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3 years ago

Three 20.0 ohm resistors are

Physics

1 answer:

V125BC [204]3 years ago

5 0

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

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What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?

eimsori [14]

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

4 0

3 years ago

A radium atom has 88 protons and 226 neutrons. How many electrons does it have?

Ymorist [56]

Answer:

No. of electrons = 88

Explanation:

Given that,

No. of protons = 88

No. of neutrons = 226

We need to find the number of electrons it have in a radium atom.

In an atom, the number of protons is equal to the number of electrons. There are 88 protons in this case.

Hence, there are 88 electrons in the radium atom.

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3 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

ivanzaharov [21]

Answer:

Question 1

The velocity of the hare 2.2 s after it starts is 1.76 m/s

Question 2

15.1 s after the hare starts, its velocity is 3.53 m/s

Question 3

The hare travels 55.49 m before it begins to slow down.

Question 4

Once it begins to slow down, the acceleration of the hare is -1.13 m/s²

Question 5

The total time the hare is moving is 21.02 s.

Question 6

The acceleration of the tortoise is 0.28 m/s².

Explanation:

These kinematic equations apply for the hare:

for the first 4.4 seconds:

v = v0 + at

x = x0 + v0t + 1/2at²

where:

v = velocity

v0 = initial velocity

a = acceleration

t = time

x = position

x0 = initial position

from 4.4 s to 17.9 s (+13.5 s)

v = constant.

the velocity is the same as the final velocity in the first 4.4 s of the race:

v = v0 + a*4.4s

x = x0 + vt

from 17.9 s until end:

v = v0 +at

x = x0 + v0t +1/2at²

Question 1

2.2 s after the start the hare is accelerating (0.8 m/s²).

from the equation:

v = v0 +at

replacing with the data:

v = 0 m/s + 0.8 m/s² * 2.2 s = 1.76 m/s

Question 2

At 15.1 s the hare is running at constant speed. It will be the final speed reached during the first 4.4 s:

v = v0 + a*4.4s

replacing with the data:

v= 0 m/s + 0.8 m/s² * 4.4s = 3.52 m/s

Question 3

We have to find the position at time 17.9 s.

For the first 4.4 s the hare runs:

x = x0 + v0t +1/2at² = 0m + 0 m/s * 4.4 s + 1/2 * 0.8 m/s² * (4.4 s)² = 7.7 m

For the next 13.5 s, the hare runs:

x = x0 + vt

where v=v0 + a*4.4s (the final velocity of the first 4.4 s)

v = 0 m/s + 0.8 m/s² * 4.4 s = 3.52 m/s

and x0 = 7.7 m (the final position of the first sprint)

Then:

x = 7.7m + 3.54 m/s * 13.5 s = 55.49 m

Question 4

The equation of position in this part of the race is:

x = x0 + v0t +1/2at²

where

x0 is the position calculated in question 3.

v0 is the final speed of the first 4.4 s calculated in question 2.

The velocity of the hare is 0 at position x = 61 m, then:

v = v0 +at

0 = v0 +at (at x = 61 m)

-v0 = at

a = -v0/t

then replacing a = -v0/t in the equation of position and solving for t:

x = x0 + v0t + 1/2(-v0/t)*t²

x = x0 + v0t -1/2v0t

x = x0 + 1/2v0t

x - x0 / (1/2v0) = t

replacing with the data:

61 m -55.49 m / 1/2* 3.53 m/s = 3.12 s

The acceleration is then:

a = -v0/t

a = -3.53 m/s / 3.12 s = -1.13 m/s²

Question 5

The hare moves for 4.4 s accelerating, for 13.5 s at a constant speed and for 3.12 s (see question 4) slowing down.

The total time is: 4.4s + 13.5 s + 3.12 s = 21.02 s

Question 6

The tortoise runs 61.0 meters in 21.02 s (the tortoise catches the hare just when it comes to stop). The equation for the position can be written as:

x = x0 +v0t +1/2at²

x0 = 0 and v0 = 0 since the tortoise starts from rest. Then, solving for a:

2x / t² = a

replacing with the data:

2*61 m / (21.02 s)² = a

a = 0.28 m/s²

"Slow and steady wins the race"

5 0

3 years ago

26. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At

garik1379 [7]

Answer:

a) 19.96 [m/s]

b) 3.26 [s]

c) 13.02 [m]

Explanation:

Use the equations for linear motion:

$X_{f} =X_{0} + vt + \frac{1}{2} at^{2}$
$v_{f} = v_{o} + at$

Since we have motion in the horizontal and vertical directions, we need to break down the problem by solving each direction separately. The first thing they ask for is the speed of the ball when it hits the ground. This will require finding the magnitude of the velocity vector when the ball hits the ground. As we know, it will have two velocity components, one in "x" and one in "y".

X component:

$v_{fx} = v_{ox}+ a_{x} t$

We have no accelerations acting on x, therefore:

$v_{fx} = 12 + 0t\\v_{fx} = 12 [\frac{m}{s} ]$

Y component:

$v_{fy} = v_{oy}+ a_{y} t$

$v_{fy} = 16 - 9.8t$

We can´t solve this equation yet as we don´t know how long it takes the ball to hit the ground. We´ll use this equation later.

The second question is how long does the ball remain in the air. We need to find a position of the ball where we can satisfy the equation and solve it. If you think about it, right at mid-flight, the ball is at maximum height and the ball has no velocity component in y (because it just transitioned from going up to going down), we can rewrite the equation as follows:

$v_{mid} = v_{0y} + a_{y} t_{\frac{1}{2} }$
$0 = 16 - 9.8t_{\frac{1}{2} }$

Solve for time in the equation which is half of the total time the ball is in the air:

$t_{\frac{1}{2} } = \frac{-16}{-9.8} = 1.63 [s]$

And finally, multiply times 2 to get the total time the ball is in the air:

$t_{air} = 2 * t_{\frac{1}{2} } = 2 * 1.63 \\t_{air} = 3.26 [s]$

We can now go back to the first question and input the total time in the equation to find out the the velocity at which the ball hits the ground in the vertical direction:

$v_{fy} = 16 - 9.8t$
$v_{fy} = 16 - 9.8(3.26)\\ v_{fy} = -15.95 [\frac{m}{s}]$

We have the components, and all we need to do is find the root square of their squares (pythagoras theorem) to find the magnitude which equals the speed:

$v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2} }\\v_{f} = \sqrt{12^{2}+(-15.95^{2} ) } \\ v_{f} = 19.96 [m/s]$

The third question is about the maximum height attained by the ball. We know the time when the ball was at its maximum height to be 1.63 [s], and we can use equation 1 to figure out the maximum height as follows: