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3241004551 [841]
3 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]3 years ago
6 0

Answer:

The measurements are inexact

Explanation:

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What is the mass of a black hole which has the radius of the sun(696,340 km).​
Leya [2.2K]

It has a diameter of about 78 billion miles. For perspective, that's about 40% the size of our solar system, according to some estimates. And it's estimated to be about 21 billion times the mass of our sun. So there you have it, black holes can be millions of times larger than suns and planets or as small as a city.

8 0
2 years ago
2. Two people, one on Earth and the other on the Moon, try and lift 200 kg blocks. Which person
Art [367]

Answer:

The person on Earth will have to exert more force to lift their block

Explanation:

The mass of the blocks to be lifted = 200 kg

The location of the first person = On Earth

The location of the second person = On the Moon

The force a person will have to exert to lift the block = The weight of the block = The gravitational force, F, on the block which is given as follows;

F = \dfrac{G \times M_1}{R^2}  \times m_2

Where;

\dfrac{G \times M_1}{R^2}= The acceleration due to gravity on the Earth or the Moon, depending on the location of the block

m₂ = The mass of the block

Therefore, given that the acceleration due to gravity on the Earth is larger than the acceleration due to gravity on the Moon, the weight of the block on the Earth is larger than the weight of the block on the Moon, and the person on Earth have to exert more force to lift the heavier weight of the block on Earth than the person on the Moon will have to exert to lift the same block as the block has a lower weight on the Moon due to lower acceleration due to gravity on the Moon.

6 0
3 years ago
If you can answer both, please do. But if you can't, just answer one.
Setler [38]

Answer:

1.The force required to stop the shopping cart is, F = 12.25 N

Explanation:

Given data,

The mass of the shopping cart, m = 7 kg

The initial velocity of the shopping cart, u = 3.5 m/s

The final velocity of the shopping cart, v = 0 m/s

The time period of acceleration, t = 2 s

The change in momentum of the cart,

                                      p = m(u - v)

                                         = 7 (3.5 - 0)

                                         = 24.5 kg m/s

The force is defined as the rate of change of momentum. To stop the shopping cart, the force required is given by the formula

                                           F = p / t

                                               = 24.5 / 2

                                               = 12.25 N

Hence, the force required to stop the shopping cart is, F = 12.25 N

2.

We have: F = m × v/t

Here, m = 8500 Kg

v = 20 m/s

t = 10 s

Substitute their values into the expression,  

F = 8500 × 20/10

F = 8500 × 2

F = 17000 N

In short, final answer would be 17000 N

Hope this helps!!

7 0
3 years ago
A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
kap26 [50]

Answer:

a) \omega \approx 219.911\,\frac{rad}{s}, b) \alpha = 16.916\,\frac{rad}{s^{2}}, c) a_{t} = 1.776\,\frac{m}{s^{2}}, d) a_{n} = 5077.889\,\frac{m}{s^{2}}, e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) v = 23.091\,\frac{m}{s}.

Explanation:

a) The maximum angular velocity of the fan is:

\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )

\omega \approx 219.911\,\frac{rad}{s}

b) The angular acceleration of the fan is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}

\alpha = 16.916\,\frac{rad}{s^{2}}

c) The magnitude of the tangential aceleration is:

a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)

a_{t} = 1.776\,\frac{m}{s^{2}}

d) The magnitude of the centripetal acceleration is:

a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)

a_{n} = 5077.889\,\frac{m}{s^{2}}

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)

v = 23.091\,\frac{m}{s}

5 0
3 years ago
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto
daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

 

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

 

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

 

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 =  $741.68/yr

 

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0
3 years ago
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