Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
Arrange the steps in order to describe what happens to a gas when it cools.
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
why can you see the path of light in a sunbeam?
Answer:
Sunbeams are seen because of light separated from water droplets and dust and smoke particles suspended in the air. If the cloud cover only has a few small holes in it, then separate rays of light will sprinkle light in every direction so you can see sunbeams.
Problem 2: A flat belt is used with a driving pulley (diameter 4 inches) and a driven pulley (diameter 18 inches) in a open configuration. The center distance between the pulleys is 48 inches. The friction coefficient between the belt and pulley is 0.6. Determine the following: a) If the belt is initially tensioned to 50 lbs, what is the force in the belt on the tight side just before slippage (neglect the centrifugal force of the belt). b) Find the maximum torque required at the driving pulley.
Answer:
Explanation:
The two pictures attached shows the solution to the problem
Two wires, both with current out of the page, are next to one another. The wire on the left has a current of 1 A and the wire on the right has a current of 2 A. We can say that:
A. The left wire attracts the right wire and exerts twice the force as the right wire does.
B. The left wire attracts the right wire and exerts half the force as the right wire does.
C. The left wire attracts the right wire and exerts as much force as the right wire does.
D. The left wire repels the right wire and exerts twice the force as the right wire does.
E. The left wire repels the right wire and exerts half the force as the right wire does.
F. The left wire repels the right wire and exerts as much force as the right wire does.
Answer:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
Explanation:
To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:
[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]
Next, you use the formula for the magnetic force produced by the wires:
[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]
if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:
[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]
Hence, due to this result you have that:
C. The left wire attracts the right wire and exerts as much force as the right wire does.
A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30 m/s. if the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10 m/s, how fast would the stone be traveling just before it hits the ground
Answer:
The velocity just before hitting the ground is [tex]v_f = 30 m/s[/tex]
Explanation:
From the question we are told that
The initial speed is [tex]u = 10 m/s[/tex]
The final speed is [tex]v = 30 \ m/s[/tex]
From the equations of motion we have that
[tex]v^2 =u^2 + 2as[/tex]
Where s is the distance travelled which is the height of the cliff
So making it the subject of the the formula we have that
[tex]s = \frac{v^2 - u^2 }{2a}[/tex]
Where a is the acceleration due to gravity with a value [tex]a = 9.8m/s^2[/tex]
So
[tex]s = \frac{30^2 - 10^2 }{2 * 9.8 }[/tex]
[tex]s = 40.8 \ m[/tex]
Now we are told that was through horizontally with a speed of
[tex]v_x =10 m/s[/tex]
Which implies that this would be its velocity horizontally through out the motion
Now it final velocity vertically can be mathematically evaluated as
[tex]v_y = \sqrt{2as}[/tex]
Substituting values
[tex]v_y = \sqrt{(2 * 9.8 * 40.8)}[/tex]
[tex]v_y = 28.3 \ m/s[/tex]
The resultant final velocity is mathematically evaluated as
[tex]v_f = \sqrt{v_x^2 + v_y^2}[/tex]
Substituting values
[tex]v_f = \sqrt{10^2 + 28.3^2}[/tex]
[tex]v_f = 30 m/s[/tex]
Match these items.
1 . pls help
asteroids
between Mars and Jupiter
2 .
fission
ice, dust, frozen gases
3 .
energy
sun's atmosphere
4 .
fusion
ability to do work
5 .
corona
splitting atoms
6 .
comets
the combining of atomic nuclei to form one nucleus
Answer:
Here's your answer :
Asteroids - Between mars and JupiterFission - splitting atomsEnergy - Sun's atmosphereFusion - The combining of atomic nuclei to form one nucleusCorona - Ability to do workComets - Ice, dust, frozen gaseshope it helps!
