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Eva8 [605]
2 years ago
9

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 29.0 mph in time 1.10 s .A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 29.0 mph in time 1.10 s , how long would it take to go from zero to 58.0 mph ?
Physics
1 answer:
bekas [8.4K]2 years ago
6 0

Answer:

2.2 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Converting mph to m/s

29\ mph=29\times 0.44704=12.96\ m/s

58\ mph=58\times 0.44704=25.93\ m/s

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{12.96-0}{1.1}\\\Rightarrow a=11.78\ m/s^2

Considering this acceleration to be constant

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{25.93-0}{11.78}\\\Rightarrow t=2.20\ s

Time it would take to go from zero to 58.0 mph is 2.2 seconds

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Answer:

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Solution:

As per the question:

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Now,

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This energy density equals the kinetic energy supplied by the field.

Thus

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\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}

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v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s

3 0
2 years ago
The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in
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Answer: the correct answer is 7.8026035971 x 10^(-13) joule

Explanation:

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m = 4.00260 u + 222.01757 u = 226.02017 u .

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Delta E = 4.87 MeV

Converting  4.87 MeV to Joules

1 joule [J] = 6241506363094 mega-electrón voltio [MeV]

4 mega-electrón voltio = 6.40870932 x 10^(-13) joule

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5 0
2 years ago
A force of 200 N is applied on small piston of a pascal press. What would be the
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Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}

Where;

F₂ is the force on big piston

F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N

Therefore, the force applied on the big piston is 1306.67 N

3 0
1 year ago
If the top circuit has an oscillation frequency of 1000 Hz, the frequency of the bottom circuit is:_______.
kiruha [24]

Answer:

1410 Hz

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5 0
2 years ago
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ipn [44]

Answer:

ball clears the net

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t = time of travel

X = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

X = v_{ox} t \\7 = v_{ox} t\\t = \frac{7}{v_{ox}}       Eq-1

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v_{oy} = initial velocity = v_{o} Sin\theta = 20 Sin5 = 1.74 ms^{-1}

t = time of travel

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Y = Final position of the ball at time "t"

a_{y} = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

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Since Y > 1 m

hence the ball clears the net

7 0
2 years ago
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