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Alexus [3.1K]
3 years ago
15

A rocket powered sled is accelerating along a straight, level track with a constant acceleration of magnitude 20 m/s2. By how mu

ch will the speed of the sled change in 3.00 seconds
Physics
1 answer:
Dafna1 [17]3 years ago
7 0

Answer:

the change in speed of the sled after 3.0 s is 60 m/s.

Explanation:

Given;

constant acceleration of the sled, a = 20 m/s²

change in time of motion, dt = 3.0 s

The change in speed of the sled after 3.0 s is calculated as;

a = \frac{dv}{dt} \\\\dv = a \times dt\\\\dv = 20 \ \times \ 3\\\\dv = 60 \ m/s

Therefore, the change in speed of the sled after 3.0 s is 60 m/s.

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A person riding on a bus hears the horn of a car that the bus is approaching. After the bus passes the car and is traveling away
Grace [21]

Answer:

option (c)

Explanation:

According to the doppler's effect, the frequency of sound decreases.

If there is a relative motion between the source and the observer, the frequency of sound changes apparently.

the formula of the doppler effect is given by

f' =\frac{v-v_{o}}{v}}f

where, f is the real frequency, v be the speed of sound, vo be the speed of the observer, here, the bus is the observer.

So, the value of f' < f

4 0
3 years ago
Just like Coke is a type of soda, __________ is a type of ___________.
gladu [14]

Answer:

  • osmosis diffusion

Explanation:

ok it's correct

8 0
3 years ago
What kind of potential energy allows you to throw a ball
Aleks [24]
Force.................
3 0
3 years ago
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Given a wind turbine with blades that sweep out a 10 m diameter circle, and a wind speed of 2 m/s, approximately what is the max
Mars2501 [29]

Answer:

223.55 W

Explanation:

v = Velocity of wind = 10 m/s

r = Radius of circle = 5 m

S = Swept area

\rho = Density of air = 1.2 kg/m³

C_p = Power coefficient = 0.593

P=C_p\frac {1}{2}}\rho Sv^{3}\\\Rightarrow P=0.593\times \frac{1}{2}1.2 \pi \times 5^2\times 2^{3}\\\Rightarrow P=223.55\ W

The maximum possible power that can be produced by the turbine is 223.55 W

8 0
4 years ago
A wye-connected load has a voltage of 480v applied to it. What is the voltage drop across each phase
rodikova [14]

Answer:

Y_A=277.128 \angle 30v

Y_B=277.128 \angle (-150)v

Y_C=277.128 \angle (90)v

Explanation:

From the question we are told that

Voltage V_L_L =480v

Generally in a case of Y_connection V_p_ h is mathematical represented as

V_p_h=\frac{V_l_l}{\sqrt{3}} \angle (\phi-30)v

Generally voltage drop across phase A

Y_A=\frac{408}{\sqrt{3}} \angle -(0-30)

Y_A=277.128 \angle 30v

Generally voltage drop across phase B

Y_B=277.128 \angle (-30-120)

Y_B=277.128 \angle (-150)v

Generally voltage drop across phase C

Y_C=277.128 \angle (-30+120)

Y_C=277.128 \angle (90)v

3 0
3 years ago
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