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pishuonlain [190]

2 years ago

12

A car slows down from 21 m/s to rest in a distance of 63m. Assuming the car has a constant acceleration, calculate the time it t

ook to come to a stop.

1 answer:

vladimir2022 [97]2 years ago

5 0

Answer:

-3.5 m/s²

Explanation:

Initial Velocity = 21m /s
Final velocity = 0m/s
Distance = 63 m .
Acclⁿ = ?

<u>We </u><u>know</u><u> </u><u>that</u><u> </u><u>:</u><u>-</u><u> </u>

$\longrightarrow$ Stopping distance = u²/2(-a)

$\longrightarrow$ 63m = (21m/s)² / -2a

$\longrightarrow$ a = - 21 * 21 / 63 * 2 m/s²

$\longrightarrow$ a = - 3.5 m/s²

<em>*</em><em>*</em><em>Edits</em><em> </em><em>are</em><em> </em><em>welcomed</em><em>*</em><em>*</em>

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Answer:

$\frac{0.065}{r}$

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

$v=\sqrt{\mu gr}....................(1)$

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$v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)$

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

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3 years ago

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

So

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

So

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

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2 years ago

A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant the

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(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

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W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0 to 3) = 64.28 J

Learn more about work done here: brainly.com/question/25573309

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Slightly raising your body temperature while increase oxygen and blood circulation throughout your body

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