Answer:
C = 0.025F
Explanation:
Charge =q=3×[tex]10^{-6}[/tex]C
Voltage=V=120V
Q=CV
C=Q/V
=3×[tex]10^{-6}[/tex]/120
=1/40×[tex]10^{-6}[/tex]
C = 0.025F
C=2.5 [tex]10^{-4}[/tex]C 0r 25 ×[tex]10^{-3}[/tex]F
A body with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time, each of duration 4s. Determine the initial velocity and acceleration of the moving body.
Answer:
Explanation:
Average velocity in the 24 m interval is 24 / 4 = 6 m/s
Average velocity in the 64 m interval is 64 / 4 = 16 m/s
There is a 4 second interval between the two points where average velocity equals actual velocity
a = Δv/t = (vf - vi) / t = (16 - 6) / 4 = 2.5 m/s²
s = v₀t + ½at²
24 = v₀(4) + ½(2.5)4²
4v₀ = 24 - 20
v₀ = 1 m/s
Not asked for but the velocity at the end of the first segment and beginning of the second segment is 11 m/s and final velocity is 21 m/s
(c) It is suggested that one side of the copper sheet cools to a lower temperature than the
other side.
Explain why this does not happen.
[2]
Answer:
Explanation:
The word "sheet" implies that the copper is quite thin.
Copper is also a very good conductor of heat.
Therefore, with a very short heat flow distance to cover and a high rate of heat transmission, temperature differences on either side of the sheet are almost instantaneously eliminated by heat flow.
refers to a fear of being trapped in a crowded, public place.
Answer:
Agrophobia
Explanation:
an anxiety disorder in which someone feals anxious and scared in a public place
Which combination of three concurrent forces acting on a body could not produce equilibrium?
1
1 N, 3N, EN
2
2 N, 2N, 2N
.
3.
3 N, 4N, EN
4.
4N, 4N, 5N
All the three concurrent forces acting on a body will not produce equilibrium.
The given parameters:
1. 1 N, 3 N and 5 N
2. 2N, 2N and 2 N
3. 3N, 4N and 5 N
4. 4N, 4N and 5 N
Concurrent forces lie on the same plane and their line of action pass through a common point.
A body under concurrent forces is in equilibrium if the resultant of the forces on the body is zero.
[tex]\Sigma F = 0\\\\F_1 + F_2 + F_3 = 0\\\\F_1 + F_ 2 = - F_3[/tex]
where;
[tex]F_3[/tex] is the equilibrant force
First set of concurrent forces;
[tex]1 \ N \ + \ 3\ N = 4 \ N\\\\F_ 3 = 5 \ N\\\\5 \ N > 4 \ N[/tex]
Second set of concurrent forces;
[tex]2 \ N \ + \ 2 \ N = 4 \ N\\\\F_ 3 = 2 \ N\\\\4 \ N > 2 \ N[/tex]
Third set of concurrent forces;
[tex]3 \ N \ + \ 4 \ N = 7 \ N\\\\F_ 3 = 5 \ N\\\\7 \ N > 5 \ N[/tex]
Fourth set of concurrent forces;
[tex]4 \ N \ + \ 4 \ N = 8 \ N\\\\F_ 3 = 5 \ N\\\\8 \ N > 5 \ N[/tex]
Thus, we can conclude that all the three concurrent forces acting on a body will not produce equilibrium.
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1. A 10 kg rocket blast off the ground with an applied force of 150 N, calculate the net force
of your rocket
The net force on the rocket at the given applied force is 52 N.
The given parameters:
Mass of the rocket, m = 10 kgApplied force, F = 150 NThe net force on the rocket is calculated by applying Newton's second law of motion as follows;
F(net) = F(up) - F(down)
F(net) = 150 - mg
F(net) = 150 - (10 x 9.8)
F(net) = 150 - 98
F(net) = 52 N
Thus, the net force on the rocket at the given applied force is 52 N.
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If 90J of energy are available for every 30C of charge, what is the potential difference?
Enter your answer as a number
V
Answer:
3 v
Explanation:
I need help with this question!
Answer:
Explanation:
Conservation of momentum
In the x direction
2(15) + 2(-10) = 2(-5) + 2(vBfx)
vBfx = 10 m/s
In the y direction
2(30) + 2(5) = 2(20) + 2(vBfy)
vBfy = 15 m/s
vBf = 10i + 15j
KEi = ½(2)(15² + 30²) + ½(2)(-10² + 5²) = 1250 J
KEf = ½(2)(-5² + 20²) + ½(2)(10² + 15²) = 750 J
KEf - KEi = 750 - 1250 = -500 J
How large is the tension in a rope that is being used to accelerate a 100 kg box upward at 2m/s2?
