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2 years ago

If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?

2 answers:

6 0

Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop of a velocity vs time describes acceleration.
If we have the straight line on the graph it means that the slope is always the same whereas the non-linear graphs has a variable slope that changes depending on your point in the graph.
To conclude - if your graph is not a straight line it has variable acc at many points.

Morgarella [4.7K]2 years ago

3 0

I accidentally answered the wrong one. Im sorry (edit)

Explanation:

You might be interested in

Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

amm1812

-- Equations #2 and #6 are both the same equation,
and are both correct.

-- If you divide each side by 'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by 'frequency', you get Equation #3,
which is also correct.
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.

7 0

3 years ago

Read 2 more answers

If you and 3 friends are playing tug of war and are pulling with a force of 35 N and your teammate with 37 N, while your opponen

mash [69]

72N and 75N. no. hope it helps you

3 0

3 years ago

A car with a mass of 1.1 × 103 kilograms hits a stationary truck with a mass of 2.3 × 103 kilograms from the rear end. The initi

snow_lady [41]

Answer:

The velocity of the truck after this elastic collision is 15.7 m/s

Explanation:

It is given that,

Mass of the car, $m_1=1.1\times 10^3\ kg$

Mass of the truck, $m_2=2.3\times 10^3\ kg$

Initial velocity of the car, $u_1=22\ m/s$

Initial velocity of the truck, u₂ = 0

After the collision the velocity of the car is, v₁ = -11 m/s

Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$1.1\times 10^3\times 22+0=1.1\times 10^3\times (11)+2.3\times 10^3\times v_2$

$v_2=15.7\ m/s$

So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).

4 0

2 years ago

A cycle track is 500 metres long. Amy completes 10 laps. She travelled at an average speed of 12.5 metres per second. She puts o

yanalaym [24]

The fast lap is irrelevant to the question, because it didn't happen
until after the 9 laps that you're interested in.

To be perfectly technical about it, we don't actually have enough
information to answer the question. You told us her average speed
for 10 laps, but we don't know anything about how her speed may
have changed during the whole 10 laps. For all we know, maybe
she took a nap first, and then got up and drove 10 laps at the speed
of 125 metres per second. That would produce the average speed
of 12.5 metres per second and we would never know it Why not ?
That's only 280 miles per hour. Bikes can do that, can't they ?

IF we can assume that Amy maintained a totally steady pace through
the entire 10 laps, then we could say that her average for 9 laps was
also 12.5 metres per second.

5 0

3 years ago

A 50.1 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resi

Varvara68 [4.7K]

Answer:

The total distance is 16.9 m

Explanation:

We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.

$W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m$

7 0

2 years ago