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KiRa [710]
3 years ago
13

A 3 kg toy car with a speed of 7 m/s collides head-on with a 2 kg car traveling in the opposite direction with a speed of 5 m/s.

If the cars are locked together after the collision with a speed of 2.20 m/s, how much kinetic energy is lost?
Physics
1 answer:
Mademuasel [1]3 years ago
6 0

Answers:

kinetic energy lost = 86.4J

Explanation:

let Kf be the kinetic energy after the collision and Ki be the kinetic energy before the collision. let the 3kg car be 1 and 2kg car be 2.

Kf = K1(f) + K2(f)

Ki = K1(i) + k2(i)

loss in kinetic energy = Kf - Ki

                                 = 1/2(3)(2.20)^2 + 1/2(2)(2.20)^2 - 1/2(3)(7)^2 - 1/2(2)(-5)^2

                                 = 12.1 - 98.5

                                 = -86.4 J

therefore, the kinetic energy lost in the collision is 86.4 J.

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A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
Which statement is correct?
Sedbober [7]

Answer:

The answer is "Choice C ".

Explanation:

The relationship between the E and V can be defined as follows:

\to E= -\Delta V

Let,

\to E= \frac{\delta V}{\delta x}

When E=0

\to   \frac{\delta V}{\delta x}=0

v is a constant value

Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.

7 0
3 years ago
An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t
Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

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At a second point downstream

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We need to calculate the depth

Using Bernoulli theorem

A_{1}V_{1}=A_{2}V_{2}

Put the value into the formula

16.0\times3.8\times2.8=16.5\times x\times 11.0

x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}

x=0.938\ m

Hence,  The depth of the water at this point is 0.938 m.

7 0
3 years ago
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