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777dan777 [17]
3 years ago
7

Suppose a car is accelerating so that its speed is increasing. First, describe the line that you would plot on a speed-time grap

h for the motion of the car. Then describe the line that you would plot on a distance-time graph.
Physics
1 answer:
Mandarinka [93]3 years ago
8 0

Think Critically: Suppose a car is accelerating so that its speed is increasing. Describe the plotted line of the motion of the car on a distance-time graph. The plot would be a curved line sloping upward whose slope is increasing.

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What is true of a reaction that has reached equilibrium?
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When a reaction has reached equilibrium, the rate of the forward reaction<span> equals the rate of the reverse </span>reaction<span>, such that the concentrations of reactants and products remain fairly stable, in a chemical </span>reaction<span>. Hope this answers the question. Have a nice day.</span>
4 0
4 years ago
(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?
VashaNatasha [74]

Answer:

\boxed {3.43 x 10^{3}}

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as v=\frac {d}{t} where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

d=vt

Substituting 343 m/s for d and 10 s for t then

d= 343\times10= 3430= 3.43 x 10^{3}

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the lights on Galaxy watch shifted of uniform amounts to what the rate in of the spectrum,regardless of the distance from Earth

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3 years ago
An object is 1.0 cm tall and its erect image is 5.0 cm tall. what is the exact magnification?
ikadub [295]
The exact magnification of the objects is calculated by dividing the cinema. We calculate it by diving the erect image size by the object size. From the given above, we find the exact magnification by dividing 5.0 cm by 1.0 cm. Thus, the answer would be 5. 
7 0
3 years ago
A tube of mercury with resistivity 9.84 × 10 -7 Ω ∙ m has an electric field inside the column of mercury of magnitude 23 N/C tha
slava [35]

Answer:

The current through the tube is 73.39A.

Explanation:

The relationship between the resistivity \rho, the electric field E, and the current density J is given by

\rho = \dfrac{E}{J}

This equation can be solved for J to get:

J = \dfrac{E}{\rho}

Since the current is I = J\cdot A

I= J\cdot A  = \dfrac{E}{\rho} \cdot A

Now, for the tube of mercury \rho = 9.84*10^{-7}\: \Omega \cdot m, E = 23N/C, and the area is A = \pi r^2 = \pi (1.0*10^{-3}m)^2 = 3.14*10^{-6}m^2; therefore,

I= \dfrac{23N/C}{9.84*10^{-7}\Omega\cdot m } *3.14*10^{-6}m^2

\boxed{I = 73.39A.}

Hence, the current through the mercury tube is 73.39A.

5 0
3 years ago
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