Answer:
Explanation:
The total distance is how far you walk from the starting point.
Distance through west = 18.0m
Distance through north = 25.0m
Total distance covered = 18.0+25.0m
Total distance covered = 43.0m
This means that I am 43.0m from the starting point
Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:
[tex]d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m[/tex]
The direction of your displacement is 30.81m
Direction is gotten according to the formula;
[tex]\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0[/tex]
Note that the direction to the west is negative, that is why the x is -18.0m
The distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
Given-
Distance travel through the west is 18 m.
Distance travel through the north is 25 m.
Distance from starting point-
To know the total distance, add both the covered distance. Thus total distance x is,
[tex]x=18+25[/tex]
[tex]x=43[/tex]
Hence, the distance from the starting point is 43 m.
The displacement vector-
Displacement is calculated as the shortest distance between starting and final point. This shortest distance can be calculated using the Pythagoras theorem which states that in a right-angled triangle, the square of the hypotenuse [tex]d[/tex] is equal to the sum of the squares of the other two sides. Therefore,
[tex]d^2=18^2+25^2[/tex]
[tex]d^2=324+625[/tex]
[tex]d^2=949[/tex]
[tex]d=\sqrt{949}[/tex]
[tex]d=30.81[/tex]
The displacement vector is 30.81 m.
The Direction of displacement-The direction of displacement [tex]\theta[/tex] with these two sides can be calculated with the formula,
[tex]\theta=tan^{-1}\dfrac{25}{-18}[/tex]
Here due to the west direction(opposite side), the sign is taken negatively.
[tex]\theta=tan^{-1}(-1.389)[/tex]
[tex]\theta=-60.27^o[/tex]
For the other quarter,
[tex]\theta=180-60.27=119.7^o[/tex]
Hence, the distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
For more about the displacement, follow the link below-
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In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by:
A) applying a known force and measuring it's acceleration.
B) measuring the volume.
C) weighing it.
Answer:
A) By applying a known force, and measuring it's acceleration.
Explanation:
This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.
Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.
What are the standard international (si) units of distance
Answer:
meter
Explanation:
Answer: The International System of Units is a system of measurement based on 7 base units
Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.
ionic bonds form when electrons?
Answer:
when the electron transferred permanently to another atom
Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years
Answer:
T = 1.4696 10⁴ years
Explanation:
For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law
F = ma
G m M / r² = m a_c = m v² / r
G M / r = v²
the speed of the circular orbit is
v = 2π r / T
we substitute
G M / r = 4π² r² / T²
T² = (4π² / G M) r³
Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse
T² = (4π² /GM) a³
the constant is worth
(4π² / GM) = 2.97 10⁻¹⁹ s² / m³
let's reduce the distance to SI units
AU is the distance from the Earth to the Sun
a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)
a = 8.976 10¹³ m
T² = 2.97 10⁻¹⁹ (8.976 10¹³)³
T² = 21.4786 10²²
T = 4.63 10¹¹ s
Let's reduce to years
T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)
T = 1.4696 10⁴ years
Weight of a body becomes greater at the pole than that at the equator . why ?
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *
A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s
Answer:
D. 10 m/s
Explanation:
Calculate the force a 75 kg high jumper must exert in order to produce an acceleration that is 3.2 times the acceleration due to gravity.
Answer:
Explanation
According to Newton's second law of motion,
F = ma
m is the mass
a is the acceleration
If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g
The formula becomes;
F = m(3.2g)
F = 3.2mg
m= 75kg
g = 9.81m/s²
F = 3.2(75)(9.81)
F = 2,354.4N
Hence the force exerted by the jumper is 2,354.4N
A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?
Answer:
Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.
Explanation:
To find the distance at which the first package will land we need to calculate the time:
[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]
Where:
Y(f) is the final position = 0
Y(0) is the initial position = 160 m
V(0y) is initial speed in "y" direction = 0
g is the gravity = 9.81 m/s²
t is the time=?
[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]
[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]
Now we can find the distance of the first package:
[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]
Then, after 2 seconds the distance traveled by plane is (from the initial position):
[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]
Now, the distance of the second package is:
[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]
The distance between the packages is:
[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]
Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.
I hope it helps you!
Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)
Answer:
III) [tex]d_{1}+ d_{2}=d_{t}[/tex]
Explanation:
I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5 with the the y being any other value.
II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)
III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]
(i) using coordinate grid notation to represent d₁, d₁ = 5y
(ii) using coordinate grid notation to represent d₂, d₂ = 5x + 5y
(ii) The sum of d₁ and d₂ is written as 5x + 10y
In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.
Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.
