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Afina-wow [57]
3 years ago
15

Describing the motion of an object can be difficult to do and using graphs help make motion easier to understand. Motion is a ch

ange in position measured by distance and time. Speed tells us the rate at which an object moves. Speed graphs are distance-time graphs; sometimes people label them as position-time graphs as well. Time is plotted on the X-axis and distance or position on the Y-axis of a line graph.
A distance-time graph tells us how far an object has moved with time.
• The steeper the graph, the faster the motion.
• A horizontal line means the object is not changing its position. It is not moving, and it is at rest.
• A downward sloping line means the object is returning to the start.

Use the text and the graphs above to choose all of the correct answers concerning speed?

In both graphs, speed will be measured in meters/second.
In graph A, the object travelled a distance of 40 meters in 4 seconds
The total distance travelled by the object in graph B is greater than the object in graph A.
In graph B, the object is at rest for 5 seconds.
The object in graph B was moving at a greater speed than the object in graph A

Physics
2 answers:
olga_2 [115]3 years ago
7 0

In both graphs, speed will be measured in meters/second.

In graph A, the object travelled a distance of 40 meters in 4 seconds

In graph B, the object is at rest for 5 seconds.

These are the correct answers

rewona [7]3 years ago
4 0

Answer:

Explanation:

1. In both graphs, speed is measured in meters/ second. The time measured in seconds is plotted on the X-axis while the distance measured in meters is plotted on the Y-axis.

2. Looking at the graph, you could identify the distance of 40 meters on the Y-axis and trace it down to time at 4 seconds on the X-axis.

3. The total distance travelled by the object in graph A is greater than the total distance travelled by the object in graph B. Graph B has a total distance of 25 m while graph A has a total distance of 50 m.

4. In graph B, the object is at rest at 5 seconds. From 5 seconds, it maintained a horizontal line indicating that it was static.

5. Graph A was moving at a greater speed than Graph B. Graph A had a speed of 10 m/s while Graph B had a speed of 5 m/s.

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8 0
3 years ago
You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30°C to
earnstyle [38]

The correct answer to the question is : D) 352.6 m/s.

CALCULATION :

As per the question, the temperature is increased from 30 degree celsius to 36 degree celsius.

We are asked to calculate the velocity of sound at 36 degree celsius.

Velocity of sound is dependent on temperature. More is the temperature, more is velocity of sound.

The velocity at this temperature is calculated as -

                            V = 331  + 0.6T m/s

                               = 331 + 0.6 × 36 m/s

                               = 331 + 21.6 m/s

                               = 352.6 m/s.

Here, T denotes the temperature of the surrounding.

Hence, velocity of the sound will be 352.6 m/s.

5 0
3 years ago
Read 2 more answers
Light is traveling through the different media shown. In which medium does light travel fastest?
bija089 [108]
Light travelling in a vacuum is the fastest thing in the universe. The speed would be 2.99x10^8 m/s. The answer to this question is 'vacuum', where light can travel the fastest. I hope this helps you. You're welcome!
8 0
4 years ago
Read 2 more answers
Options are:<br>a)4Cn<br>b)5Cn<br>c)6 Cn<br>d)3 Cn<br>​
nasty-shy [4]

Answer:

Option B. 5 nC

Explanation:

From the question given above, the following data were obtained:

Capicitance (C) = 100 pF

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Next, we shall determine the quantity of charge. This can be obtained as follow:

Capicitance (C) = 1×10¯¹⁰ F

Potential difference (V) = 50 V

Quantity of charge (Q) =?

Q = CV

Q = 1×10¯¹⁰ × 50

Q = 5×10¯⁹ C

Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:

1 C = 1×10⁹ nC

Therefore,

5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C

5×10¯⁹ C = 5 nC

Thus, the quantity of charge is 5 nC

3 0
3 years ago
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
Gala2k [10]
It’s not in English can you change it please
3 0
3 years ago
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