What must x be so that the handle end of the bat remains at rest as the bat begins to move? (Hint: Consider the motion of the center of mass and the rotation about the center of mass. Find x so that these two motions combine to give v=0 for the end of the bat just after the collision. Also, note that integration of equation ∑τ⃗ =dL⃗ dt gives ΔL=∫t1t2(∑τ)dt. )
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]
The bat length is [tex]L_b = 0.900 \ m[/tex]
The distance of the bat's center of mass to the handle end is [tex]z_c = 0.600 \ m[/tex]
The moment of inertia of the bat is [tex]I = 0.0530 \ kg \cdot m^2[/tex]
The objective of the solution is to find x which is the distance from the handle of the bat to the point where the baseball hit the bat
Generally the velocity change at the end of the bat is mathematically represented as
[tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]
Where [tex]\Delta v_c[/tex] is the velocity change at the center of the bat which is mathematically represented as
[tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]
We are told that the impulse is J so
[tex]\Delta v_c = \frac{J}{m_b }[/tex]
And [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as
[tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]
Now we have that
[tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero and the impulse will be 1
So
[tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]
=> [tex]x = \frac{I}{m_b z_c} + m_b[/tex]
substituting values
[tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]
[tex]x = 0.710 \ m[/tex]
A person drives north 6 blocks, then turns west, and drives 6 blocks. The driver then turns south and drives 6 blocks. How could the driver have made the distance shorter while maintaining the same displacement?
Answer:
Considering that there is no obstructions, he could go west from the start.
Explanation:
Three packing crates of masses, M1 = 6 kg, M2 = 2 kg
and M3 = 8 kg are connected by a light string of
negligible mass that passes over the pulley as shown.
Masses M1 and M3 lies on a 30o
incline plane which
slides down the plane. The coefficient of kinetic friction
on the incline plane is 0.28.
Determine the acceleration of the system.
Answer:
a = 2.5 m / s²
Explanation:
This is an exercise of Newton's second law, in this case we fix a coordinate system with the x axis parallel to the plane with positive direction
Let's write the second law for bodies in the inclined plane
W₁ₓ + W₃ₓ - fr = (m₁ + m₃) a
N₁ - [tex]W_{1y}[/tex] + N₃- W_{3y} = 0
N₁ + N₃ = W_{1y} + W_{3y}
let's use trigonometry to find the weight components
sin 30 = Wₓ/ W
Wₓ = W sin 30
cos 30 = W_{y} / W
W_{y} = W cos 30
we substitute
N₁+ N₃ = W₁ cos 30 + W₃ cos 30
W₁ₓ + W₃ₓ - μ (m₁ + m₃) g cos30 = (m₁ + m₃) a
a = (m₁g sin 30 + m₃g sin 30 - μ (m₁ + m₃) g cos 30) / (m₁ + m₃)
a = g sin 30 - μ g cos30
let's calculate
a = 9.8 sin 30 - 0.28 9.8 cos 30
a = 4.9 - 2,376
a = 2.5 m / s²
A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2300 kg SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.75, and the cars slide to a halt at a point 5.54 m west and 6.19 m south of the impact point. How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?
Answer:
Explanation:
momentum of sedan of 1600 kg = 1600x v , where v is its velocity
momentum of suv of 2300 kg = 2300 x u where u is its velocity .
force of friction = ( 1600 + 2300 ) x 9.8 x .75 ( fiction = μ mg )
= 28665 N
distance by which friction acted = √ (5.54² + 6.19²)
= 8.3 m
work done by friction
= 28665 x 8.3
= 237919.5 J
Total kinetic energy of cars = work done by friction
1/2 x 1600 x v² + 1/2 x 2300 u² = 237919.5
16 v² + 23 u² = 4758.4
1600 x v / 2300 u = 6.19 / 5.54
v / u = 1.6
v = 1.6 u
putting this equation in fist equation
40.96 u² + 23 u² = 4758.4
= 63.96 u² = 4758.4
u² = 74.4
u = 8.62 m /s
v = 13.8 m /s
1. A block of metal of mass 2kg is resting on
a frictionless floor. It is struck by a jet
releasing water at the rate of 1kg/sec at a
speed of 5ms-1. What will be the initial
acceleration of the block?