(i) Using coordinate grid to represent d₁ = (0, 5)
[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]
(ii) Using coordinate grid to represent d₂ = (5, 5)
[tex]d_2 = 5x + 5y[/tex]
(iii) The total vector is written as;
[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]
Learn more here: https://brainly.com/question/17212749
A motorboat is a lot heavier than a pebble. Why does the boat float?
Answer:
The boat has more buoyancy
Explanation:
A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?
Answer:
The displacement of the canoe is 1.46 m
Explanation:
Given that,
Mass of canoe = 61 kg
Mass of man = 86.5 kg
Length = 4 m
Let the the displacement of the canoe is x'
We need to calculate the displacement of the man
Using formula of displacement
[tex]x=x_{2}-x_{1}[/tex]
Put the value into the formula
[tex]x=4-(0.75+0.75)[/tex]
[tex]x=2.5\ m[/tex]
We need to calculate the displacement of the canoe
Using conservation of momentum
[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]
[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]
[tex]86.5\times2.5=(61+86.5)\times x'[/tex]
[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]
[tex]x'=1.46\ m[/tex]
Hence, The displacement of the canoe is 1.46 m
The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The original voltage is [tex]V_o[/tex]
The new voltage is [tex]V =\frac{V_o}{2}[/tex]
The capacitance is [tex]C = 150\ nF = 150 *10^{-9} \ F[/tex]
The first resistance is [tex]R_i = 26 \Omega[/tex]
The second resistance is [tex]R_E = 200 \Omega[/tex]
Generally the equivalent resistance is
[tex]R_e = R_1 + R_E[/tex]
=> [tex]R_e = 26 +200 [/tex]
=> [tex]R_e = 226 \ \Omega [/tex]
Generally the time constant is mathematically represented as
[tex]\tau = RC[/tex]
=> [tex]\tau = 226 * 150 *10^{-9}[/tex]
=> [tex]\tau = 3.39 *10^{-5} \ s [/tex]
Generally the voltage is mathematically represented as
[tex]V = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]\frac{V_o}{2} = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]0.5 = e^{-\frac{t}{\tau} }[/tex]
=> [tex]ln(0.5) = {-\frac{t}{ 3.39 *10^{-5} } }[/tex]
=> [tex]ln(0.5) * 3.39 *10^{-5} = -t [/tex]
=> [tex]t = 2.35*10^{-5} \ s [/tex]
m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.
Answer:
T₂ = 49.3°C
Explanation:
Applying law of conservation of energy to the system we get the following equation:
Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water
E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)
where,
E = Energy Supplied by Resistor = 100 KJ = 100000 J
m = mass of copper tank = 13 kg
C = Specific Heat of Copper = 385 J/kg.°C
T₂ = Final Temperature of Copper Tank
T₁ = Initial Temperature of Copper Tank = 27°C
T'₂ = Final Temperature of Water
T'₁ = Initial Temperature of Water = 50°C
m' = Mass of Water = 4 kg
C' = Specific Heat of Water = 4179.6 K/kg.°C
Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂
Therefore,
(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)
100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J
100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂
(1071055 J)/(21723.4 J/°C) = T₂
T₂ = 49.3°C
It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Answer:
102 m
Explanation:
Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Let the stopping distance be equal to S.
According to the definition of speed,
Speed = distance / time.
make time the subject of the formula
Time = distance / speed
then, the equivalent time is:
48.96 / 12 = S / 25
Cross multiply
12S = 48.96 x 25
12S = 1224
S = 1224 / 12
S = 102 m
Therefore, the stopping distance is 102 m
which statement is correct about the strength of forces?
-Electrostatic forces are exactly 10 times stronger than gravitational forces.
-Electrostatic forces are exactly 10 times weaker than gravitational forces.
-Electrostatic forces are trillions of times stronger than gravitational forces.
-Electrostatic forces are trillions of times weaker than gravitational forces.
Answer:
Thanks!!!!! adding this so it doesn’t get deleted.
Explanation:
1. Electrostatic forces are trillions of times stronger than gravitational forces. 2. normal force and friction 3. contact forces 4. The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate. 5. The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.
The correct statement over here is that electrostatic forces are trillions of times stronger than gravitational forces. Hence, option C is correct.
What is an Electrostatic Force?One of the basic forces in the cosmos is electrostatic force. In the universe, there are four basic forces. These include gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Under the umbrella of electromagnetic force is electrostatic force. Two charges placed apart are subject to the electrostatic force. The size of each charged and the separation between them determines how much electrostatic force there will be.
When two charges of the same type are brought together, whether positive or negative, they repel one another. It is known as the electrostatic force of repelling when it operates among two charges that are similar.
