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3 years ago

Velocity is the slope of the acceleration vs. time graph. A.) True B.)False (apex)

2 answers:

8 0

Velocity vs. time graph shows the acceleration as a slope whereas displacement vs. time graph shows the velocity as a slope. So, the given statement is false.

Answer: Option B

Explanation:

To understand the acceleration graphically, consider the x axis of the graph as the run and the y axis as the velocity rise. Now, as we all know that,

$\text {Acceleration}=\frac{\text {Change in velocity }}{\text {Time interval}}$

We can estimate this through the graph. let's draw the motion of an object with time if it's velocity is changing in every second by 4 m/s. Now if we draw this on graph, we will see that there is a slope between the two corresponding values of time and velocity. This slope defines the acceleration for the object with time.

Now, in the same way, if we draw a distance and time graph respective to the y and x axis; we'll get a slope which defines the velocity of the object i.e. change in distance with time.

Hence, with a velocity vs time graph, we get the slope for acceleration whereas with the distance and time graph, we get the slope for velocity. So both the cases, we see there is no velocity slope on an acceleration and time graph. Hence the statement is false.

Y_Kistochka [10]3 years ago

4 0

Answer:

its true for apex

Explanation:

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(4.3 x 10-2 )4.950 x 105) O 21.285 x 103 O 2.13 x 104 O 21 x 103 O 2.1 x 104

Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

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3 years ago

If two micro coulomb of charge is flowing in a circuit for 5 minutes. What is the amount of current in the circuit?

Murrr4er [49]

Answer:

6.67×10¯⁹ A

Explanation:

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

Quantity of electricity (Q) = 2 μC

Time (t) = 5 mins

Current (I) =?

Next, we shall convert 2 μC to C. This can be obtained as follow:

1 μC = 1×10¯⁶ C

Therefore,

2 μC = 2 μC × 1×10¯⁶ C / 1 μC

2 μC = 2×10¯⁶ C

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 secs

Therefore,

5 min = 5 min × 60 sec / 1 min

5 mins = 300 s

Finally, we shall determine the current in the circuit. This can be obtained as follow:

Quantity of electricity (Q) = 2×10¯⁶ C

Time (t) = 300 s

Current (I) =?

Q = It

2×10¯⁶ = I × 300

Divide both side by 300

I = 2×10¯⁶ / 300

I = 6.67×10¯⁹ A

Thus, the current in the circuit is 6.67×10¯⁹ A

4 0

2 years ago

Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y

Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density $\rho = 1000 kg/m^3$ , and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool

$W = mg = \rho V g = 1000 * 469.8 * 9.81 = 4608738 N$

The area of the bottom

A = 30 * 8.7 = 261 square meters

Therefore the pressure is its force over unit area

$P = F/A = 4608738 / 261 = 17658 N/m^2$ or 17.66 kPa

7 0

3 years ago

A 23.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position o

lara [203]

Answer:

Explanation:

We use the harmonic motion position equation:

$x(t) = A\cos(\omega t+\phi)$

where A = 0.350 and for t = 0

$x(0) A = A\ cos(\phi)$

so: $\phi = 0$

and also:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{4.10} = 1.532 rad/s$

so we have:

x(t)=0.350cos(1.532 t)

For t = 3.403 s

x(3.403)=0.350cos(1.532 (3.403)) = 0.348 m

3 0

3 years ago

Please help.

LenKa [72]

Mu wildest guess is c 1500

7 0

3 years ago

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