Answer:
Assume that [tex]\rm g= 9.81\; N\cdot kg^{-1}[/tex]; [tex]\rho(\text{Water}) = \rm 1000\;kg\cdot m^{-3}[/tex].
Density of the disk: approximately [tex]\rm 2.19\times 10^{3}\; kg\cdot m^{-3}[/tex].
Weight of the disk: approximately [tex]\rm 245\;N[/tex].
Buoyant force on the disk if it is submerged under water: approximately [tex]\rm 112\; N[/tex].
The disk will sink when placed in water.
Explanation:
Convert the dimensions of this disk to SI units:
Diameter: [tex]d = \rm 25\; inches = (25\times 0.3048)\; m = 0.762\;m[/tex].Thickness [tex]h = \rm 2.5\; cm = (2.5\times 0.01)\; m = 0.025\;m[/tex].The radius of a circle is 1/2 its diameter:
[tex]\displaystyle r = \rm \frac{1}{2}\times 0.762\;m = 0.381\; m[/tex].
Volume of this disk:
[tex]V(\text{disk}) = \pi\cdot r^{2}\cdot h = \pi\times 0.381^{2}\times 0.025 \approx 0.0114009\; m^{3}[/tex].
Density of this disk:
[tex]\displaystyle \rho(\text{disk}) = \frac{m}{V} = \rm \frac{25\; kg}{0.0114009\; m^{3}} = 2.19\times 10^{3}\;kg\cdot m^{-3}[/tex].
[tex]\rho(\text{disk}) >\rho(\text{water})[/tex] indicates that the disk will sink when placed in water.
Weight of the object:
[tex]W(\text{disk}) = m\cdot g = \rm 25\times 9.81 = 245.25\; N[/tex].
The buoyant force on an object in water is equal to the weight of water that this object displaces. When this disk is submerged under water, it will displace approximately [tex]\rm 0.0114009\; m^{3}[/tex] of water. The buoyant force on the disk will be:
[tex]\begin{aligned}F(\text{buoyant force}) &= W(\text{Water Displaced}) \\& = \rho\cdot V(\text{Water Displaced})\cdot g\\ & = \rm 1\times 10^{3}\; kg\cdot m^{-3}\times 0.0114009\; m^{3}\times 9.81\; N\cdot kg^{-1}\\ &\approx \rm 112\; N\end{aligned}[/tex].
The size of this disk's weight is greater than the size of the buoyant force on it when submerged under water. As a result, the disk will sink when placed in water.
A small frictionless cart is attached to a wall by a spring. It is pulled 14 cm from its rest position, released at time tequals0, and allowed to roll back and forth for 5 seconds. Its position at time t is s equals 14 cosine left parenthesis pi t right parenthesis. a. What is the cart's maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart's speed then?
Answer:given below
Explanation:
Cart is pulled 14 cm from mean position
and its position is given by
[tex]x=14cos\left ( \pi t\right )[/tex]
therefore its velocity is acceleration is given by
[tex]v=-14\pi sin\left ( \pi t\right )[/tex]
[tex]a=-14\pi ^2cos\left ( \pi t\right )[/tex]
[tex]\left ( a\right ) cart\ max.\ speed\ is[/tex]
[tex]v_{max}=14\pi at\ t=0.5sec[/tex]
and its position is x=0
acceleration at t=0.5sec
a=0
[tex]\left ( b\right )[/tex]
[tex]a_{max}=14\pi ^2 cm/s^2[/tex]
at t=0,1,2 sec
at t=0
x=14 cm
v at t=0
v=0 cm/s
for t=1 sec
x=-14 cm i.e. 14 cm behind mean position
v=0 m/s
The cart's maximum speed is 14π m/s and occurs at the equilibrium point in its oscillation at integer multiples of the half period. At these moments, the magnitude of the acceleration is 14π² m/s². The magnitude of the acceleration is greatest when the cart is at the extreme points of its oscillation, where its speed is zero.
Explanation:The cart, undergoing simple harmonic motion, has a maximum speed when it passes through equilibrium - the midpoint of its oscillating path. From the equation for the straightforward harmonic motion, we know that the speed v of the cart is given by the derivative of s(t) = 14cos(πt). The derivative of this function is v(t) = -14πsin(πt), and the maximum speed occurs when sin(πt) = ±1, which gives a maximum speed of ±14π m/s.
