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andrew11 [14]
3 years ago
5

Which of these components is present in this circuit schematic?

Physics
2 answers:
Alchen [17]3 years ago
7 0

Answer:

the answer is d for plato users

Explanation:

Svetlanka [38]3 years ago
4 0
Its resistor :}...............

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You serve a volley ball with a mass of 1.5 kg. The ball leaves your hand at 15m/s. The ball has how much kinetic energy?
pav-90 [236]

Answer:

the answer is 168.75 J

5 0
3 years ago
As an aid in working this problem, consult Interactive Solution 3.41. A soccer player kicks the ball toward a goal that is 20.0
alekssr [168]

Answer:

V=14.9 m/s

Explanation:

In order to solve this problem, we are going to use the formulas of parabolic motion.

The velocity X-component of the ball is given by:

Vx=V*cos(\alpha)\\Vx=15.7*cos(31^o)=13.5m/s

The motion on the X axis is a constant velocity motion so:

t=\frac{d}{Vx}\\t=\frac{20.0}{13.5}=1.48s

The whole trajectory of the ball takes 1.48 seconds

We know that:

Vy=Voy+(a)*t\\Vy=15.7*sin(31^o)+(-9.8)*(1.48)=-6.42m/s

Knowing the X and Y components of the velocity, we can calculate its magnitude by:

V=\sqrt{Vx^2+Vy^2} \\V=\sqrt{(13.5)^2+(-6.42)^2}=14.9m/s

6 0
3 years ago
A piston-cylinder device initially contains 0.3 kg of nitrogen gas at 160 kPa and 140oC. The nitrogen is now expanded isothermal
irakobra [83]

Answer:

<h2>Work done by the gas is given as</h2><h2>W = 1.72 \times 10^4 J</h2>

Explanation:

As we know that the process is isothermal so here work done is given as

W = nRTln(\frac{P_1}{P_2})

here we know that

n = \frac{w}{M}

n = \frac{300}{28} = 10.7 moles

now we have

W = 10.7 (8.31)(273 + 140) ln(\frac{160}{100})

so we have

W = 1.72 \times 10^4 J

4 0
3 years ago
Marina walked 2km in half an hour, what was her average speed during her walk?
ad-work [718]

Answer:

4km/hr

Explanation:

6 0
3 years ago
dado los vectores a=3.5 b=4 c=6.5 , si se sabe que a+b+c=0 (resultante) cual es el angulo formado por A y B?
lawyer [7]

cos(A) = (b2 + c2 − a2) / 2bc = (16+6.5^2-3.5^2)/(2*4*6.5) = 0.88461538  so A = cos−1 (0.88461538) = 27.79 Deg cos(B) = (c2 + a2 − b2)/2ca = (3.5^2+6.5^2-16)/(2*3.5*6.5) = 0.84615385  so B = cos−1 (0.84615385) = 32.2 Deg

6 0
3 years ago
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