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3 years ago

11

A circuit has two batteries and a resistor as shown. In the figure, the terminal voltage of the 12 V battery is closest to _____

.
a. 15.6 V
b. 13.2 V
c. 8.4 V
d. 10.8 V
e. 9.6 V

1 answer:

Andrei [34K]3 years ago

7 0

Answer:

a 15.6voltz

i Hope this helps

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Which change occurs when an atom undergoes decay?

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<span>When an atom undergoes decay its nucleus becomes unstable because it does not have a balance of neurons and protons. This means the atoms try to find a confirmation in which it is stable. In doing so, it is possible that the atom changes its element, and it frees a huge amount of energy in order to balance itself. This is the base of radioactivity.</span>

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3 years ago

Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc

svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

$v=v_{i}+gt\\$

Also get the equation of displacement

$S=v_{i}t+(1/2)gt^{2}$

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

$v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\ And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m$

For t₂

$v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\ And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m$

For t₃

$v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\ And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m$

For t₄

$v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\ And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m$

For t₅

$v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\ And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m$

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4 years ago

(1 point) A frictionless spring with a 6-kg mass can be held stretched 0.2 meters beyond its natural length by a force of 50 new

guapka [62]

Answer:

x(t) = 0.077cos(6.455t)

Explanation:

If the spring can be stretched 0.2 m by a force of 50 N, then the spring constant is:

k = 50 / 0.2 = 250 N/m

The equation of simple harmonic motion is as the following:

$x(t) = Acos(\omega t - \phi)$

where $\omega = \sqrt{k/m} = \sqrt{250 / 6} = 6.455$

We also know that the initial velocity is 0.5 m/s, which is also the maximum speed at the equilibrium:

$v_{max} = A\omega$

$A = v_{max}/\omega = 0.5 / 6.455 = 0.077 m$

$\phi = 0$ is the initial phase

Therefore, the position of the mass after t seconds is

x(t) = 0.077cos(6.455t)

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3 years ago

What is the mass of 1.5 in newtons?

Thepotemich [5.8K]

Answer:

0.15 kg

Explanation:

F = mg

1.5N = m(9.8 m/s^2)

m = 0.15kg

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3 years ago

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The Correct answer: 18

They are also known as families because they are in the same column.

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3 years ago