Answer:
The acceleration is [tex]a = 2.5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the metal block is [tex]m_b = 2 \ kg[/tex]
The mass flow rate of the water is [tex]\r m = 1\ kg/s[/tex]
The speed of the water of the water release is [tex]v_w = 5 m/s[/tex]
Generally according to the law of conservation of linear momentum
[tex]p_i = p_f[/tex]
Now [tex]p_i[/tex] is the initial momentum of the system which mathematically represented as
[tex]p_i = m_w * v_w + m_b * v_b[/tex]
Now [tex]m_w[/tex] is the mass of water at the point of contact with the block which can be deduced as [tex]m_w = 1 \ kg[/tex]
Now since at initial the block is at rest
[tex]v_b = 0 \ m/s[/tex]
So
[tex]p_i = 1 * 5[/tex]
[tex]p_i = 5 \ kgm/ s[/tex]
And [tex]p_f[/tex] is the final momentum of the system which mathematically represented as
[tex]p_f = m_w * v__{fw} } + m_b * v__{fb}}[/tex]
So [tex]v__{fw} }[/tex] is the final velocity of water which is zero due to the fact that when the water hits the block it losses its momentum and eventually the velocity becomes zero
So
[tex]5 = 2 * v__{fb }[/tex]
Thus [tex]v__{fb }} = \frac{5}{2}[/tex]
[tex]v__{fb }} = 2.5 \ m/s[/tex]
Thus
[tex]p_f = 2.5 * 2[/tex]
[tex]p_f = 5 \ kgm /s[/tex]
Now the average momentum change is
[tex]p_a = \frac{p_i +p_f}{2}[/tex]
[tex]p_a = \frac{5+5}{2}[/tex]
[tex]p_a =5 kgm/s[/tex]
Now the force acting on the block is
[tex]F = \frac{p_a }{t}[/tex]
and from the question the initial movement of the block took 1s as it is a mass of water moving at a rate of 1kg/s that caused the first movement of the block
So
[tex]F= \frac{5}{1}[/tex]
[tex]F= 5 \ N[/tex]
Now the acceleration is
[tex]a = \frac{F}{m_b}[/tex]
=> [tex]a = \frac{5}{2}[/tex]
[tex]a = 2.5 \ m/s^2[/tex]
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is what is the kinetic energy of the block?
Complete Question
A 0.025-kg block on a horizontal frictionless surface is attached to an ideal massless spring whose spring constant is 150 N/m. The block is pulled from its equilibrium position at x = 0.00 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the horizontal x-axis. When the displacement is x = 0.024 m, what is the kinetic energy of the block?
Answer:
The kinetic energy is [tex]KE = 0.4368\ J[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m= 0.025\ kg[/tex]
The spring constant is [tex]k = 150 N/m[/tex]
The length of first displacement is [tex]x_1 = 0.80 \ m[/tex]
The length of first displacement is [tex]x_2 = 0.024 \ m[/tex]
At the [tex]x_2[/tex] the kinetic energy is mathematically evaluated as
[tex]KE = \Delta E[/tex]
Where [tex]\Delta E[/tex] is the change in energy stored on the spring which is mathematically represented as
[tex]\Delta E = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
=> [tex]KE = \frac{1}{2} k (x_1 ^2 - x_2^2)[/tex]
Substituting value
[tex]KE = \frac{1}{2} * 150 * (0.08^2 - 0.024^2)[/tex]
[tex]KE = 0.4368\ J[/tex]
By which process does the heat from the Sun reach the Earth? (AKS 4b DOK 1) *
A bowler releases a 7.0kg bowling ball from rest to a final velocity of 8.0m/s. What is the magnitude of the change in momentum of the bowling ball?
Answer:
56 kg m/s
Explanation:
Δp = mΔv
Δp = (7.0 kg) (8.0 m/s − 0 m/s)
Δp = 56 kg m/s
Is mercury (the planet) rocky or gaseous(meaning relating to or having the characteristics of a gas.)
Answer:
Mercury is rocky
Explanation:
Answer:
Rocky
Explanation:
It has no atmosphere so it cannot hold gas.
