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3 years ago

I need help answering this question in the photo (middle)

Physics

1 answer:

GalinKa [24]3 years ago

3 0

I believe it is because of weight if Timmy is larger and bigger than Maria that would mean he would stop slower just because of his bodyweight pushing on the back of the skateboard while Maria is all those skinny and she doesn’t have as much weight as she can go farther

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It is 6.00 km from your home to the physics lab. as part of your physical fitness program, you could run that distance at 10.0 k

kirill115 [55]

1.) It is 6.00km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0km/hr (which uses up energy at the rate of 700W ), or you could walk it leisurely at 3.00km/hr (which uses energy at 290 W). A.)Which choice would burn up more energy? running or walking? b.)How much energy (in joules) would it burn? c.)Why is it that the more intense exercise actually burns up less energy than the less intense one? Follow 2 answers Report Abuse Answers billrussell42 Best Answer: running, at 10 km/hour for 6 km is 6 km / 10 km/hour = 0.6 hour or 36 min energy used is 700 watts or 700 joules/s x 36 min x 60s/min = 1.512e6 joules or 1.5 MJ walking, at 3 km/hour for 6 km 6 km / 3 km/hour = 2 hour or 120 min energy used is 290 watts or 290 joules/s x 120 min x 60s/min = 1.872e6 joules or 1.8 MJ C) should be obvious PS, this has nothing to do with potential energy. billrussell42 Â· 5 years ago 0 Thumbs up 1 Thumbs down Report Abuse Comment Simon van Dijk I assume the watt consumption is per hour. Then running 6km at 10.0 km/h results in 700*6/10 = 420 w.h and walking in 290*6/3 = 580 w.h So walking would burn up more energy (kwh) b) 1 kilowatt hour = 3 600 000 joules so 420 wh = 0.42 kwh = 1.51.10^6 joule c) when you put more effort in making the distance your energy is used more efficient. Simon van Dijk Â· 5 years ago 0 Thumbs up 2 Thumbs down Report Abuse Comment

7 0

2 years ago

You ride your bike around the neighborhood block at a constant speed 9 km/h. What changes?

Stolb23 [73]

Nothing will change, it's still going at 9 km/h.

Hope this helps.

5 0

3 years ago

How far did the object travel by the end of eight seconds, according to the graph above?

Ulleksa [173]

Answer:

The object traveled 4 cm by the end of eight seconds. Correct: A)

Explanation:

Speed vs Time Graph

In a speed-time graph, speed is plotted on the vertical axis and time is plotted on the horizontal axis. If the graph is a horizontal line, the speed is constant, if the line is sloped up, the speed is increasing and the acceleration is positive and constant, and if the line is sloped down, the speed is decreasing and the acceleration is negative and constant.

The distance traveled by the object can be found by calculating the area under the graph and above the x-axis.

The graph provided shows two different zones: the first 4 seconds, the speed is constant at 1 cm/s, and the last 4 seconds, the speed is zero, i.e. the object is not moving.

The area behind the first zone is a rectangle of height 1 cm/s and base 4 sec, thus the distance is 1 * 4 = 4 centimeters.

The second zone corresponds to an object at rest, thus no distance is traveled.

The object traveled 4 cm by the end of eight seconds.

A) Correct. As shown above

B) The distance traveled is 4 cm. Incorrect

C) The distance traveled is 4 cm. Incorrect

D) The distance traveled is 4 cm. Incorrect

6 0

3 years ago

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at

vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

$g=-9.8 m/s^2$ (acceleration of gravity)

The initial vertical velocity is

$u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s$

where the negative sign means the direction is downward.

The vertical position of the ball is given by

$y(t) = h + u_y t + \frac{1}{2}gt^2$

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

$0 = 4.50 -1.96t-4.9t^2$

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

$v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s$

where u = 3.05 m/s is the initial speed and $\theta$ the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

$d=v_x t$

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

$d=(2.34)(0.78)=1.83 m$

7 0

3 years ago

Compare the Summer Solstice with the Autumn Equinox. Justify your response in two or more complete sentences.

ryzh [129]

Midway between the two solstices we have equinoxes – Vernal Equinox in March and Autumnal Equinox in September. ... After this time, the Earth's northern axis is tilted more and moretowards ... Then on Summer Solstice, the Sun will reach its farthest north position in the sky

4 0

3 years ago

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