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erik [133]
4 years ago
9

A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s. What is th

e average speed of the object?
Physics
1 answer:
GaryK [48]4 years ago
6 0
To find the average speed, we must calculate the total distance covered by the object and then divide it by the total time of the motion.

The total distance covered by the object is
S=2m+80m=82 m
while the total time of the motion is
t=2s+40 s=42 s

Therefore, the average speed of the object is
v= \frac{S}{t}= \frac{82 m}{42 s}=1.95 m/s
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After hitting a long fly ball that goes over the right fielder's head and lands in the outfield, the batter decides to keep goin
jarptica [38.1K]

Answer:

a) 633.39 J

b) 0.28

Explanation:

a)The kinetic energy of the player = \frac{1}{2} mv^{2}

Work done by friction = energy change of the player

                                    = \frac{1}{2} 67(4.35)^{2} = 633.9 J

b) Assuming the frictional force stays constant,

Work done by friction = Frictional force×distance

Frictional force = kinetic friction(μ)×normal reaction

Normal reaction = weight = mass×gravitational acceleration ( g=10m/s2 )

Combining these equations

633.9  =  F×3.4 ⇒ F = 186.44 N

F = μmg ⇒ μ = F/mg

                     = 186.44/670

                      = 0.28

5 0
3 years ago
You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a d
gulaghasi [49]

Answer: 94.6 N

Explanation:

P = F d / t

82 = F (1.3) / 1.5

F = (82 x 1.5) / 1.3

F = 94.6N

4 0
3 years ago
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports
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3 years ago
Calculate the average force with which the Sun pulls the Earth. It will help to know
algol13

Answer:

7.9 x 10^21 pound-force

Explanation:

The average distance between the Earth and sun is 150 trillion meters, or 1.5 x 10^11 meters. The mass of the sun is 1.99 x 10^30 kilograms, while the Earth weighs in at 6.0 x 10^24 kilograms. The gravitational constant is 6.67 x 10^-11 meter^3 / (kilogram - second^2). So the Earth and sun pull on each other with a force equal to 3.52 x 10^22 newtons. The newton is a unit of force equal to a kilogram-meter/second^2. One newton is equal to 0.22 of the rarely used English unit called pound-force, so 3.52 x 10^22 newtons is 7.9 x 10^21 pound-force.

4 0
3 years ago
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
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