10) Two students want to use a 12-meter long rope to create standing waves. They first measure the speed at which a single wave pulse moves from one end of the rope to another and find that it is 36 m/s. What frequency must they vibrate the rope at to create the second harmonic
Answer:
To create a second harmonic the rope must vibrate at the frequency of 3 Hz
Explanation:
First we find the fundamental frequency of the rope. The fundamental frequency is the frequency of the rope when it vibrates in only 1 loop. Therefore,
f₁ = v/2L
where,
v = speed of wave = 36 m/s
L = Length of rope = 12 m
f₁ = fundamental frequency
Therefore,
f₁ = (36 m/s)/2(12 m)
f₁ = 1.5 Hz
Now the frequency of nth harmonic is given in general, as:
fn = nf₁
where,
fn = frequency of nth harmonic
n = No. of Harmonic = 2
f₁ = fundamental frequency = 1.5 Hz
Therefore,
f₂ = (2)(1.5 Hz)
f₂ = 3 Hz
An electric dipole consists of a positive and a negative charge of equal magnitude. Consider an electric dipole with each charge having a magnitude of 1 × 10−6 C. The negative charge is located at (3 cm, 0) and the positive charge is located at (−3 cm, 0). Calculate the electric field from each charge at the points A through E, described below. Use symmetry as much as possible! Using the scale 1 cm = 105 N/C, draw the vector to represent the magnitude and direction of the electric field from each charge. (When entering angle values, enter a number greater than or equal to 0° and less than 360° measured counterclockwise from the +x-axis.) • A = (−13 cm, 0) • B = (−3 cm, 10 cm) • C = (0, 10 cm ) • D = (3 cm, 10 cm) • E = (13 cm, 0) For the negative charge:
Answer:
Explanation:
To find the electric field you use the equation for an electrostatic electric field:
[tex]E=k\frac{q_1q_2}{r^2}[/tex]
r: distance in which E is calculated, from each charge
In the of a dipole you have two contributions to E:
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
where E1 is the electric field generated by the first charge and E2 by the second one.
A. (-13 cm, 0):
First you calculate the vectors E1 and E2:
[tex]E_1=(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13-0.03)^2}\hat{j}\\\\E_1=350781.25N/C\\\\E_2=-(8.98*10^9)\frac{(1*10^{-6}C)}{(-0.13+0.03)^2}\hat{j}\\\\E_2=-989000N/C[/tex]
Then, you sum both contributions:
[tex]\vec{E}=-547218.75N/C\hat{j}[/tex]
B. (-3cm, 10cm):
[tex]r_1=\sqrt{(0.06)^2+(0.1)^2}=0.116m\\\\\theta=tan^{-1}(\frac{0.06}{0.1})=30.96\°\\\\r_2=0.1m\\\\E_1=(8.98*10^9Nm^2/C)\frac{(1.6*10^{-6}C)}{(0.116m)^2}[cos(30.96\°)\hat{i}+sin(30.96\°)\hat{j}]\\\\E_1=[-915646\hat{i}-549306.42\hat{j}]N/C\\\\\theta=(90-30.96)+180=239.04\°\\\\[/tex]
the last angle is calculated again because the vector direction is measured from the +x axis.
and for the second vector:
[tex]E_2=(8.98Nm^2/C)\frac{1.6*10^{-6}C}{(0.1m)^2}\hat{j}\\\\E_2=1436800N/C\hat{j}[/tex]
the total E is:
[tex]\vec{E}=[-915646\hat{i}+887493.58\hat{j}]N/C[/tex]
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram-meter^3 and the density of silicon in other units: 2.33 gram-centimeter^3. You decide to convert the density of silicon into units of kilogram-meter^3 to perform the comparison.
By which combination of conversion factors will you multiply 2.33 gram-centimeter^3 to perform the unit conversion?
Answer:
Explanation:
To convert gram / centimeter³ to kg / m³
gram / centimeter³
= 10⁻³ kg / centimeter³
= 10⁻³ / (10⁻²)³ kg / m³
= 10⁻³ / 10⁻⁶ kg / m³
= 10⁻³⁺⁶ kg / m³
= 10³ kg / m³
So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³
2.33 gram / cm³
= 2.33 x 10³ kg / m³ .
While watching a movie a spaceship explodes and there is a loud bang and flash of light. What is wrong with this scene? Explain how you know using evidence and scientific reasoning from the lesson.
Pendulum clock. Your friend is trying to construct a clock for a craft show and asks you for some advice. She has decided to construct the clock with a pendulum. The pendulum will be a very thin, very light wooden bar with a thin, but heavy, brass ring fastened to one end. The length of the rod is 80 cm and the diameter of the ring is 10 cm. She is planning to drill a hole in the bar to place the axis of rotation 15 cm from one end. She wants you to tell her the period of this pendulum.
