What does the slope of the line on this velocity versus time graph represent?

1 answer:

8 0

What do you mean when you say "this" graph ? I don't see any graph.

First of all, I'm sure you don't have a "velocity versus time" graph.
You may have a "speed versus time" graph, that doesn't show any
information about the direction of the motion.

As you look at your graph ... a feat that's impossible for me ...

-- If the line is horizontal, then speed is not changing. If the direction
of motion is also not changing, then there is zero acceleration. Sadly,
the graph most likely doesn't carry any information about the direction.
It's possible that the speed is constant but the direction is changing.
Then acceleration isn't zero, but you can't tell that from the graph.

-- If the line is proceeding upward from left to right, whether it's straight
or curving, then there is positive acceleration.

-- If the line is proceeding downward from left to right, whether it's straight
or curving, then there is negative acceleration, or what some people might
call "positive deceleration".

You might be interested in

A student measured the specific heat of water to be 4.29 J/g.Co. The

leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%$

So, the student's percent error is 2.63 %.

7 0

3 years ago

on a very muddy football field, a 120 kg linebacker tackles an 75 kg halfback. immediately before the collision, the linebacker

DIA [1.3K]

<u>Momentum</u>
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.

Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum

Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum

Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude

!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s

If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]

The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.

6 0

3 years ago

Explain why the driver's reaction time affects the thinking distance.

Lemur [1.5K]

Answer:

The thinking distance depends on the reaction time of the driver which could can affected by alcohol, distractions and tiredness. A faster speed increases both thinking distance, increasing the total stopping distance.

<h2><em>I hope this is helpful. Would appreciate if you add me as brainliest.</em></h2>

7 0

3 years ago

An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux

zhenek [66]

Answer:

$\phi=1.56\times 10^{-5}\ Wb$