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
ionic bonds form when electrons?
Answer:
when the electron transferred permanently to another atom
If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer
Answer:
The charge pass through your hair dryer is 3000 C.
Explanation:
Given that,
Power = 1200 W
Voltage = 120 V
Flow time = 5 min
We need to calculate the current
Using formula of power
[tex]P=VI[/tex]
[tex]I=\dfrac{P}{V}[/tex]
Put the value into the formula
[tex]I=\dfrac{1200}{120}[/tex]
[tex]I=10\ A[/tex]
We need to calculate the charge pass through your hair dryer
Using formula of current
[tex]I=\dfrac{Q}{t}[/tex]
[tex]Q=It[/tex]
Put the value into the formula
[tex]Q=10\times5\times60[/tex]
[tex]Q=3000\ C[/tex]
Hence, The charge pass through your hair dryer is 3000 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
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!
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]
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
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 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.
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]
what is the force produced on a body of 30kg mass when a body moving with the velocity of 26km/hr is acceleted to gain the velocity of 54 km/hr in 4 sec
Answer:
F = 58.35 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.
We have to convert speeds from kilometers per hour to meters per second
[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]
[tex]v_{f}=v_{o}+(a*t) \\[/tex]
where:
Vf = final velocity = 15 [m/s]
Vi = initial velocity = 7.22 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Note: the positive sign of the above equation is because the car increases its speed
15 = 7.22 + (a*4)
a = 1.945 [m/s^2]
Now we can use the Newton's second law:
F = m*a
F = 30*1.945
F = 58.35 [N]
help me get the answer in Physical Science.
Answer:
lithium
Explanation:
I took physical science 2 years ago and passed with an A
If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.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:
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?
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.
An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?
Answer:
(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]
Explanation:
Given that,
The frequency of FM radio station, f = 93.4 MHz
(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]
(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]
Hence, this is the required solution.
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
7. A 1,500-N force is applied to a 1,000-kg car. What is the car's acceleration?
Answer:
1.5m/s^2
Explanation:
Answer:
1.5 m/s2. accerelation =force ÷mass
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]
the diagram shows a contour map. letter a through k are reference points on the map. which points are located at the same elevation above sea level?
Answer:
K and I
Explanation:
Contour maps use lines that represent spaces in a map that have the same elevation, this means that all the lines should be continuous and closed, in this case, we are not able to see the full extent of most of the lines, but since the points are located in different lines we can assume that they are at different heights, so since only point K and point I are on the same line, we know that these two points are at the same height.
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.
the peripheral nervous system is responsible for both sending and receiving signals to and from the brain
Answer:
its true trust me
Explanation:
Answer: true
Explanation: edge
A motorboat is a lot heavier than a pebble. Why does the boat float?
Answer:
The boat has more buoyancy
Explanation:
Jumping on a trampoline cause you to fly up in the air. What type of newton’s law is it ?
Answer:
The Third law
Explanation:
For every action there is an equal and opposite reaction.
Answer:
First Law
Explanation:
An object at rest (not moving) will stay at rest unless an unbalanced force acts on it.
An object in motion will stay in motion (in a straight line and at a constant speed) unless an unbalanced force acts on it.
You jump down on a trampoline and fly up in the air as a result.
21. Prediction: If you were to measure the current at points A, B and C, how do you think the values would compare? Why? 22. Prediction: If you were to measure the potential differences across these bulbs (what the voltmeter measures) how do you think the values will compare to each other and to the potential difference across the battery pack or the power supply? Why?
Answer:
hello your question is incomplete attached below is the complete question
21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Explanation:
The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C
The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)
hence the voltage in the battery will be equal to the voltage across each bulb
Weight of a body becomes greater at the pole than that at the equator . why ?
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:
https://brainly.com/question/3611251
During a thunderstorm the electric field at a certain point in the earth's atmosphere is 1.07 105 N/C, directed upward. Find the acceleration of a small piece of ice of mass 1.08 10-4 g, carrying a charge of 1.05 10-11 C.
Answer:
The acceleration of a small piece of ice is 10.40 m/s².
Explanation:
The electric force is given by:
[tex]F = Eq[/tex]
Where:
E is the electric field = 1.07x10⁵ N/C
q is the charge = 1.05x10⁻¹¹ C
The electric force is equal to Newton's second law:
[tex] Eq = ma [/tex]
Where:
m is the mass = 1.08x10⁻⁴ g = 1.08x10⁻⁷ kg
a is the acceleration
Hence, the acceleration is:
[tex] a = \frac{Eq}{m} = \frac{1.07 \cdot 10^{5} N/C*1.05 \cdot 10^{-11} C}{1.08 \cdot 10^{-7} kg} = 10.40 m/s^{2} [/tex]
Therefore, the acceleration of a small piece of ice is 10.40 m/s².
I hope it helps you!
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.