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Snezhnost [94]
3 years ago
6

Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase you

r speed by 8.2 mi/h , the time to go one mile decreases by 12 s . what is your original speed?
Physics
2 answers:
Finger [1]3 years ago
6 0
<h3>Answer:</h3>

Your original speed was 45.668 mph.

<h3>Explanation:</h3>

To find seconds per mile, divide 3600 by speed in miles per hour. Let x represent the original speed. Then 3600/x is the original time per mile.

The new time per mile is 3600/(x +8.2), and that is 12 seconds less than the original time per mile:

... 3600/(x +8.2) = 3600/x -12

Multiplying by the product of the denominators, we have ...

... 3600(x) = 3600(x +8.2) -12(x)(x +8.2)

Dividing by 12 and putting this into standard form, we have

... x^2 +8.2x -300(8.2) = 0

... x = (-8.2 +√(8.2² -4(-300·8.2)))/2 = -4.1 +√2476.81 . . . . solved using the quadratic formula

... x ≈ 45.6676 . . . mi/h

Your original speed was about 45.67 mph.

_____

<em>Check</em>

Your original time per mile was about 3600/45.66756 ≈ 78.8306 seconds. Your shorter time per mile is about 3600/53.86756 ≈ 66.8306 seconds, which is 12 seconds less.

djverab [1.8K]3 years ago
5 0
From the first condition
1 = vo t

From the second condition
1 = 8.2 (t - 12/60)

The time it takes to travel 1 mile in the second condition is
t = 0.32 hr

The original speed is
1 = vo (0.32)
vo = 3.13 mi/h
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kicyunya [14]

Answer:

a) 0.049 m

b) Yes, increase

Explanation:

Draw a free body diagram.

In the y direction, there are three forces acting on the feeder.  Two vertical components of the tension forces in each rope pulling up, and weight force pulling down.

Apply Newton's second law to the feeder in the y direction.

∑F = ma

2Ty − mg = 0

Ty = mg/2

Let's say the distance the rope sags is d.  The trees are 4m apart, so the feeder is 2m horizontally from either tree.  Using Pythagorean theorem, we can find the length of the rope on either side:

L² = 2² + d²

L = √(4 + d²)

Using similar triangles, we can write a proportion using the forces and distances.

Ty / T = d / L

Substitute:

(mg/2) / T = d / √(4 + d²)

Solve for d:

Td = mg/2 √(4 + d²)

T² d² = (mg/2)² (4 + d²)

T² d² = (mg)² + (mg/2)² d²

(T² − (mg/2)²) d² = (mg)²

d² = (mg)² / (T² − (mg/2)²)

d = mg / √(T² − (mg/2)²)

Given m = 2.4 kg and T = 480 N:

d = (2.4) (9.8) / √(480² − (2.4×9.8/2)²)

d = 0.049 m

b) If a bird lands on a feeder, this will increase the tension in the rope to support the bird's weight.

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3 years ago
Objects in free fall accelerate due to ______
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Answer:

The correct answer is A.

Objects in free fall accelerate due to <u>gravity</u>.

Explanation:

Momentum can help an object to keep its state of motion at a constant velocity when no external force is applied. It can never accelerate the object.

According to the laws of motion, we know that acceleration is produced in a body only when a Force is applied in the direction of motion of body.

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Answer:

An egg raised above the ground has potential energy due to the force of gravity. When dropped, the egg's potential energy is converted into the kinetic energy of motion. :)

Explanation:

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3 years ago
Important changes in the musculoskeletal system due to physical activity seem to begin at _______ minutes of activity a week. A.
baherus [9]
I think the answer is a
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When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming
Zielflug [23.3K]

COMPLETE QUESTION:

<em>When the magnetic flux through a single loop of wire increases by </em>30 Tm^2<em> , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of </em><em>2.5 ohms </em><em>, (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?</em>

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf \varepsilon is

(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }

and since from Ohm's law

\varepsilon  = IR,

then equation (1) becomes

(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.

(a).

We are told that the change in magnetic flux is \Phi_B = 30Tm^2,  the current induced is I = 40A, and the resistance of the wire is R = 2.5\Omega; therefore, equation (2) gives

(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

which we solve for \Delta t to get:

\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},

\boxed{\Delta t = 0.3s},

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been I =20A, then equation (2) would give

(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },

\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},

\boxed{\Delta t = 0.6 s\\}

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux \dfrac{\Delta \Phi_B}{\Delta t} is greater than in part b, and therefore , the current in (a) is greater than in (b).

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3 years ago
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