The maximum speed is attained at integer multiples of the half period (1/2, 3/2, 5/2 seconds, and so on). At this moment, the cart is at the equilibrium position, s=0. Acceleration a(t) is given by the second derivative of the position function s(t), which is a(t) = -14π²cos(πt). The magnitude of the acceleration at the time of maximum speed is 14π² m/s².
In the case of the maximum magnitude acceleration, this is attained at the turning points of the oscillation when cos(πt) = ±1. At these moments, the speed of the cart is zero.
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Substances A and B, initially at different temperatures, come in contact with each other and reach thermal equilibrium. The mass of substance A is twice the mass of substance B. The specific heat capacity of substance B is twice the specific heat capacity of substance A.
Which statement is true about the final temperature of the two substances once thermal equilibrium is reached?1) The final temperature will be closer to the initial temperature of substance B than substance A.2) The final temperature will be exactly midway between the initial temperatures of substances A and B.3) The final temperature will be closer to the initial temperature of substance A than substance B.
Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.
Explanation:
[tex]heat_{absorbed}=heat_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)][/tex] .................(1)
where,
q = heat absorbed or released
[tex]m_A[/tex] = mass of A = 2x
[tex]m_B[/tex] = mass of B = x
[tex]T_{final}[/tex] = final temperature = z
[tex]T_A[/tex] = temperature of A
[tex]T_2[/tex] = temperature of B
[tex]c_A[/tex] = specific heat capacity of A = y
[tex]c_B[/tex] = specific heat capacity of B = 2y
Now put all the given values in equation (1), we get
[tex]2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)][/tex]
[tex]2z=T_B+T_A[/tex]
[tex]z=\frac{T_B+T_A}{2}[/tex]
Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.
When two substances with different temperatures come into contact and reach thermal equilibrium, heat flows until they reach the same final temperature. In this case, the final temperature will be closer to the initial temperature of substance B because substance B requires more heat to raise its temperature and has a smaller mass. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.
Explanation:When two substances at different temperatures come into contact with each other and reach thermal equilibrium, heat flows from the hotter substance to the cooler substance until they reach the same final temperature. In this case, substance A has twice the mass of substance B but substance B has twice the specific heat capacity of substance A. The specific heat capacity determines how much heat is needed to raise the temperature of a substance. Since substance B requires more heat to raise its temperature and has a smaller mass, it will be able to absorb more heat from substance A than substance A can absorb from substance B.
This means that the final temperature will be closer to the initial temperature of substance B than substance A. The specific heat capacity ratio determines the proportion of heat absorbed or released by the substances.
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easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.
Answer:
1.034 m above the floor
Explanation:
The location of center of body for a compound body, when the weights are given is calculated as:
[tex]\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}[/tex]
where,
[tex]\bar x[/tex] is the center of gravity of the entire body
W = weight of the individual body
x = center of gravity of the individual body
Thus on substituting the values we get,
[tex]\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}[/tex]
or
[tex]\bar x = \frac{691.83}{669}[/tex]
or
[tex]\bar x =1.034m[/tex]
Hence, the center of gravity of the entire body lies 1.034 m above the floor
Julie and Eric row their boat (at a constant speed) 63 miles downstream for 7 hours helped by the current. Rowing at the same rate, the trip back against the current takes 9 hours. Find the rate of the boat in still water.
Answer:
Boat speed = 8 miles/hr
Explanation:
Let the speed of the boat be U
Let the speed of the current be V.
Therefore, the downstream speed is U+V
and the upstream speed is U-V
Now we know that Speed = distance / time
Therefore, downstream speed, U+V = 63 / 7
U+V = 9 miles/hr ------(1)
Upstream speed, U-V = 63 / 9
U-V = 7 miles/hr --------(2)
Therefore subtracting (2) from (1), we get
( U+V) - ( U-V ) = 9-7
2V = 2
V = 1 miles/hr
Therefore the speed of the current is V = 1 mile/hr
Now from (1) we get
U+V = 9
U+1 = 9
U = 8
Therefore, the speed of the boat is U = 8 miles/hr
Displacement is the slope of a velocity v. time graph.
A. True
B. False
Answer:
B. False
Explanation:
Acceleration is the slope of a velocity vs. time graph.
Displacement is the area under a velocity vs. time graph.
