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Mila [183]
2 years ago
9

A sports car starts from rest it covers a distance of 900 m to attain a speed of 80m s determine the acceleration of the car and

the time required to reach this speed
Physics
1 answer:
Lady bird [3.3K]2 years ago
5 0

The acceleration of the car and the time required to reach this speed will be 8 m/s² and 10 sec.

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u=0  m/s

Final velocity, v= 80 m/sec

Distance travelled,s =900m

From Newton's third equation of motion;

v²=u²+2as

a =(v²-u²)/2s

Substitute the given values;

a = (80²-0)/2 ×900

a = 6400/1800

a=8 m/s²

The time required to reach this speed is found in Newton's first equation of motion as;

v = u+at

Substitute the given values;

80 = 0 + 8t

t=80/8

t = 10 sec

Hence, the acceleration of the car and the time required to reach this speed will be 8 m/s² and 10 sec.

To learn more about acceleration, refer to the link brainly.com/question/2437624

#SPJ1

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A solid conducting sphere of radius 2.00 cm has a charge of 6.77 μC. A conducting spherical shell of inner radius 4.00 cm and ou
madreJ [45]

Answer:

Part a)

E = 0

Part b)

E = 6.77 \times 10^7 N/C

Part c)

Electric field inside the conductor is again zero

E = 0

Part d)

E = 8.52 \times 10^6 N/C

Explanation:

Part a)

conducting sphere is of radius

R = 2 cm

so electric field inside any conductor is always zero

So electric field at r = 1 cm

E = 0

Part b)

Now at r = 3 cm

By Gauss law

E = \frac{kq}{r^2}

E = \frac{(9\times 10^9)(6.77 \muC)}{0.03^2}

E = 6.77 \times 10^7 N/C

Part c)

Again when we use r = 4.50 cm

then we will have

Electric field inside the conductor is again zero

E = 0

Part d)

Now at r = 7 cm

again by Gauss law

E = \frac{kQ}{r^2}

E = \frac{(9\times 10^9)(6.77\mu C - 2.13\mu C)}{0.07^2}

E = 8.52 \times 10^6 N/C

5 0
3 years ago
PHYSICS CIRCUIT QUESTION PLEASE HELP!! 20 Points!
dimulka [17.4K]
This really calls for a blackboard and a hunk of chalk, but
I'm going to try and do without.

If you want to understand what's going on, then PLEASE
keep drawing visible as you go through this answer, either
on the paper or else on a separate screen.

The energy dissipated by the circuit is the energy delivered by
the battery.  We'd know what that is if we knew  I₁ .  Everything that
flows in this circuit has to go through  R₁ , so let's find  I₁  first.

-- R₃ and R₄ in series make 6Ω.
-- That 6Ω in parallel with R₂ makes 3Ω.
-- That 3Ω in series with R₁ makes 10Ω across the battery.
--  I₁ is  10volts/10Ω  =  1 Ampere.

-- R1:  1 ampere through 7Ω ... V₁ = I₁ · R₁ = 7 volts .

-- The battery is 10 volts. 
    7 of the 10 appear across R₁ .
   So the other 3 volts appear across all the business at the bottom.

-- R₂:  3 volts across it = V₂. 
           Current through it is  I₂ = V₂/R₂ = 3volts/6Ω = 1/2 Amp.

-- R3 + R4:  6Ω in the series combination
                     3 volts across it
                     Current through it is I = V₂/R = 3volts/6Ω = 1/2 Ampere

--  Remember that the current is the same at every point in
a series circuit.  I₃  and  I₄  must be the same 1/2 Ampere,
because there's no place in the branch where electrons can
be temporarily stored, no place for them to leak out, and no
supply of additional electrons.

-- R₃:  1/2 Ampere through it = I₃ .
           1/2 Ampere through 2Ω ... V₃ = I₃ · R₃ = 1 volt

-- R₄:  1/2 Ampere through it = I₄
           1/2 Ampere through 4Ω ... V₄ = I₄ · R₄ = 2 volts

Notice that  I₂  is 1/2 Amp, and (I₃ , I₄) is also 1/2 Amp.
So the sum of currents through the two horizontal branches is 1 Amp,
which exactly matches  I₁  coming down the side, just as it should.
That means that at the left side, at the point where R₁, R₂, and R₃ all
meet, the amount of current flowing into that point is the same as the
amount flowing out ... electrons are not piling up there.

Concerning energy, we could go through and calculate the energy
dissipated by each resistor and then addum up.  But why bother ?
The energy dissipated by the resistors has to come from the battery,
so we only need to calculate how much the battery is supplying, and
we'll have it.

The power supplied by the battery  = (voltage) · (current)

                                                         =  (10 volts) · (1 Amp) = 10 watts .

"Watt" means "joule per second".
The resistors are dissipating 10 joules per second,
and the joules are coming from the battery.

             (30 minutes) · (60 sec/minute)  =  1,800 seconds

             (10 joules/second) · (1,800 seconds)  =  18,000 joules  in 30 min

The power (joules per second) dissipated by each individual resistor is

                       P  =  V² / R
             or
                       P  =  I² · R ,

whichever one you prefer.  They're both true.

If you go through the 4 resistors, calculate each one, and addum up, you'll
come out with the same 10 watts / 18,000 joules total. 

They're not asking for that.  But if you did it and you actually got the same
numbers as the battery is supplying, that would be a really nice confirmation
that all of your voltages and currents are correct.
7 0
3 years ago
IF it possible for an object to move for 10 seconds at a high speed and end up with an average velocity of zero? true or false
marishachu [46]

True, if you move something forward at 100 miles an hour but your on something moving backwards 100 miles an hour you up staying in the same location, aka zero velocity.

4 0
3 years ago
Read 2 more answers
Because light travels in a straight line and casts a shadow, Isaac Newton hypothesized that light is
KonstantinChe [14]

A stream of particles

3 0
3 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
3 years ago
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