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3 years ago

Plzz answer this questions will mark as brainlist

Physics

1 answer:

EastWind [94]3 years ago

4 0

Answer:

1. 19.28 secs

2. 154.22 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

Force (F) = 1000 N

Mass (m) = 1200 Kg

Time (t) =..?

Distance (s) =...?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Force (F) = 1000 N

Mass (m) = 1200 Kg

Acceleration (a) =.?

Force (F) = mass (m) x acceleration (a)

F = ma

1000 = 1200 x a

Divide both side by 1200

a = 1000/1200

a = 0.83 m/s²

Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²

1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Time (t) =.?

v = u + at

0 = 16 + (–0.83 x t)

0 = 16 – 0.83t

Rearrange

0.83t = 16

Divide both side by 0.83

t = 16/0.83

t = 19.28 secs.

Therefore, the time taken for the car to halt is 19.28 secs.

2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:

Initial velocity (u) = 16 m/s

Final velocity (v) = 0

acceleration (a) = – 0.83 m/s²

Distance (s) =..?

v² = u² + 2as

0 = 16² + (2 x –0.83 x s)

0 = 256 – 1.66s

Rearrange

1.66s = 256

Divide both side by 1.66

s = 256/1.66

s = 154.22 m

Therefore, the distance travelled by the car before coming to rest is 154.22 m.

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2 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

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Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

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$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

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$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

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$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

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3 years ago

Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average

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Answer:

reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

Explanation:

given data

electricity consume = 14000 kWh

fuel consume = 900 gal

CO₂ produced of fuel = 26.4 lbm/gal

CO₂ produced of electricity = 1.54 lbm/kWh

oil and electricity usage = 21 percent

to find out

the reduction in the amount of CO₂ emissions

solution

we calculate the amount of CO₂ produce here that is

amount of CO₂ produce = ( electricity consume×CO₂ produce electricity + fuel consume × CO₂ consume fuel ) ........................1

put here value

amount of CO₂ produce = ( 14000 × 1.54 + 900 × 26.4 )

amount of CO₂ produce = 45320 lbm/yr

we know reduction is 21%

reduction in amount of CO₂ produced is

reduction in CO₂ produced = 45320 × 21%

reduction in CO₂ produced = 9517.2 lbm per year

so reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

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Plzz answer this questions will mark as brainlist ​

Plzz answer this questions will mark as brainlist