Answer:
5 miles per hour.
Explanation:
10miles / 2hours = 5 mph
please.
I mark u brainliest answer
Answer:
destroy
Explanation:
cuz if anyhow throw rubbish, it will affect/destroy the world environment
Answer:
bathing regularly 2pollute 3 throwing waste in land water water pollution
How does climate change?
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
A plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again. The apparent weight of the gas is the difference between these two masses. The gas is squeezed out of the bag to determine its volume by the displacement of water. What is the actual weight of the gas
The actual weight of the gas = apparent weight + weight.
The actual weight = [tex]W_{A}[/tex] + W
Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.
If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]
The gas is squeezed out of the bag to determine its volume by the displacement of water. Since
density = mass / volume
The density of water is 1000 kg/[tex]m^{2}[/tex]
we can get the mass of the gas by making m the subject of the formula.
W = mg
The actual weight of the gas = apparent weight + weight
That is,
The actual weight = [tex]W_{A}[/tex] + W
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Using the expression for the total energy of this system, it is possible to show that after the switch is closed, d2qdt2=−kq, where k is a constant. Find the value of the constant k.
The value of the constant K is [tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
According to Kirchhoff's loop rule, the total algebraic sum of potential differences in any loop, combining voltage provided by voltage sources as well as resistive components, must equal zero.
Thus, the relation for Kirchhoff's loop rule can be expressed as:
[tex]\mathbf{\dfrac{q}{c}- L\dfrac{dI}{dt} = 0}[/tex]
We all know that the current in the nonconstant charge flow can be written as:
[tex]\mathbf{I = \dfrac{dq}{dt}}[/tex]
∴
Replacing the current (I) into Kirchhoff's loop rule, we have:
[tex]\mathbf{ L\dfrac{d}{dt} ( \dfrac{dq}{dt})= -\dfrac{q}{c}}[/tex]
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -\dfrac{q}{Lc} \ \ ---(1)}[/tex]
From the given question, when the switch is closed
[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -kq\ \ ---(2)}[/tex]
Then, the charges on the capacitor start to b, resulting in the rise of the current in the circuit.
∴
By equating both equations (1) and (2);
[tex]\mathbf{K = \dfrac{1}{LC}}[/tex]
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the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?
Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
PLEASEEEEEEEE HELPPPPPPP
Answer: 62.25
Explanation: F = ma
F = 7.5 * 8.3
F = 62.25
Thus, the answer is 62.25 Newtons.
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This drawing of the field lines between these two charged plates is correct.
True or false
Answer:
True
Explanation:
Since positive and negative forces are both opposites of each other they attract. But if the forces are the same where both are positive or negative thye repel do to being the same charges in that case the picture would show lines moving in opposite directions.
2. Match the term to the definition
1. Reversibility
a. A break between activities during a workout.
2. Overload
3. Progression
4 Duration
5. Rest
b. Length of an individual workout.
c. The Improvement seen when you are training will be lost if you stop
training.
d. Moving forward through a sequence to gradually build up the body.
e Number of training sessions per week.
f. Level of difficulty of your workout.
9. Improvements will only be seen in the areas that you work on.
h The time between workouts that allows your body to heal and adapt to
the stress from your workout.
6. Recovery
7. Frequency
8. Specificity
1. Increase the amount of stress you put on your body.
9. Intensity
For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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7. A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m. If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm, then A) what is the maximum spring potential energy of the simple harmonic motion
At the equilibrium position of the spring, the mass has kinetic energy
1/2 (1.00 kg) (0.0400 m/s)² = 0.000800 J = 0.800 mJ = 800 µJ
and when the spring is maximally compressed, this kinetic energy is converted into potential energy.
The maximum spring potential energy of the simple harmonic motion is approximately 7.48 millijoules.
To find the maximum spring potential energy (Us) in simple harmonic motion, we need to determine the maximum displacement (A) of the mass from its equilibrium position.
In simple harmonic motion, the maximum displacement is equal to the amplitude of the oscillation.
The formula for maximum displacement (A) is given by:
A = v0 / ω
where v0 is the initial velocity (4.00 cm/s) and ω is the angular frequency of the oscillation.
The angular frequency (ω) is given by:
ω =[tex]\sqrt{(k / m)[/tex]
where k is the spring constant (9.00 N/m) and m is the mass (1.00 kg).
Substitute the values:
ω = [tex]\sqrt{(9.00 N/m / 1.00 kg)} = \sqrt{9.00 rad/s[/tex] ≈ 3.00 rad/s
Now, calculate the maximum displacement:
A = 4.00 cm/s / 3.00 rad/s ≈ 1.33 cm
Next, calculate the maximum spring potential energy:
Us =[tex](1/2) * k * A^2[/tex]
Us = (1/2) * 9.00 N/m * [tex](1.33 cm)^2[/tex]≈ 7.48 mJ
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--The compleete Question is, A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m. If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm, then what is the maximum spring potential energy of the simple harmonic motion--
The passage of a wave through a medium is...?
Answer:
Transmission
Explanation:
It’s the passage of electromagnetic waves through a medium.
The passage of a wave through a medium is known as transmission.
What is the transmission of a wave?The transmission of waves may be characterized as a wave traveling through a medium or into a new medium. Not all waves transmit through the same materials. It significantly depends on the type of wave, the type of material, and even the frequency of the wave.
Transmission is generally the passage of electromagnetic radiation through a medium. While Reflection is the process by which electromagnetic radiation is returned either at the boundary between two media (surface reflection) or at the interior of a medium (volume reflection).
Waves can go through things. Waves go through water, light can pass through the glass, and sound can pass through walls. The material the wave passes through is called a medium. A single disturbance passing through a medium is called a pulse.