[tex]\\ \sf\Rrightarrow T=F[/tex]
[tex]\\ \sf\Rrightarrow T=ma[/tex]
[tex]\\ \sf\Rrightarrow T=100(2)[/tex]
[tex]\\ \sf\Rrightarrow T=200N[/tex]
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
13. In which of the following climates does chemical weathering generally occur most
rapidly?
A. Canda
B. Greenland
C. Brazil
D. Algeria
A 70 kg man is running up the stairs which is 3m high in 2s.(a)How much work is done by the man?(b)What is the power exerted by the man? (Use g = 10ms 2)
Explanation:
m = 70 kg
s = 3m
t =2s
g = 10 m/s²
(a)How much work is done by the man?
W = Fs
= mg X s
= 70 x 3 x 10
= 210 x 10
= 2100 Joule
(b)What is the power exerted by the man?
P = W/t
= 2100/2
P = 1050 Watt
13) What property of matter would be measured with this piece of equipment?
A) The mass of an apple
hing
)
By the temperature of a room
The volume of water in a glass.
D) The length of a piece of string.
Answer:
a
Explanation:
i took the test 100%
12- Calculate the power when a force of 60N moves an object over a distance of 0.6 km in 20
minutes
A. 100watts
B. 6,000 watts
C. 0.25watts
D. 30 watts
Hi there!
To solve, we must begin by calculating the total WORK done on the object.
W = F · d (Force · displacement)
Plug in the given values. Remember to convert km to m:
1 km = 1000 m
0.6 km = 600 m
W = 60 · 600 = 36000 J
Now, we can solve for power:
P = W/t
Convert minutes to seconds:
1 min = 60 sec
20 min = 1200 sec
P = 36000/1200 = 30 W ⇒ Choice D.
That’s easy please tell me!
i need help with the problem below
Answer:
Explanation:
a) F = ma
a = F/m
a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s
b) t = v/a
t = 200 / 7.2 x 10⁻⁶
t = 2.8 x 10⁷ s about 10½ months
c) v² = u² + 2as
s = (v² - u²) / 2a
s = (200² - 0²) / (2( 7.2 x 10⁻⁶))
s = 2.8 x 10⁹ m nearly 7 times around the earth
And all this assumes NO FRICTION.
what is the speed of a boy moving around a circular park of radius 6m if he goes around the park in 20s
Answer:
1.884 meters per second
Explanation:
s = d/t
d = C = 2πr = 2π(6) = 37.68 m
t = 20 s
s = (37.68 m)/(20 s)
s = 1.884 m/s
The speed of the boy is 1.88 m/s while he is moving around a circular park of radius 6m and taking time 20 seconds.
What is speed?Speed can be defined as a measurement of how quickly the distance an object traveled changes with time. Speed is a scalar quantity as it has magnitude but no direction as a unit of measurement.
An object that moves quickly and with high speed, traveling a lot of ground in a lesser time.
Given, the radius of the circular park, r = 6m,
The time taken to cover one round of the park, t = 20 s
The total distance traveled by the boy is equal to the circumference of the circular park, so
Circumference of park, C = d = 2πr = 2× 3.14 × 6 =37.68 m
The speed of the boy, S = d/t
S = 37.68 /20
S = 1.88 m/s
Therefore, the speed of the boy is equal to 1.88 m/s.
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Lisa runs up 4 flights of stairs in 22 seconds. She weighs 510 Newtons. If each flight rises 310 cm: (a) What is her change in potential energy? (b) What average power (watts) was required during the 22 s?
The change in potential energy is 6324 Joules while the power is 263.5 Watts.
Potential energy is the amount of energy possessed by a body due to its position relative to another object.
a) Potential energy = weight * height = 510 N * 4 flight * 3.1 m per flight
Potential energy = 6324 Joules
b) Power = Energy / time = 6324 Joules / 22 seconds = 263.5 Watts
The change in potential energy is 6324 Joules while the power is 263.5 Watts.
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1. The acceleration equation of a certain particle is a=2t, after 4s, its velocity reaches 20m/s, then the initial velocity of the particle movement is ) ms B Sms 2.
Answer:
Change in velocity is the integral of all the differential accelerations acting over the period of acceleration.
[tex]20 = u + \int\limits^4_0 {2t} \, dt[/tex]
20 = u + t²[tex]\left \{ {{4} \atop {0}} \right.[/tex]
20 = u + 4² - 0²
20 = u + 16
u = 4 m/s
Would it be possible to predict the speeds that a coaster will reach before it’s ever placed on the track?
