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VikaD [51]
4 years ago
6

4)Sarah averages 30 miles per hour driving from Orlando to Tampa. What must her average speed be on the return trip if the avera

ge speed for the whole round trip was 40 miles an hour
Physics
1 answer:
vichka [17]4 years ago
3 0

Answer:

y = 60 mph

Explanation:

given,

average speed of Sarah driving from Orlando to Tampa = 30 mph

average speed of return = ?

whole trip average speed = 40 mph

now,

avg\ speed = \dfrac{total\ distance}{total\ time}

Let x be the distance from Orlando to Tampa

y be the speed of return

avg\ speed = \dfrac{x + x}{\dfrac{x}{30}+\dfrac{x}{y}}

avg\ speed = \dfrac{2}{\dfrac{1}{30}+\dfrac{1}{y}}

40= \dfrac{2\times 30 y}{30+y}

 60  + 2 y = 3 y

    y = 60 mph

hence, speed of the return is equal to 60 mph.

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Answer:

e. Object X has traveled four times as far as object Y.

Explanation:

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d=ut+\frac{1}{2}at^2

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t is the time

a is the acceleration

The two objects in the problem have same initial velocity, u = 0 (since they start from rest), so we can rewrite the equation as

d=\frac{1}{2}at^2

We see that the distance covered is proportional to the square of the time. In this problem, the two objects X and Y have same acceleration, but object X accelerates for twice the time: since d \propto t^2, this means that the distance covered by X will be 2^2 = 4 times higher that the distance covered by object Y.

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
4 years ago
Convert the number from scientific into standard notation: 5.9 x 10-2
guapka [62]
Move the decimal point to:
Left : (if the exponent of ten is a negative number -) ... OUR CASE HERE (-2)
or to
Right : (if the exponent is positive +).

You should move the point as many times as the exponent indicates.
Do not write the power of ten anymore.

So, standard form is:
Two points to the left {Exponent of Ten is Negative (-2)}
0.059 ... (without the 10)
6 0
3 years ago
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