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3 years ago

In the decreasing order of magnitude, which of the following is correct?

Physics

1 answer:

Sergio [31]3 years ago

5 0

Answer:

b. Static > sliding > rolling friction.

Explanation:

Static friction is greater than sliding friction. It takes more force to get an object to start sliding than to keep it sliding.

Sliding friction is greater than rolling friction. There are fewer points of contact for a round surface compared to a flat one.

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A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?

Delvig [45]

Answer:

u =30 m/s

v = 35 m/s

t = 5 secs

Explanation:

a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

8 0

2 years ago

Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?

Alik [6]

Answer:

Below

Explanation:

You can use this equation to find the distance :

distance = velocity x time

distance = (26.7)(3.06)

= 81.702 m

Rounding to 3 sig figs

= 81.7 m

Hope this helps

6 0

2 years ago

Damage to the frontal lobes of the brain is a likely result of what

ad-work [718]

binge drinking alcohol

8 0

3 years ago

Read 2 more answers

A calcium bromide compound is represented by which formula?

ipn [44]

Answer:

Calcium bromide is the name for compounds with the chemical formula CaBr2(H2O)x.

4 0

3 years ago

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A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored o

Nataly_w [17]

Explanation:

The given data is as follows.

$C_{1} = 1.50 \times 10^{-6} F$

$C_{1} = 3.50 \times 10^{-6} F$

Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

$\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

$\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

= $0.945 \times 10^{-6} F$

C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

Q = CV

= $1.06 \times 10^{-6} F \times 2.50 V$

= $2.65 \times 10^{-6} C$

= $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.

5 0

3 years ago