Answer:
The time period for this pendulum is 1.68 seconds
Explanation:
Solution
Given that:
The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum
Now,
In this case, the length becomes:
L= 80 - 15+5
L = 70 cm
The time period = T = 2π √L/g
T = 2* 3.14 *√0.7/9.8
= 1.68 seconds
Note: Kindly find an attached work to the part of the solution of the given question
A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat
Answer:
0.99m
Explanation:
Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:
[tex]v=\lambda f=\lambda\frac{1}{T}=(110m)\frac{1}{8.3s}=13.25\frac{m}{s}[/tex]
the relative velocity is:
[tex]v'=13.25m/s-5m/s=8.25\frac{m}{s}[/tex]
This velocity is used to know which is the distance traveled by the boat after 20 seconds:
[tex]x'=v't=(8.25m/s)(20s)=165m[/tex]
Next, you use the general for of a wave:
[tex]f(x,t)=Acos(kx-\omega t)=Acos(\frac{2\pi}{\lambda}x-\omega t)[/tex]
you take the amplitude as 2.0/2 = 1.0m.
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{8.3s}=0.75\frac{rad}{s}[/tex]
by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:
[tex]f(165,20)=1.0m\ cos(\frac{2\pi}{110m}(165)-(0.75\frac{rad}{s})(20s))\\\\f(165,20)=0.99m[/tex]
What happens to the brightness of the lightbulb when its resistance is increased? *
Answer:
It will be dimmer than before
Answer:For parallel connection,the brightness would be dimmer, while for series connection it would be brighter
Explanation:
For parallel connection,resistance and brightness are inversely proportional.meaning as resistance increases, brightness decreases.
For series connection,resistance and brightness are directly proportional. Meaning as the resistance increases, brightness also increases.
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/s2. How long after car A begins to accelerate does it take car A to overtake car B?
Answer:
The taken is [tex]t_A = 19.0 \ s[/tex]
Explanation:
Frm the question we are told that
The speed of car A is [tex]v_A = 22 \ m/s[/tex]
The speed of car B is [tex]v_B = 29.0 \ m/s[/tex]
The distance of car B from A is [tex]d = 300 \ m[/tex]
The acceleration of car A is [tex]a_A = 2.40 \ m/s^2[/tex]
For A to overtake B
The distance traveled by car B = The distance traveled by car A - 300m
Now the this distance traveled by car B before it is overtaken by A is
[tex]d = v_B * t_A[/tex]
Where [tex]t_B[/tex] is the time taken by car B
Now this can also be represented as using equation of motion as
[tex]d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300[/tex]
Now substituting values
[tex]d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
Equating the both d
[tex]v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
substituting values
[tex]29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
[tex]7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300[/tex]
[tex]7 t_A =1.2 t_A^2 - 300[/tex]
[tex]1.2 t_A^2 - 7 t_A - 300 = 0[/tex]
Solving this using quadratic formula we have that
[tex]t_A = 19.0 \ s[/tex]
The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only a small amount of fuel. Suppose the probe, which has a mass of 474 kg is travelling at an initial speed of 275 m/s. No forces act on it except the 5.60 x 10⁻² N thrust from the engine. This external force is directed PARALLEL to the displacement. The displacement has a magnitude of 2.42 x 10⁹ m. {PART A} Calculate the INITIAL kinetic energy of the probe [2 marks] {PART B} Find the work done BY THE ENGINE on the space probe [2 marks] {PART C} Calculate the FINAL KINETIC ENERGY of the probe (Hint W=∆E) [2 marks] {PART D} Determine the final speed of the probe, assuming that its mass remains constant [3 marks]
Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s
In the figure calculates the acceleration of the block friction not today
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
The figure shows a crane whose weight is 12.5 kN and center of gravity in G. (a) If the crane needs to suspend the 2.5kN drum, determine the reactions on the wheel in A and B when the boom is in the position shown.(b) Considering the same situation illustrated, what is the value of the maximum weight that the crane can suspend without tipping over?
Answer:
(a) Ra = 9.25 kN; Rb = 5.75 kN
(b) 26.7 kN
Explanation:
(a) Draw a free-body diagram of the crane. There are four forces:
Reaction Ra pushing up at A,
Reaction Rb pushing up at B,
Weight force 12.5 kN pulling down at G,
and weight force 2.5 kN pulling down at F.
