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shtirl [24]
3 years ago
6

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

Physics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

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The specific heat of copper is 0.385 J/g•°C. Which equation would you use to calculate the amount of heat needed to raise the te
Pavlova-9 [17]

Answer:

Q = 0.75 g x 0.897 J/g•°C x 22°C

Explanation:

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In this problem, the mass of the aluminium is m=0.75 g. The specific heat of aluminium is C=0.897 J/g•°C, while the increase in temperature is

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So, the equation becomes

Q = 0.75 g x 0.897 J/g•°C x 22°C

4 0
3 years ago
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A 0.150-kg cart that is attached to an ideal spring with a force constant (spring constant) of 3.58 N/m undergoes simple harmoni
SVETLANKA909090 [29]

Answer:

E = 0.01 J

Explanation:

Given that,

The mass of the cart, m = 0.15 kg

The force constant of the spring, k = 3.58 N/m

The amplitude of the oscillations, A = 7.5 cm = 0.075 m

We need to find the total mechanical energy of the system. It can be given by the formula as follows :

E=\dfrac{1}{2}kA^2

Put all the values,

E=\dfrac{1}{2}\times 3.58\times (0.075)^2\\\\=0.01\ J

So, the value of total mechanical energy is equal to 0.01 J.

3 0
3 years ago
Two soccer players start from rest, 36 m apart. They run directly toward each other, both players accelerating. The first player
Cerrena [4.2K]
A) Both players are moving by uniformly accelerated motion, and we can write the position at time t of each of the two players as follows:
x_1(t)= \frac{1}{2}a_1 t^2
x_2(t)=d- \frac{1}{2}a_2 t^2
where
a_1 = 0.58 m/s^2 is the acceleration of the first player
a_2=0.42 m/s^2 is the acceleration of the second player
d=36 m is the initial distance between the two players
and where I put a negative sign in front of the acceleration of the second player, since he's moving in the opposite direction of the first player.

The time t at which the two players collide is the time t at which x_1 = x_2, therefore:
\frac{1}{2}a_1 t^2 = d- \frac{1}{2}a_2 t^2
from whic we find
t= \sqrt{ \frac{2d}{a_1+a_2} }= \sqrt{ \frac{2 \cdot 36 m}{0.58 m/s^2+0.42 m/s^2} }=8.5 s

b) We can use the equation of x_1(t) to find how far the first player run in t=8.5 s:
x_1(t)= \frac{1}{2}a_1 t^2= \frac{1}{2}(0.58 m/s^2)(8.5 s)^2=21.0 m
8 0
3 years ago
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