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san4es73 [151]
3 years ago
9

A red car and a blue car can move along the same straight one-lane road. Both cars can move only at one speed when they move (e.

g., 60 mph). The driver of the red car sounds his horn. In which one of the following situations does the driver of the blue car hear the highest horn frequency?a. Both cars are moving at the same speed, and they are moving apart.b. Both cars are moving in the same direction at the same speed.c. Both cars are moving at the same speed, and they are moving toward each other.d. The red car is moving toward the blue car, which is stationary.e. The blue car is moving toward the red car, which is stationary.
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:c

Explanation:

When both cars move towards each other with same speed , apparent frequency will be highest

this can also be explained by Doppler frequency Formula

f'=f\left ( \frac{v+v_o}{v-v_s}\right )

where f'=Apparent\ frequency

f=Original\ frequency

v_o=velocity\ of\ observer

v_s=velocity\ of\ source

v=velocity\ of\ sound

as denominator is smaller than Numerator therefore apparent frequency will be greater

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Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

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Explanation:

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A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same
Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

-  175 m/s · sin 36.7° /  - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s ·  sin 36.7 - 1/2 · 9.8 m/s² · t

-  175 m/s ·  sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

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