Sum of moments about B in the counterclockwise direction:
∑τ = Iα
-Ra (0.66 m + 0.42 m + 2.52 m) + 12.5 kN (2.52 m + 0.42 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
-Ra (3.6 m) + 12.5 kN (2.94 m) − 2.5 kN (1.38 m) = 0
Ra = 9.25 kN
Sum of moments about A in the counterclockwise direction:
∑τ = Iα
Rb (0.66 m + 0.42 m + 2.52 m) − 12.5 kN (0.66 m) − 2.5 kN ((3.6 m + 0.9 m) cos 30° + 0.66 m + 0.42 m) = 0
Rb (3.6 m) − 12.5 kN (0.66 m) − 2.5 kN (4.98 m) = 0
Rb = 5.75 kN
Alternatively, you can use sum of the forces in the y direction as your second equation.
∑F = ma
Ra + Rb − 12.5 kN − 2.5 kN = 0
Ra + Rb = 15 kN
9.25 kN + Rb = 15 kN
Rb = 5.75 kN
However, you must be careful. If you make a mistake in the first equation, it will carry over to this equation.
(b) At the maximum weight, Ra = 0.
Sum of the moments about B in the counterclockwise direction:
∑τ = Iα
12.5 kN (2.52 m + 0.42 m) − F ((3.6 m + 0.9 m) cos 30° − 2.52 m) = 0
12.5 kN (2.94 m) − F (1.38 m) = 0
F = 26.7 kN
Potassium is a crucial element for the healthy operation of the human
body. Potassium occurs naturally in our environment (and thus our
bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-
41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical
human body contains about 3.0 grams of Potassium per kilogram of body
mass.
1. How much Potassium-40 is present in a person with a mass of 80
kg?
2. If, on average, the decay of Potassium-40 results in 1.10 MeV of
energy absorbed, determine the effective dose (in Sieverts) per year
due to Potassium-40 in an 80-kg body. Assume an RBE of 1.2. The
half-life of Potassium-40 is 1.28 x 10° years.
Which of the following statements are characteristics of magnetic fields? Select all that apply.
Magnetic fields point from the north pole to the south pole of a magnet.
The earth's magnetic field has no effect on the electron rays coming from the sun.
An example of the Biot-Savart law is the effect of the earth's maghytic field on the electron rays coming from the sun.
The north pole of a magnet will be attracted to the south pole of the earth.
If a bar magnet is cut in half two magnets with like poles will be created.
Answer:
Magnetic fields point from the north pole to the south pole of a magnet.
An example of the Biot-Savart law is the effect of the earth's maghytic field on the electron rays coming from the sun.
The north pole of a magnet will be attracted to the south pole of the earth.
If a bar magnet is cut in half two magnets with like poles will be created
Explanation:
The magnetic field of Earth is due to the presence of iron in the core of the Earth.
The metal emits the magnetic waves from it and the North and South pole of the planet.
Both the poles emit the magnetic rays which create magnetic sheet around it. The Earth acts like a magnet bar if which is cut into two half, the planet will act like two magnets. Also, Biot Savarts's law states that the magnetic field does not affect the electron rays coming from the Sun.
Thus, the selected options are correct.
Answer:
ACDE
Explanation:
Which theory states that deviance results not only from what people do, but also from how others respond to those actions?
A.
strain theory
B.
labeling theory
C.
differential association theory
D.
control theory
E.
conflict theory
The theory states that deviance results not only from what people do but also from how others respond to those actions are labeling theory. Hence, the option B is correct.
What is labeling theory?This theory states that deviance and conformity result not so much from what people do but also from how others respond. It is called labeling theory. Eg: Skipping school, and underage drinking.
It also suggests that any deviance results in how society responds to certain behaviors. It defines the behavior of human beings influenced by other members of society.
It also notes that a person is made to act in a negative way by the manner in which society identifies him. If a person is identified as a criminal then he involves in the criminal activities.
Conflict theory refers to the theory linking deviance to social inequality. Anticipating the consequences of a person's behavior is control theory.
Hence, the correct option is B) labeling theory.
To learn more about the labeling theory:
https://brainly.com/question/31366596
#SPJ3
What spectacles are required for reading purposes by a person whose near point is 2.0m
Answer:Convex lens spectacles is required for reading purpose..
Explanation:
I don't say you have to mark my ans as brainliest but if it has really helped you please don't forget to thank me...
