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2 years ago

A 1200 kg sports car accelerates from 0 m/s to 30 m/s in 10 s. What is the average power of the engine?

Physics

2 answers:

DIA [1.3K]2 years ago

5 0

Answer:

3600N

Explanation:

Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s

ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

trapecia [35]2 years ago

3 0

Answer:

P = 54000 W or P = 54 kW

Explanation:

P = Work/time

W = Force * distance

first, find acceleration to find net force using kinematics.

v = vo + at

30 = 0 + a(10)

a = 3 m/s^2

Fnet = ma = 1200(3) = 3600N = Force

Next, find the distance traveled using kinematics.

∆x = vo(t) + 1/2at^2

∆x = 0(10) + 1/2(3)10^2

∆x = 150 m

Now use the force and distance you just calculated to find the work done.

W = Force*distance

W = 3600N * 150m

W = 540000

Calculate power.

P = W/t

P = 540000/10

P = 54000 W

P = 54 kW

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tino4ka555 [31]

Answer:

λ = 5.85 x 10⁻⁷ m = 585 nm

f = 5.13 x 10¹⁴ Hz

Explanation:

We will use Young's Double Slit Experiment's Formula here:

$Y = \frac{\lambda L}{d}\\\\\lambda = \frac{Yd}{L}$

where,

λ = wavelength = ?

Y = Fringe Spacing = 6.5 cm = 0.065 m

d = slit separation = 0.048 mm = 4.8 x 10⁻⁵ m

L = screen distance = 5 m

Therefore,

$\lambda = \frac{(0.065\ m)(4.8\ x\ 10^{-5}\ m)}{5\ m}$

Now, the frequency can be given as:

$f = \frac{c}{\lambda}$

where,

f = frequency = ?

c = speed of light = 3 x 10⁸ m/s

Therefore,

$f = \frac{3\ x\ 10^8\ m/s}{5.85\ x\ 10^{-7}\ m}\\\\$

5 0

3 years ago

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

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3 years ago

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3 years ago

Read 2 more answers

A long iron bar lies along the x-axis and has current of I = 16.4 A running through it in the +x-direction. The bar is in the pr

Gre4nikov [31]

Answer:

B = 8.0487mT

Explanation:

To solve the exercise it is necessary to take into account the considerations of the Magnetic Force described by Faraday,

The magnetic force is given by the formula

$F =BILsin\theta$

Where,

B = Magnetic Field

I = Current

L = Length

$\theta =$ Angle between the magnetic field and the velocity, for this case are perpendicular, then is 90 degrees

According to our data we have that

I = 16.4A

F = 0.132N/m

As we know our equation must be modificated to Force per length unit, that is

$\frac{F}{L} = BI sin(90)$

Replacing the values we have that

$0.132 = 16.4 (1) B$

Solving for B,

$B = \frac{0.132}{16.4}$

$B = 8.0487mT$

8 0

3 years ago