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lyudmila [28]
3 years ago
15

Starting from rest, a car takes 2.4 s to travel the first 15 m. Assuming a constant acceleration, how long will it take the car

to travel the next 15 m?
Physics
1 answer:
Nadusha1986 [10]3 years ago
3 0

Answer:

The time it will take the car to travel the next 15 m is 1 s.

Explanation:

Given;

initial velocity of the car, u = 0

time of motion, t = 2.4 s

initial distance traveled, d = 15 m

determine the constant acceleration of the car;

d = ut  +  ¹/₂at²

15 = 0  +  (0.5 x 2.4²)a

15 = 2.88a

a = 15 / 2.88

a = 5.21 m/s²

The velocity of the car at the end of the first 15 m

v - u + at

v = 0 + 5.21 x 2.4

v = 12.5 m/s

The time to cover the next 15 m;

d₂ = vt  + ¹/₂at²

15 = 12.5t  +  (0.5 x 5.21)t²

15 = 12.5t  +  2.61t²

0 = 2.61t²  +   12.5t   -  15

Use the formula method to solve the above quadratic equation.

a = 2.61

b = 12.5

c = -15

t = \frac{-b \ \ +/-   \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-12.5 \ \ +/-   \ \ \sqrt{12.5^2 - (4\times 2.61 \times -15)} }{2\times 2.61}\\\\t = 0.9937 \ s \ \ \ or \ \ \ -5.78 \ s\\\\time \ can \ only \ be \ positive;\\\\t \approx \ 1\ s

Therefore, the time it will take the car to travel the next 15 m is 1 s.

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The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider M_{1} and M_{2} as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

                      \text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}

As gravitational constant G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}, M_{1} = 20 kg and  M_{2} = 100 kg, while d = 2.6 m, then

                    \text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}

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                   \text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}

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7 0
3 years ago
2 pts
const2013 [10]

Answer:

I think the answer 1

Explanation:

im probably wrong too i dont know

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supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistan
saw5 [17]

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

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ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

<u>T₂ = 95.56°C</u>

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