM_sM_p}{r}

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0
3 years ago
A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr
weqwewe [10]

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

\phi = 0.85 rad = 48.72°

(a). Impedance of the circuit is given by

Z = \frac{V}{I}

Z = \frac{150}{2.75}

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }

Put the values of Z & \phi in above formula we get

R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

3 0
3 years ago
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