Answer:
For a velocity versus time graph how do you know what the velocity is at a certain time?
Ans: By drawing a line parallel to the y axis (Velocity axis) and perpendicular to the co-ordinate of the Time on the x axis (Time Axis). The point on the slope of the graph where this line intersects, will be the desired velocity at the certain time.
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How do you know the acceleration at a certain time?
Hence,
By dividing the difference of the Final and Initial Velocity by the Time Taken, we could find the acceleration.
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How do you know the Displacement at a certain time?
Ans: As Displacement equals to the area enclosed by the slope of the Velocity-Time Graph, By finding the area under the slope till the perpendicular at the desired time, we find the Displacement.
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Answer:
due to condection of heat from pan in thermal energy according to 9th
Explanation:
Answer:
V is approximately = 23m/s
Explanation:
Kinetic energy = ½ mv²
Where m= mass = 0.450kg
V= velocity =?
K. E = 119J
Therefore
K. E = ½ mv²
Input values given
119= ½ × 0.450 × v²
Multiply both sides by 2
119 ×2 = 2 × 1/2 × 0.450 × v²
238= 0.450v²
Divide both sides by 0.450
238/0.450 = 0.450v²/0.450
v² = 528.89
Square root both sides
Sq rt v² = sq rt 528.89
V = 22.998m/s
V is approximately = 23m/s
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To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:
Where,
= Mass of each object
= Initial Velocity of Each object
= Final Velocity
Our values are given as
Replacing we have that
Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s
Answer:
The balloon will continue to expand and eventually burst.
Explanation:
Simply, the reason for this is because the density of the atmosphere decreases gradually as you increase in altitude closer to space. This means that the air on the outside of the balloon can't provide enough pressure over the surface of the balloon in order to counteract the gas on the inside of the balloon from expanding.
