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mario62 [17]
3 years ago
11

The earth's orbital angular speed in rad/s due to its motion around the sun

Physics
1 answer:
Anna71 [15]3 years ago
5 0

           (2π rad/year) x (1 yr/365 days) x (1 day / 86,400 seconds)

       =    (2π) / (365 x 86,400)        rad/sec

       =        0.000 000 2  radian/sec

       =          0.2 microrad/sec
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What services do plants provide? Select the three that apply.
Irina18 [472]

The answer to this is B, C, and D. hope this helped

6 0
2 years ago
You have a 100 ohm resistor. How
sp2606 [1]

Answer:

R2 = 300 Ohms

Explanation:

Let the two resistors be R1 and R2 respectively.

RT is the total equivalent resistance.

Given the following data;

R1 = 100 Ohms

RT = 75 Ohms

To find R2;

Mathematically, the total equivalent resistance of resistors connected in parallel is given by the formula;

RT = \frac {R1*R2}{R1 + R2}

Substituting into the formula, we have;

75 = \frac {100*R2}{100 + R2}

Cross-multiplying, we have;

75 * (100 + R2) = 100R2

7500 + 75R2 = 100R2

7500 = 100R2 - 75R2

7500 = 25R2

R2 = 7500/25

R2 = 300 Ohms

4 0
2 years ago
A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera
Fantom [35]
In order to accelerate the dragster at a speed v_f = 100 m/s, its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
W= K_f - K_i = K_f =  \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is -W_f:
W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W

And since 1 horsepower is equal to 746 W, we can rewrite the power as
P=4.08 \cdot 10^6 W \cdot  \frac{1 hp}{746 W} =547 hp



3 0
3 years ago
Read 2 more answers
A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.
sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east  Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35°  Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be  

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

             = 1.10 + 1.4(-0.57)

             = 1.10 -0.798

              = 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

7 0
2 years ago
Pouvez vous m'aider dans ce devoir sil vous plait.​
ahrayia [7]

in english please

i don't understand actually

8 0
3 years ago
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