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storchak [24]
3 years ago
6

If a car is moving on a highway at 70 km/hr, going South and then turns to travel 70 km/hr East. What happens to the cars speed

and velocity?
(A)Speed - changing, Velocity - constant


(B)Speed - constant, Velocity - changing


(C)Speed - constant, Velocity - constant


(D)Speed - changing, Velocity - changing
Physics
1 answer:
frozen [14]3 years ago
6 0

Answer:

(B)Speed - constant, Velocity - changing

Explanation:

The magnitude of the velocity ( speed ) has not changed as car is traveling constantly at 70 km/h. But we know the velocity is also having a direction component, and the car which was traveling South is now traveling East. As direction is changed, velocity is changed but as magnitude of velocity has not changed, the speed remain unchanged.

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astraxan [27]
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6 0
2 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 (\frac{v_0}{v_0-v})

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 (\frac{v_0}{v_0-v})

300 = f_0 (\frac{343}{343-v})

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 (\frac{v_0}{v_0-v/2})

290 = (343)(\frac{v_0}{343-v/2})

290*343-145v = 343f_0

Dividing the two expression we have,

\frac{(300*343) - 300v}{290*343-145v} = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0
3 years ago
As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o
MrRa [10]

Answer:

18.6012339739 A

Explanation:

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A

The current necessary to produce this field is 18.6012339739 A

7 0
3 years ago
A fish swims 12.0 m in 5.0 s. It swims the first 4.0 m in 2.0 s, the next 3.0 m in 1.2 s, and the last 5.0 m in 1.8 s. What is t
xz_007 [3.2K]
Use the distance swan and the time elapsed in that interval.

Average velocity = distance / time

Average velocity = [4.0 m + 3.0m] / 3.2 s = 2.1875 m/s 
7 0
3 years ago
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How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2
soldier1979 [14.2K]
∑F = ma = (90 kg)(1.2 m/s²) = 108 N = 100 N (1 significant digit)

3 0
3 years ago
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