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3241004551 [841]
2 years ago
11

The distance vs. time graph of a car moving at constant speed should be a straight line. Why do the data points in the graph plo

tted from observing the motion not fall right on the line? A graph with 7 points and a line segment representing a straight line. Horizontal axis is labeled Time, in seconds, and extends from 0 to 7. Vertical axis is labeled Distance, in meters, and extends from 0 to 50. Approximate coordinates of plotted points are begin ordered pair 1 comma 8 end ordered pair comma begin ordered pair 2 comma 13 end ordered pair comma begin ordered pair 3 comma 21 end ordered pair comma begin ordered pair 4 comma 28 end ordered pair comma begin ordered pair 5 comma 35 end ordered pair comma begin ordered pair 6 comma 39 end ordered pair comma begin ordered pair 7 comma 49 end ordered pair. Line segment has positive slope, and it begins at the point begin ordered pair 0 comma 0 end ordered pair, and it ends at the point begin ordered pair 7 comma 49 end ordered pair. The measurements are inexact. Data points can never look like they fall right on a straight line. The distance is not exactly proportional to time even if the speed is exactly constant. The distance and time are proportional only if the correct units of meters and seconds are used.
Physics
1 answer:
iragen [17]2 years ago
6 0

Answer:

The measurements are inexact

Explanation:

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Answer:

b) the height the ball bounces

Explanation:

the control variable is the variable that you change yourself. since you change the height that the ball bounces from we know this is the answer

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2 years ago
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Is everyone in your class able to hear a quiet sound equally well?
Sergio039 [100]

Answer:

No

Explanation:

Loudness describes how people perceive sound (see loudness). ... If people could hear equally well at all frequencies, the contour lines would be flat because the same measured sound intensity would be perceived to be equally loud regardless of the sound frequency. In fact, people do not hear as well at low frequencies.

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3 years ago
A projectile is launched into the air with an initial speed of 40 m/s and a launch angle of 20° above the horizontal. The projec
miv72 [106K]

Explanation:

V=40m/s

Vy=V.sina=40.sin20=40 . 0.342=13.68m/s

Vx=V.cosa=40.cos20=40 . 0.766=30.64m/s

Projectile travels during 5 seconds and the ramge becomes:

x=V.t=30.64 . 5=153.2m

7 0
2 years ago
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2. The coil rotates in a uniform magnetic f
Reptile [31]

Answer:

331.75 V

Explanation:

Given:

Number of turns of the coil, N = 40 turns

Area, A = 0.06 m²

Magnetic Field, B = 0.4 T

Frequency, f = 55 Hz

                           Maximum induce emf, E₀ = NABω

but ω = 2πf

                           Maximum induce emf, E₀ = NAB(2πf₀)

                           Maximum induce emf, E₀ = 2πNABf₀

Where;

N is number of turns of the coil

A is area

B is magnetic field

ω is the angular velocity

f is the frequency

                                     E₀ = 2 × π × 40 × 0.06 × 0.4 × 55

                                     E₀ = 342.81 V

The maximum induced emf is 331.75 V

6 0
2 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

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

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
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