A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on the ball and shell, respectively, are (1) +4q, 0; (2) –6q, +10q; (3) +16q, –12q. Rank the situations according to their charge on (a) the inner surface of the shell and (b) the outer surface, most positive first
Answer:
Part a)
2) > 1) > 3)
Part b)
1) = 2) = 3)
Explanation:
Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.
So here we can say
1)+ 4q, 0
so inner surface has charge - 4q and outer surface charge is +4q
2) -6q , +10q
so inner surface charge is +6q, outer surface charge is +4q
3) +16q , -12q
so inner surface charge is -16q, outer surface charge is +4q
Part a)
situations in which inner surface charge is arranged in decreasing order is given as
2) > 1) > 3)
Part b)
Situations in which outer surface charge is arranged in decreasing order is given as
1) = 2) = 3)
Situation (1) has the most positive charge on the inner surface of the shell, while situation (2) has the most positive charge on the outer surface.
Explanation:To rank the situations according to their charge on the inner and outer surfaces of the metallic spherical shell, we need to consider the net charges on the small charged ball and the shell. Let's analyze each situation:
For situation (1) with a net charge of +4q on the ball and 0 charge on the shell: (a) The inner surface of the shell has a charge of +4q, and (b) the outer surface has a charge of 0.For situation (2) with a net charge of -6q on the ball and +10q on the shell: (a) The inner surface of the shell has a charge of -6q, and (b) the outer surface has a charge of +10q.For situation (3) with a net charge of +16q on the ball and -12q on the shell: (a) The inner surface of the shell has a charge of +16q, and (b) the outer surface has a charge of -12q.Therefore, ranking the situations according to the charge on the inner surface would be: (1), (3), (2) - from most positive to least positive. For the outer surface, the ranking would be: (2), (1), (3) - from most positive to least positive.
Model rocket engines are rated by their thrust force and by the impulse they provide. You can use this information to determine the time interval at which the engines fire. Two rocket engines provide the same impulse. The first engine provides 6 N of thrust for 2 s; the second provides 4 N of thrust.
For how long does this second engine fire?
Answer: 3 seconds
Explanation:
Since they provide the same impulse,
Impulse= Ft
F1 = 6N
t1 = 2s
F2= 4N
t2= ?
F1= Force of first engine
t1= time elapsed by first engine
F2= force of second engine
t2= time elapsed by second engine
F1t1 = F2t2
6 × 2 = 4t2
t2= 12/4
t2 = 3 seconds
This second engine fires for 3 s
[tex]\texttt{ }[/tex]
Further explanationAcceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
thrust of first engine = F₁ = 6 N
elapsed time of first engine = t₁ = 2 s
thrust of second engine = F₂ = 4 N
Asked:
elapsed time of second engine = t₂ = 2 s
Solution:
We will use Newton's Law of Motion to solve this problem as follows:
[tex]\Sigma F = ma[/tex]
[tex]\Sigma F = m \Delta v \div t[/tex]
[tex]\Sigma F = I \div t[/tex]
[tex]\boxed {I = \Sigma F \times t}[/tex] → Impulse Formula
[tex]\texttt{ }[/tex]
Two rocket engines provide the same impulse :
[tex]I_1 = I_2[/tex]
[tex]\Sigma F_1 \times t_1 = \Sigma F_2 \times t_2[/tex]
[tex]6 \times 2 = 4 \times t_2[/tex]
[tex]12 = 4 \times t_2[/tex]
[tex]t_2 = 12 \div 4[/tex]
[tex]\boxed {t_2 = 3 \texttt{ s}}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Dynamics
The force of gravity on an object varies directly with its mass. The constant of variation due to gravity is 32.2 feet per second squared. Which equation represents F, the force on an object due to gravity according to m, the object’s mass?F = 16.1mF = F = 32.2mF =
Answer:
F = 32.2m
Explanation:
The force of gravity on an object is given by:
[tex]F=mg[/tex]
where
m is the mass of the object
g is the acceleration due to gravity
Here we have:
- An object of mass m
- The acceleration of gravity is expressed as [tex]g=32.2 ft/s^2[/tex]
Therefore, substituting into the formula above, we find that the force of gravity on the object is
[tex]F=m\cdot 32.2 = 32.2m[/tex]
Answer:
F = 32.2m
Explanation:
If an object is propelled upward from a height of 128 feet at an initial velocity of 112 feet per second, then its height h after t seconds is given by the equation h equals negative 16 t squared plus 112 t plus 128. After how many seconds does the object hit the ground? Round to the nearest tenth of a second.
