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Ipatiy [6.2K]
3 years ago
11

A race car has a mass of 820 kg. It starts from rest and travels 50.0m in 3.0s. The car is uniformly accelerated during the enti

re the time. How big is the net force acting on the car?
Physics
1 answer:
Contact [7]3 years ago
7 0
A=DELTAv/DELTAt=50/3
f=ma=820.50/3
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The brakes application to a car produce an acceleration of 6ms2 in the opposite direction to the motion .If the car takes 2 seco
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Answer:12 meter

Explanation:

acceleration(a)=6m/s^2

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2 years ago
11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
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a =  \frac{v - u}{t}

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u = initial velocity

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the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

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the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

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3 years ago
What happens when electromagnetic waves cause a disturbance in electric
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2 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
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