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jeyben [28]
3 years ago
12

Two cars start a race with initial velocity 7ms-1 and 4ms-1 respectively.Their acceleration are 0.4ms-2 and 0.5ms-2 respectively

.If the two cars finish the race at the same time,what is the distance of the track?
Physics
1 answer:
Ronch [10]3 years ago
7 0

The distance of the track is 600 m

Explanation:

The two cars move by uniformly accelerated motion, so we the distance they cover after a time t is described by the following suvat equation

d=ut+\frac{1}{2}at^2

where

u is the initial velocity

a is the acceleration

For car 1, we have

d_1 = u_1 t + \frac{1}{2}a_1 t^2

where

u_1 = 7 m/s is the initial velocity of car 1

a_1 = 0.4 m/s^2 is the acceleration of car 1

So the equation can be rewritten as

d_1 = 7t + 0.2t^2

For car 2, we have

d_2 = u_2 t + \frac{1}{2}a_2 t^2

where

u_2 = 4 m/s is the initial velocity of car 2

a_2 = 0.5 m/s^2 is the acceleration of car 2

So the equation can be rewritten as

d_2= 5t + 0.25t^2

The two cars finish the race at the same time: this means that they cover the same distance in the same time t, so we can write

d_1 = d_2\\7t + 0.2t^2=5t + 0.25t^2

And solving for t, we find

2t - 0.05t^2= 0\\t(2-0.05t)=0

Which gives two solutions:

t = 0 (initial instant)

t = 40 s

Therefore, the distance of the track is

d_1 = 7t +0.2t^2 = 7\cdot 40+0.2\cdot 40^2=600 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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At a time t = 2.80 s , a point on the rim of a wheel with a radius of 0.230 m has a tangential speed of 51.0 m/s as the wheel sl
bonufazy [111]

Answer:

Part A) the angular acceleration is α= 44.347 rad/s²

Part B) the angular velocity is 195.13 rad/s

Part C)  the angular velocity is 345.913 rad/s

Part D ) the time is t= 7.652 s

Explanation:

Part A) since angular acceleration is related with angular acceleration through:

α = a/R = 10.2 m/s² / 0.23 m =   44.347 rad/s²

Part B) since angular acceleration is related

since

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(3.4 s - 2.8 s) = 44.88 m/s

since

ω = v/R = 44.88 m/s/ 0.230 m = 195.13 rad/s

Part C) at t=0

v = v0 + a*(t-t0) =  51.0 m/s + (-10.2 m/s²)*(0 s - 2.8 s) = 79.56 m/s

ω = v/R = 79.56 m/s/ 0.230 m = 345.913 rad/s

Part D ) since the radial acceleration is related with the velocity through

ar = v² / R → v= √(R * ar) = √(0.23 m  * 9.81 m/s²)= 1.5 m/s

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v = v0 + a*(t-t0) → t =(v -  v0) /a + t0 = ( 1.5 m/s - 51.0 m/s) / (-10.2 m/s²) + 2.8 s = 7.652 s

t= 7.652 s

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3 years ago
If the primary of a transformer were connected to a dc power source,
QveST [7]

Answer:

D. only briefly while being connected or disconnected.

Explanation:

As we know that transformer works on the principle of mutual inductance

here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil

So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.

Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always

So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output

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D. only briefly while being connected or disconnected.

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wite a paragraph explaining the difference between things that have matter and things that don't have matter.
mihalych1998 [28]

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Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

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\lambda is the wavelength

For the radio wave in this problem,

\lambda = 0.3 m

c=300,000,000 m/s = 3\cdot 10^8 m/s

Therefore, the frequency is:

f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz

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