Explanation:
The equation of motion of an object is given by :
[tex]h(t)=-16t^2+112t+128[/tex]
Where
t is the time in seconds
We need to find the time when the object hits the ground. When the object hits the ground, h(t) = 0
So,
[tex]-16t^2+112t+128=0[/tex]
[tex]-t^2+7t+8=0[/tex]
On solving above equation using online calculator, t = 8 seconds. So, the object hit the ground after 8 seconds. Hence, this is the required solution.
A rock is thrown into a still pond. The circular ripples move outward from the point of impact of the rock so that the radius of the circle formed by a ripple increases at the rate of 3 feet per minute. Find the rate at which the area is changing at the instant the radius is 4 feet. When the radius is 4 feet, the area is changing at approximately nothing square feet per minute. (Type an integer or a decimal. Round to the nearest thousandth as needed.)
Answer:
Rate of change of area is [tex]75.398ft^{2}/sec[/tex]
Explanation:
[tex]Area=\pi r^{2}\\\\\frac{d(Area)}{dt}=\frac{\pi r^{2}}{dt}=2\pi r\frac{dr}{dt}[/tex]
Applying values we get [tex]\frac{d(Area)}{dt}=2\pi 4\times 3=75.398ft^{2}/sec[/tex]
When the temperature goes up 3^\circ on the Cantor scale, it goes up 8^\circ on the Frobenius scale. On both scales, 18^\circ is the same temperature. How many Frobenius degrees are equal to 30^\circ Cantor?
Answer:
50°
Explanation:
It is given that the both scales are the same at 18˚
Now,
at 30˚ on the Cantor scale, the temperature is 30˚ − 18˚ = 12˚ above
the 18˚ mark.
also, it is given that with every 3˚ increase in the temperature on the Cantor scale is there is an 8˚ increase on the Frobenius scale,
mathematically, we can write it as (by unitary method)
3˚ increase in the temperature on the Cantor = 8˚ increase on the Frobenius scale
or
1˚ increase in the temperature on the Cantor = (8/3)˚ increase on the Frobenius scale
thus, for x˚ increase in the temperature on the Cantor = ((8/3)˚ × x) increase on the Frobenius scale
hence, for 12° increase we have
12˚ increase in the temperature on the Cantor = ((8/3)˚ × 12) increase on the Frobenius scale
or
12˚ increase in the temperature on the Cantor = 32° increase on the Frobenius scale
hence, the final reading on the Frobenius scale will be, 18˚ + 32˚ = 50˚.
A box at rest on a ramp is in equilibrium, as shown.
What is the force of static friction acting on the box? Round your answer to the nearest whole number.
_______N
What is the normal force acting on the box? Round your answer to the nearest whole number.
_______ N
Answer:
Ffs = 251 N
Fn = 691 N
Explanation:
Take the y direction to be normal to the ramp and the x direction to be parallel to the ramp.
The angle of the ramp is 20°, so the angle that the weight vector makes with the normal is also 20°. Therefore:
Fgx = Fg sin 20°
Fgy = Fg cos 20°
Sum of the forces in the x direction (parallel to the ramp):
∑F = ma
Ffs − Fgx = 0
Ffs = Fgx
Ffs = Fg sin 20°
Ffs = 735 sin 20°
Ffs ≈ 251
Sum of the forces in the y direction (normal to the ramp):
∑F = ma
Fn − Fgy = 0
Fn = Fgy
Fn = Fg cos 20°
Fn = 735 cos 20°
Fn ≈ 691
To answer that question we need to apply equations of movement ( from Newton´s laws )
In equilibrium:
∑F = 0 or ∑Fx = 0 ; ∑Fy = 0
Solution is:
a) F(sf) = 251 [N]
b) Fn = 691 [N]
From the attached drawings we can see: ( Body free diagram)
∑ Fₓ = F(sf) - Pₓ = 0 where P = m×g = 735 [N] ( the weigth)
and Pₓ = P× cos20°
Then F(sf) = Pₓ × sin 20°
F(sf) = m×g×cos20° = 735× 0.34202 [N]
F(sf) = 251.3847
F(sf) = 251 [N]
rounding to the nearest number
F(sf) = 691 [N]
∑ Fy = 0
∑ Fy = Fn - Py = 0 Py = P×cos20° Py = m×g×cos20°
Py = 735×0.939693 [N] Py = [N]
Fn = Py = 690.674 [N]
rounding to the nearest whole number
Fn = 691 [N]
Related question Brainly.com/question/ 11544804
Two airplanes leave an airport at the same time and travel in opposite directions. One plane travels 87 km/h faster than the other. If the two planes are 11,865 km apart after 7 hours, what is the rate of each plane?
