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Eva8 [605]
3 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]3 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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A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou
lakkis [162]

Answer:

C. while the magnet is moving

Explanation:

Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

8 0
3 years ago
In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.
HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is 41.9^{\circ}

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

s=u_y t + \frac{1}{2}at^2 (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

u_y = u sin \theta is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

\theta=40.0^{\circ} is the angle of projection

So

u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s

a=g=-9.8 m/s^2 is the acceleration due to gravity (downward)

Substituting the numbers, we get

-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

d=u_x t

where

u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

d=(9.19)(1.778)=16.34 m

B)

In this second case,

\theta=42.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s

So the equation for the vertical motion becomes

4.9t^2-8.1t-1.80=0

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s

So, the range of the shot is

d=u_x t = (8.85)(1.851)=16.38 m

C)

In this third case,

\theta=45^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s

So the equation for the vertical motion becomes

4.9t^2-8.5t-1.80=0

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s

So, the range of the shot is

d=u_x t = (8.49)(1.925)=16.34 m

D)

In this 4th case,

\theta=47.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s

So the equation for the vertical motion becomes

4.9t^2-8.8t-1.80=0

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s

So, the range of the shot is

d=u_x t = (8.11)(1.981)=16.07 m

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is \theta=42.5^{\circ}.

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

s-u sin \theta t + \frac{1}{2}gt^2=0

The solutions of this quadratic equation are

t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}

This is the time of flight: so, the horizontal range is

d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta

It can be found that the maximum of this function is obtained when the angle is

\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})

Therefore in this problem, the angle which leads to the maximum range is

\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
a 25-N net force is applied to a rolling cart and pruduces an acceleration of 5 m/s 2 what’s the cart mass
nignag [31]

Answer:

The mass of the cart is 5 kg

Explanation:

You divide 25 by 5 and get 5. Have a great day! :D

<em>The Equation:</em>

25/5 = 5

8 0
2 years ago
A type of energy embodied in oscillating electric and magnetic fields is called
yaroslaw [1]

Answer: Electromagnetic radiation

Explanation:

Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

To understand it better:

This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.

It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.

4 0
3 years ago
you place a beaker of water on a hot plate and heat it up eventually it starts to boil and you see bubbles for me you also seeme
Nuetrik [128]
It's a physical change. the water is changing it's physical state

8 0
3 years ago
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