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jeka57 [31]
2 years ago
8

A sports car accelerates from 0 to 30 mph in 1.5 s. How long would it take to accelerate from 0 to 60 mph, assuming the power of

the engine to be independent of velocity and neglecting air drag?
Physics
1 answer:
Crank2 years ago
8 0

Answer:

6 s

Explanation:

given,

Sports car accelerate from 0 to 30 mph in 1.5 s

time taken to accelerate  0 to 60 mph = ?

The power of the engine is independent of velocity and neglecting friction

power =

P = constant  

the kinetic energy for 60 mph larger than this of 30 mph

 = \dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

 = \dfrac{v_1^2}{v_2^2}

 = \dfrac{60^2}{30^2}

 = 4

gain in kinetic energy  = P x t

time = 4 x 1.5

       = 6 s

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A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th
mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

6 0
2 years ago
Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .
grandymaker [24]
Effect of temperature.

"If the temperature of the substance is increased then the rate of chemical reaction is also increased because the kinetic energy is greater."

Effect of surface area.

"If the surface area is increased then the rate of reaction is increased because there will be more active sites for the reaction to occur. 
3 0
2 years ago
Read 2 more answers
The mass of Mars is 6.42 × 1023 kg. Its moon Phobos is 9378 kilometers away from Mars and has a mass of 1.06 × 1016 kg. What is
dangina [55]

Answer:

5.16 × 1021 N

Hope this helps!

4 0
2 years ago
Read 2 more answers
Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
Yuri [45]

Answer:

Q = 12.5 kJ

Explanation:

The expression to use to calculate Heat is:

Q = H° * n

Where:

Q: heat (J or kJ)

H°: enthalpy of reaction (kJ/mol)

n: moles

Now, as it was stated in the comments, the question is incomplete, and here is the missing part:

Given:

2A + B  A2B (1)

H° = – 25.0 kJ/mol

2A2B  2AB + A2 (2)

H° = 35.0 kJ/mol

With these two reactions, we can calculate the heat.

Now, with the above two reactions, we need to get the general reaction (The one the question is giving), so, let's use (1) and (2) and do the sum of them:

2A + B -------> A2B   H°1 = -25 kJ/mol

2A2B --------> 2AB + A2   H°2 = 35 kJ/mol

Now, we sum both equations, we can see that one A2B cancels out with one A2B from equation 2, so, the equation gives:

2A + B + 2A2B -------> 2AB + A2

And the enthalpy, it's just summed:

H°3 = -25 + 35 = 10 kJ/mol

Now with this value we can calculate heat:

Q = 10 * 2.5 = 25 kJ

However, in the reaction we have 2A, so it's not 1:1 mole ratio, but instead is 1:2, so this result we have to divide it between 2 so:

<u>Q = 25 / 2 = 12.5 kJ</u>

3 0
2 years ago
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