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Harlamova29_29 [7]
3 years ago
8

3. When you graph the motion of an object, you put ____ on the horizontal axis and ____ on the axis.   a.  speed, time   b.  dis

tance, time   c.  time, speed   d.  time, distance
Physics
2 answers:
Papessa [141]3 years ago
7 0
The answer is B. distance , Time
Mrac [35]3 years ago
5 0
(c) and (d) can both be correct. It just depends on whether you're graphing speed or distance.
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Select the correct answer.
Vesna [10]

Answer:

C

Explanation:

Radiant= list onto solar panels, Electric= solar into power, Radiant= Electric into light

8 0
3 years ago
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Can someone help me with physics;((
vladimir2022 [97]

Answer: vf= 51 m/s and d= 112 m

Explanation: solution attached

4 0
3 years ago
Compute the work performed when 32 pounds is lifted 10 feet.
Murljashka [212]
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)
7 0
3 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

The initial velocity is 50mi/h (v_1=50)

When it stops the final velocity is (v_2=0)

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s =  2500/(185644.54)

s=0.0134


So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0
3 years ago
Lab: Motion
I am Lyosha [343]

Answer: In this lab we wanted to know how motion can be described. So the hypothesis is if the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration. My prediction is that cars travel faster on higher tracts. So the heighten the track was intentionally manipulated. So it is the independent variable the speed of the car is the dependent variable. The speed at the first quarter checkpoint is 1.09 m/s. The speed at the second quarter checkpoint is 1.95 m/s. The speed at the third quarter checkpoint is 2.373.36 m/s. The speed at the finish line is  2.803.00 m/s. The average speed increases as the height increases.

The cars on the higher track travel farther than the cars on the lower track, in the same time.

This means that the cars on the higher track have a greater average speed than those on the lower track. This is demonstrated by the

slope of the higher track line being greater than the slope of the lower track line.

Explanation: put it in notes then send it to files to compress it to submit it.

5 0
3 years ago
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