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inn [45]
3 years ago
14

The table shows the changes in velocity for several different cars. Each change takes place over a 10-second interval. Which car

s are either speeding up or slowing down?
Physics
2 answers:
sesenic [268]3 years ago
6 0
Car B, Car C, And Car E Are either speeding up or slowing down
erica [24]3 years ago
3 0

Answer:

B, C, E are the right answers

Explanation:

I got this question on a test and got these as the right answers

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Which of the following is the correct name for CCl4
Ira Lisetskai [31]

Answer: Carbon tetrachloride Or Tetrachloromethane

Explanation: Carbon tetrachloride is an important nonpolar covalent compound. You determine its name based on the atoms present in the compound.

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3 years ago
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Which sentence in the passage describes retrograde motion of a planet
Soloha48 [4]

Answer:

what is the passage? I don’t see it

Explanation:

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3 years ago
A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the poli
NNADVOKAT [17]

Answer:74.33 feet per sec

Explanation:

Given

Police car is 50 feet side of road

Red car is 140 feet up the road

Distance between them is decreasing at the rate of 70 feet per sec

From figure

x^2+y^2=z^2

z=\sqrt{140^2+50^2}

z=148.66 feet

as the red car is moving

therefore its velocity magnitude is given by

\frac{\mathrm{d} x}{\mathrm{d} t}

2x\times \frac{\mathrm{d} x}{\mathrm{d} t}+0=2z\frac{\mathrm{d} z}{\mathrm{d} t}

x\frac{\mathrm{d} x}{\mathrm{d} t}=z\frac{\mathrm{d} z}{\mathrm{d} t}

140\times \frac{\mathrm{d} x}{\mathrm{d} t}=148.66\times 70

\frac{\mathrm{d} x}{\mathrm{d} t}=74.33 feet\ per\ sec

6 0
4 years ago
A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?
dedylja [7]

Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

E=\frac{1}{2} CV^2=\frac{1}{2C} Q^2.

Substituting numerical value

E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J

7 0
3 years ago
. During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from th
mixer [17]

Answer:

106.7 N

Explanation:

We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:

F \Delta t = m (v-u)

where

F is the average force

\Delta t is the duration of the collision

m is the mass of the ball

v is the final velocity

u is the initial velocity

In this problem:

m = 0.200 kg

u = 20.0 m/s

v = -12.0 m/s

\Delta t = 60.0 ms = 0.06 s

Solving for F,

F=\frac{m(v-u)}{\Delta t}=\frac{(0.200 kg) (-12.0 m/s-20.0 m/s)}{0.06 s}=-106.7 N

And since we are interested in the magnitude only,

F = 106.7 N

5 0
3 years ago
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