Yes, it's possible to predict the speeds that a coaster will reach before it’s
ever placed on the track.
This is usually calculated with the potential energy which is
Potential energy = m g h
where m is mass, g is acceleration due to gravity and h is height.
The given formula above is used in predicting the speeds that a coaster will
reach before it’s ever placed on the track.
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Explain how the removal of heat energy affects the speed of the particles in a substance
Answer:
The removal of heat energy slows the speed of particles
Explanation:
When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.
g A 24-gg bullet strikes and becomes embedded in a 1.50-kgkg block of wood placed on a horizontal surface just in front of the gun. Part A If the coefficient of kinetic friction between the block and the surface is 0.23, and the impact drives the block a distance of 9.5 mm before it comes to rest, what was the muzzle speed of the bullet
The muzzle speed of the bullet before the collision is 415.3 m/s.
The given parameters:
Mass of the bullet, m₁ = 24 gMass of the wood, m₂ = 1.5 kgCoefficient of kinetic friction, μk = 0.23Distance traveled by the block before stopping, d = 9.5 mApply the principle of work-energy theorem to determine the final velocity of the block-bullet system;
[tex]F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s[/tex]
Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;
[tex]m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s[/tex]
Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.
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Please answer the question in the picture
Physical Science A 2021-2022
Why does increasing the number of trials increase confidence in the results of the experiment?
Answer:
Increasing the number of trials reduces the impact of any one imprecise measurement. … To increase the number of attempts, you can find an average result for the experiment, as well as find and discrepancies as human error if you perform an experiment several times.
Explanation:
hope it helps :)
Answer:
It is because the increase in the number of trials reduces the impact of any one imprecise measurement. Using an average value for data points provides a better representation of the true value.
A small ball with a mass of 0.6 kg and a velocity of 12 m/s hits another ball with the same mass. The firs forward and hits a third ball with a mass of 0.2 kg. If the system is closed, what is the velocity of the third ball?
4 m/s
36 m/s
30 m/s
1.44 m/s
A 2N and an 6N force pull on an object to the right and a 4N force pulls on the object to the left. If the object has a mass of 0.25 kg what is its acceleration?
Answer:
[tex]16m/s^{2}[/tex]
Explanation:
Please ask if you have more questions!
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Answer:
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Explanation:
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VERY EASY QUESTION FOR HIGH SCHOOL STUDENTS:
Which of the following frequencies would you expect a young person to be able to hear? 500 Hz, 6000 Hz, 25000 Hz, 15 Hz, 15000 Hz
Answer:
Explanation: 6000z
when a student comned his dry hair for a long time, his hair began to stand up. Which of these has MOST LIKELY happened
His hair got charged due to continuous friction causing them to get attracted to other objects and thus standing up.
Answer:The comb and hair have become charged with opposite charges.
Explanation:When a comb is run through your hair charges pass between your hair and the comb, so the comb becomes charged either positively or negatively, and the hair oppositely charged. When the comb is brought close to paper an opposite charge is induced in the paper, and the opposite charges attract.
You hang a light in front of your house using an
elaborate system to keep the 12-kg object in static
equilibrium (Figure 1). What are the magnitudes of the
forces that the ropes must exert on the knot connecting
the three ropes if 02 = 639 and 03 = 45° ?
The magnitudes of the forces that the ropes must exert on the knot connecting are :
F₁ = 118 N F₂ = 89.21 N F₃ = 57.28 NGiven data :
Mass ( M ) = 12 kg
∅₂ = 63°
∅₃ = 45°
Determine the magnitudes of the forces exerted by the ropes on the connecting knota) Force exerted by the first rope = weight of rope
∴ F₁ = mg
= 12 * 9.81 ≈ 118 kg
b) Force exerted by the second rope
applying equilibrium condition of force in the vertical direction
F₂ sin∅₂ + F₃ sin∅₃ - mg = 0 ---- ( 1 )
where: F₃ = ( F₂ cos∅₂ / cos∅₃ ) --- ( 2 ) applying equilibrium condition of force in the horizontal direction
Back to equation ( 1 )
F₂ = [ ( mg / cos∅₂ ) / tan∅₂ + tan∅₃ ]
= [ ( 118 / cos 63° ) / ( tan 63° + tan 45° ) ]
= 89.21 N
C ) Force exerted by the third rope
Applying equation ( 2 )
F₃ = ( F₂ cos∅₂ / cos∅₃ )
= ( 89.21 * cos 63 / cos 45 )
= 57.28 N
Hence we can conclude that The magnitudes of the forces that the ropes must exert on the knot connecting are :
F₁ = 118 N, F₂ = 89.21 N, F₃ = 57.28 N
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