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valina [46]
3 years ago
11

The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

to a speed of 65 km/h. Calculate the time average F of the total force tangent to the road exerted on the tires during the 18 s. Treat the car as a particle and neglect air resistance.
Physics
1 answer:
Brrunno [24]3 years ago
4 0

Answer:

Explanation:

We shall apply concept of impulse to solve the problem  .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

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Answer:

  g ≈ 7.4 m/s²

Explanation:

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  g = GM/r² = (6.67·10^-11 × 4·10^22)/(6·10^5)^2

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6 0
3 years ago
Two engineering students, John with a weight of 96 kg and Mary with a weight of 48 kg, are 30 m apart. Suppose each has a 0.04%
djyliett [7]

Answer:

6.8370869499\times 10^{20}\ N

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

e = Charge of electron = 1.6\times 10^{-19}\ C

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Z = Atomic number of water = 18

M = Molar mass of water = 0.018 kg/mol

m = Mass of person

The charge is given by

q=imbalance\times n\times e

Total number of protons and electrons in each sphere

n=\dfrac{mN_AZe}{M}

q=imbalance\times \dfrac{mN_AZe}{M}

q_1=0.0004\times \dfrac{96\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=3699916.8\ C

q_2=0.0004\times \dfrac{48\times 6.022\times 10^{23}\times 18\times 1.6\times 10^{-19}}{0.018}\\\Rightarrow q_1=1849958.4\ C

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^{9}\times 3699916.8\times 1849958.4}{30^2}\\\Rightarrow F=6.8370869499\times 10^{20}\ N

The electrostatic force of attraction between them is 6.8370869499\times 10^{20}\ N

4 0
3 years ago
a surface recieving sound is moved from it original position to a position three times farther away from the source of the sound
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Sound intensity = 1/(r^2)

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7 0
3 years ago
Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin
Arte-miy333 [17]

Answer:

x=\dfrac{r}{\sqrt2}

Explanation:

Given that

Radius =r

Electric filed =E

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The electric filed at distance x given as

E=K\dfrac{Q}{(r^2+x^2)^{3/2}}

For maximum condition

\dfrac{dE}{dx}=0

E=K{Q}{(r^2+x^2)^{-3/2}}

\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}

For maximum condition

\dfrac{dE}{dx}=0

K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0

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3 0
3 years ago
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shutvik [7]

Answer:

c) The wavelength decreases but the frequency remains the same.

Explanation:

Light travels at different speed in different mediums.

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Or ,

n = c/v.

<u>The frequency of the light does not change but the wavelength of the light changes with change in the speed.</u>

c = frequency × Wavelength

Frequency is constant,

The formula can be written as:

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5 0
3 years ago
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