Answer:
Speed of A = 891 km/h
Speed of B = 804 km/h
Explanation:
Let the speed of aeroplane B is v, the speed of aeroplane is A is 87 km/h faster than B.
So, the speed of aeroplane A is 87 + v.
Distance traveled by A after 7 hours, d1 = (87 + v) x 7
Distance traveled by B after 7 hours, d2 = v x 7
Total distance traveled = 11865 km
So, d1 + d2 = 11865
(87 + v) x 7 + v x 7 = 11865
609 + 14 v = 11865
14 v = 11256
v = 804 km/h
So, the speed of A = 87 + 804 = 891 km/h
Speed of B = 804 km/h
Which of the following best explains why snow predictions by meteorologists are sometimes incorrect?
A. Weather data are misinterpreted.
B. Weather instruments are extremely precise.
C. Interference from the sun causes data to be collected inaccurately.
D. Local variations in weather are too small for weather instruments to replicate.
Answer:
C. Interference from the sun causes data to be collected inaccurately.
Explanation:
Snow predictions by meteorologists are sometimes incorrect because from the sun causes data to be collected inaccurately.
A rock is suspended by a light string. When the rock is in air, the tension in the string is 43.8 N . When the rock is totally immersed in water, the tension is 31.3 N . When the rock is totally immersed in an unknown liquid, the tension is 18.3 N .
Answer:
Density of unknown liquid is [tex]2047 kg/m^3[/tex]
Explanation:
When rock is suspended in air then the weight of the rock is counter balanced by the tension force in the string
So here we have
[tex]T = mg = 43.8 N[/tex]
now when the rock is immersed in water then the tension in the string is and buoyancy force due to water is counter balanced by the weight of the object
so here we have
[tex]T_1 + F_b = mg[/tex]
[tex]31.3 + F_b = 43.8[/tex]
[tex]F_b = 43.8 - 31.3 = 12.5 N[/tex]
now we have
[tex](1000)V(9.81) = 12.5[/tex]
[tex]V = 1.27 \times 10^{-3} m^3[/tex]
now when the rock is immersed into other liquid then we have
[tex]T_2 + F_b' = mg[/tex]
[tex]18.3 + F_b' = 43.8[/tex]
[tex]F_b' = 25.5 N[/tex]
now we have
[tex]\rho(1.27 \times 10^{-3})(9.81) = 25.5[/tex]
[tex]\rho = 2047 kg/m^3[/tex]
Energy transformations that involve one transformation from one type of energy to another are called...............transformations.
Answer:
It is called single transformation
Answer:
Single transformation
Explanation:
Energy can neither be created nor be destroyed, it can only covert from one form to another. Sometimes, to get a work done one form of energy only require to be transformed into another form then it is known as single transformation.
For Example, cell phone transforming electrical energy into electromagnetic energy that makes it's working feasible.
Moving vehicle converting chemical energy to kinetic energy.
You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when you are finished
Based on the provided information, the correct statement is:
b. The bottom of your front bumper must not be more than 28 inches above the pavement.
The modification to increase ground clearance in the pick-up truck involves specific regulations for the front bumper height.
The statement "The bottom of your front bumper must not be more than 28 inches above the pavement" indicates a legal restriction to maintain a certain elevation for safety and compliance reasons.
This limitation aims to prevent potential hazards, such as underride collisions, and ensures that the modified truck adheres to regulatory standards. By specifying the maximum height of the front bumper, authorities aim to strike a balance between vehicle customization and road safety.
This regulation emphasizes the importance of maintaining a reasonable height for front bumpers to mitigate risks and uphold public safety standards, reflecting the broader considerations in vehicular modifications within the legal framework.
