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Wittaler [7]
3 years ago
5

A stalled car is pushed with a force of 342N from rest. How far does the car travel in 12 seconds if it’s mass is 989kg

Physics
1 answer:
allsm [11]3 years ago
6 0

24.89m

Explanation:

Given parameters:

Force on car = 342N

Initial velocity = 0

Time taken = 12s

Mass of the car = 989kg

Unknown:

Distance covered by the car = ?

Solution:

To solve this problem, we simply use one of the appropriates equations of motions.

                     s = ut + \frac{1}{2}at²

   s is the distance

    u is the initial velocity

    t is the time taken

    a is the acceleration of the car.

To find the acceleration;

       Acceleration = \frac{Force }{mass} = \frac{342}{989}

          Acceleration = 0.35m/s²

Inputting the parameters in the equation:

           s = 0 x 12 + \frac{1}{2} x 0.35 x 12² = 24.89m

Learn more:

Velocity brainly.com/question/10962624

#learnwithBrainly

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Calculate the heat flux (in W/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°C
ExtremeBDS [4]

Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

5 0
3 years ago
A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its or
Brut [27]

A sound wave leaves the loudspeaker. As it travels, it experiences a temporary increase in wavelength and then returns to its original wavelength.  The sound wave traveled through a helium balloon (helium is less dense than air could explain this change in wavelength

The pattern of disruption brought on by energy moving away from the sound source is known as a sound wave. Longitudinal waves are what makeup sound. This indicates that the direction of energy wave propagation and particle vibrational propagation are parallel. The atoms oscillate when they are put into vibration.

A high-pressure and a low-pressure zone are created in the medium as a result of this constant back and forth action. Compressions and rarefactions, respectively, are terms used to describe these high- and low-pressure zones. The sound waves go from one medium to another as a result of these regions being transmitted to the surrounding media.

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4 0
2 years ago
1.10) For a fixed mass of gas at constant temperature, if the volume of the gas (V) increases to twice its original amount, the
Aliun [14]
B. 1/2p
From the relation
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P2=P1V1/V2
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P2=1*1/2(1)
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7 0
3 years ago
Read 2 more answers
The equation for density is mass divivded by volume.an increase in denstiy can result from all of following expect??
sineoko [7]
It would be option A (a decrease in mass with an increase in volume)
4 0
3 years ago
The electron gun in a television tube is used to accelerate electrons with mass 9.109 × 10−31 kg from rest to 3 × 107 m/s within
zaharov [31]

Answer:

Electric field, E = 40608.75 N/C

Explanation:

It is given that,

Mass of electrons, m=9.1\times 10^{-31}\ kg

Initial speed of electron, u = 0

Final speed of electrons, v=3\times 10^7\ m/s

Distance traveled, s = 6.3 cm = 0.063 m

Firstly, we will find the acceleration of the electron using third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(3\times 10^7)^2}{2\times 0.063}

a=7.14\times 10^{15}\ m/s^2

Now we will find the electric field required in the tube as :

ma=qE

E=\dfrac{ma}{q}

E=\dfrac{9.1\times 10^{-31}\times 7.14\times 10^{15}}{1.6\times 10^{-19}}

E = 40608.75 N/C

So, the electric field required in the tube is 40608.75 N/C. Hence, this is the required solution.

3 0
3 years ago
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