Therefore, the electrostatic forces are trillions of times stronger than the gravitational forces.
To know more about Electrostatic Force:
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cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?
Answer:
Displacement is 3 km North
Explanation:
Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answer:
A group of protons and neutrons that are surrounded by electrons I think that's the answer...
Explanation:
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground
Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Part D
Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three
possible combinations. In two of the combinations, the two ends are the same. In one combination, the
two ends are different. Describe the force you feel in each combination
Answer:
i. The magnetic force of repulsion.
ii. The magnetic force of attraction.
Explanation:
A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.
i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.
ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.
a car is moving eastward and speeding up. the momentum of the car is
A particle moves along a path described by y=Ax^3 and x = Bt, where tt is time. What are the units of A and B?
Answer:
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
Explanation:
From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:
[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)
[tex][l] = B\cdot [t][/tex] (Eq. 2)
Now we finally clear each constant:
[tex]A = \frac{[l]}{[l]^{3}}[/tex]
[tex]A = \frac{1}{[l]^{2}}[/tex]
[tex]B = \frac{[l]}{[t]}[/tex]
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
A 0.75 m3 rigid tank initially contains air whose density is 1.18 kg/m3. The tank is connected to a high-pressure supply link through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 4.95 kg/m3. Determine, in kg, the mass of air that has entered the tank..
Answer:
2.83kg
Explanation:
Answer:
2.83kg
Explanation:
Given
initial density = 1.18 kg/m3
Final density in the tank = 4.95 kg/m3.
Let us write the mass balance first.
Change in the mass of the system=mass of the air entering the system - Mass of air out the system
Mass that entered= M2 - M1
But DENSITY= MASS/ VOLUME
Mass= volume × Density
We can expressed the mass in terms of density since density is given in the question.
Mass that entered= (volume × density)2 - ( volume × density)1
= (V ρ)2 - (V ρ)1
But V1= V2 the volume remains the same
= ( ρ2 - ρ1)v
= (4.95 kg/m3 - 1.18 kg/m3) 0.75 m3
= 3.77× 0.75
= 2.8275kg
Mass that entered= 2.83kg
therefore, mass of air that has entered the tank= 2.83kg
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Answer:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?
Answer:
12.5 m
Explanation:
The first thing we would do is to calculate the wavelength. To do this, we use the formula
v = fλ, where
v = wave speed
f = frequency
λ = wavelength
If we make wavelength the formula, we have
wavelength = speed / frequency
Now, we substitute the values we had been given and we have
wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m
half of this said wavelength will be
= 3.50 / 2
= 1.75 m
As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,
(10 / 1.75) = 5.7
So the separation will have to be 7 half wavelengths
= (7 * 1.75) = 12.5 m
On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)
Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]
Therefore, the height of the cliff is 121.276 m
What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75
Answer:
1800 m/s
Explanation:
The equation is v = fλ
λ= 0.75
f = 2400 Hz
V = 2400 × 0.75
V = 1800 m/s
[ you did not give units for wavelength, I assumed it would be m/s]
A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?
Answer:
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) E=0
Explanation:
Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.
With this analysis we can answer the specific questions
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside
d) If it is very reasonable and this system configuration is called a Faraday Cage
e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:
29 m/s2
31 m/s2
33 m/s2
30 m/s2
32 m/s2
Explanation:
The question is incomplete. Here is the complete question.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of 30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...
Given
initial velocity of rocket u = 0m/s
final velocity of rocket = 30m/s
Height reached by the rocket = 98m
Required
upward acceleration of the rocket
Using the equation of motion below to get the acceleration a:
[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]
Hence upward acceleration of the rocket during the burn phase is closest to 5m/s²
Note that the velocity used in calculation was assumed.
A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the: __________A. capacitance
B. surface charge density on each plate
C. stored energy
D. electricfield between the two places
E. charge on each plate"
Answer: C.
Explanation:
For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.
Now, the stored energy can be written as:
E = (1/2)*Q^2/C
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.
If the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.
The given problem is based on the concept of parallel plat capacitor. For a parallel-plate capacitor where the distance between the plates is d.
The capacitance is:
C = e*A/d
here.
e is the permittivity of free space.
Since, the distance is inversely proportional then if we double the distance, the capacitance halves. Now, the stored energy can be given as,
E = (1/2)*Q^2/C
here,
Q is the charge stored in the capacitor.
Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:
E = (1/2)*Q^2*d/(e*A)
e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.
Then we can define:
K = (1/2)*Q^2/(e*A)
And now we can write the energy as:
E = K*d
So, the energy is proportional to the distance between the plates.
Thus, we can conclude that if the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.
Learn more about the energy stored in a capacitor here:
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