Therefore, the passage of a wave through a medium is known as transmission.
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Calculate the altitude above the surface of Earth, in meters, at which the acceleration due to gravity is g
Answer:
By definition the acceleration due to gravity at the surface is g:
The altitude above the surface is zero for an acceleration of g.
What’s 1 + 1 many half a window
Answer:
1
+
1
is
2
Explanation:
because if you add one to a one it would end up being 2 <3
Answer: 2
Explanation: 1+1=2
This was THE hardest question I've ever decided to answer
TRUE OR FALSE! PLZ HELP
Magnitude is a measure of how strong a force is.
A. True
B. False
Answer:
True
Explanation:
Magnitude is the "value" the greater the value the greater the force is and vice versa
Answer:
TRUE
Explanation: I did the test
Hope this helps
II) A 0.40-kg ball, attached to the end of a horizontal ord, is rotated in a circle of radius 1.3 m on a friction- less horizontal surface. If the cord will break when the tension in it exceeds 60 N, what i~ the maximum speed the ball can have
Hi there!
In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.
The net force is equivalent to the centripetal force, so:
∑F = T
mv²/r = T
Solve for v:
v = √rT/m
v = 13.96 m/s
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
Explanation:
If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k
2. If a vehicle with four wheels is traveling in a linear path through soft dirt, how many and which tires would you expect to leave tire track impressions
Given our understanding of the situation, for this vehicle with four wheels is traveling in a linear path through soft dirt, we would expect for only two wheels to leave tire track impressions.
The vehicle has four wheels, and one might think initially that it makes sense for all four wheels to create tire track impressions on the dirt. This assessment is correct, all four tires will create impressions, However, not all four tires will leave a said impression.
This is due to the fact that as the vehicle advances, the tires in the back will erase the impressions left by the front tires. Therefore, only two tire track impressions will be left.
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according to Newton's __ law, an object with no net force acting on it remains at rest or in motion with a constant veloctiy
Answer:
i think its his law of inertia
Explanation:
this law is about motion
A 260-kg object and a 560-kg object are separated by 3.10 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 49.0-kg object placed midway between them.
C is the correct answer:))))))))
A single paragraph can be selected by triple in the paragraph. ture or false
Answer:
The answer for the question should be True
A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........
If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
A car driving on level ground at 20.0 m/s slams on its brakes and skids to a halt. If the coefficient of kinetic friction between the car’s tires and the road is 0.750, how far does the car skid before stopping? How far would the car have skidded if it had been moving at 40.0 m/s?
The distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
The linear force acting on the car is opposed by the frictional force. Hence;
[tex]F=F_f[/tex]
[tex]ma = -\mu R\\ma =-\mu mg[/tex]
m is the mass of the car
a is the acceleration
[tex]\mu[/tex] is the coefficient of friction
R is the normal force
Given the following parameters
[tex]a=-\mu g[/tex]
[tex]a =-0.75(9.8)\\a=-7.35m/s^2[/tex]
Get the distance the car has skidded if it had been moving at 40.0 m/s
[tex]v^2=u^2+2as[/tex]
[tex]0^2=20^2+2(-7.35)s\\0=400-14.7s\\s=\frac{400}{14.7}\\s= 27.2m[/tex]
Hence the distance the car has skidded if it had been moving at 40.0 m/s is 27.2m
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help meee please please please
To become a healthy individual, a person must have a properly executed routine to achieve his/her fitness goals. On the contrary, in what instance does a person become unhealthy
Two ends of a steel wire of length 8m and 2mm radius are fixed to two rigid supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steels = 12x10^_6 per kelvin and Young's modules for steel =2x10^11 n/m^2
The increase in tension on the steel wire is 8,484.75 N.
The given parameters;
original length of the wire, l = 8 mradius of the wire, r = 2 mmThe area of the steel wire is calculated as follows;
[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]
The extension of the steel wire is calculated as follows;
[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]
The increase in tension on the steel wire is calculated as follows;
[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]
Thus, the increase in tension on the steel wire is 8,484.75 N.
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builder places a 3kg hammer on the top of a ladder, which is 4m above the ground. Calculate the gravitational potential energy of the hammer while on the ladder.
Answer:
[tex]E=mgh[/tex]
[tex]m=3kg[/tex]
[tex]h=4m[/tex]
[tex]g=9.8m/s^{2}[/tex]
[tex]E= 3*4*9.8=117.6J[/tex]
Explanation:
Only substitute amounts to formula.
Hope this helps ;)
Cheers :D
a. Block on a smooth incline.
A block of mass m= 3.8 kg on a smooth inclined plane of angle 36° is connected by a cord over a small frictionless
pulley to a second block of mass m2 = 3 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2
for g.
What is the acceleration of each block to 1 decimal place?
Answer:
Explanation:
F = ma
3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a
a = 1.10566...
a = 1.1 m/s²
the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.
Question 2 (4 points)
Listen
A 14,000kg freight train, coasting at 8.1m/s, strikes another car, and the two move
forward together at 3.6m/s.
What is the mass of the second car?
Paragraph
B
1
U
A
+
Answer:
Explanation:
I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was
Conservation of momentum
14000(8.1) + m(0) = (14000 + m)(3.6)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
The mass of the second car is 17500 kg.
What is law of conservation of momentum?The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.
Newton's third law of motion has a direct impact on the idea of momentum conservation.
mass of the first car = 14000 kg.
Initial speed of first car = 8.1 m/s.
Final speed of the two cars = 3.6 m/s.
From law of conservation of momentum; we can write:
Total initial momentum = total final momentum
14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
Hence, the mass of the second car is 17500 kg.
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