The probable question may be:
You have a pick-up truck that weighed 4,000 pounds when it was new. You are modifying it to increase its ground clearance. When you are finished
a. The bottom of your front bumper can be up to 32 inches above the pavement.
b. The bottom of your front bumper must not be more than 28 inches above the pavement.
c. You will no longer be required to have a rear bumper.
A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank (Fig. P11.43). The width of the box is 75.0 cm, and sand is to be distributed uniformly throughout it. The center of gravity of the nonuniform plank is 50.0 cm from the right end. What mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed just below its midpoint?
Answer:
Required mass of sand is 20 kg
Explanation:
Given:
Mass of the plank = 25 kg
Distance of the Center of gravity of the Plank from the fulcrum = [tex]\frac{2}{2}-0.50 = 0.5m[/tex]
Distance of the Center of gravity of the sand box from the fulcrum = [tex]\frac{2}{2}-\frac{0.75}{2}= 0.625m[/tex]
Balancing the torque due to the plank and the sand box with respect to the fulcrum
Torque = Force × perpendicular distance
thus, we get
(25 × g) × 0.5 = weight of sand × 0.625
where, g is the acceleration due to gravity
or
(25 × g) × 0.5 = (mass of sand × g) × 0.625
or
mass of sand = 20 kg
Hence, the required mass of the sand is 20 kg
Answer:
20 Kg mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed horizontally on a fulcrum placed just below its midpoint.
Explanation:
Use the second condition of equilibrium.
[tex]$\sum} \tau=0$[/tex]
[tex]MgL-$M g x_{c m}=0$[/tex]
[tex]$M=\frac{m x_{c m}}{L}[/tex]
[tex]=\frac{25(0.50)}{0.625}[/tex]
[tex]=20 \mathrm{~kg}$[/tex]
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Which type of energy is thermal energy a form of?
Answer:
Kinetic Energy
Explanation:
Heat energy is another name for thermal energy. Kinetic energy is the energy of a moving object. As thermal energy comes from moving particles, it is a form of kinetic energy.
A lens collects light and focuses it into a small spot. This increases the ________ of the light wave.A lens collects light and focuses it into a small spot. This increases the ________ of the light wave.
Answer:
intensity.
Explanation:
when the light collected by the lens is focused into a small spot it tends to increase the intensity of the light.
as different path of light with different intensity combines from passing through the lens it tends to make the light path and intensity coherent and after being coherent there intensity increases.
How to tell if something is an electrolyte
Answer: If it has ions, it is an electrolyte
Explanation:
Let's start by explaining that electrolytes are compounds that contain charged particles or ions, which can be cations (positive ions) or anions (negative ions).
So, it is this composition that makes an electrolytic material conduct electricity.
In this sense, the way to identify if a material is an electrolyte or not, is knowing whether it is composed of ions or not.
Given that the electromagnetic force is far stronger than gravity on a per-particle basis, why doesn't the electromagnetic force dominate the interactions of large objects like planets, stars, and galaxies?
Explanation:
The electro magnetic force is given by
F = [tex]\frac{k.q_{1}.q_{2}}{r^{2}}[/tex]
where [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are charged particles
k =Coulombs constant
r = distance between two charges
And gravitational force is given by
F = [tex]\frac{G.m_{1}.m_{2}}{r^{2}}[/tex]
where [tex]m_{1}[/tex] and [tex]m_{2}[/tex] are masses
G =Garvitation constant
r = distance between two masses
Now since the planets, stars and galaxies are electrically neutral, therefore they have zero electrical charge and so electro magnetic forces have no affect on these planets, stars and heavenly bodies.
Whereas the masses of the heavenly bodies are very large, so they are largely affected by the gravitational force since Gravitational force is directly proportional to the product of the masses of a body.
Therefore, though the electromagnetic force is stronger than the gravitational force, the electromagnetic force does not dominate the forces in the heavenly bodies as they as not electrically charged.
The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 90I I2 + I + 9 where I is the light intensity (measured in thousands of foot-candles). For what light intensity is P a maximum?
Answer:
At light intensity I = 3, is P a maximum
Explanation:
Given:
[tex]P=\frac{90I}{I^2+I+9}[/tex]
now differentiating the above equation with respect to Intensity 'I' we get
[tex]\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}[/tex]
or
[tex]\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}[/tex]
or
[tex]\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}[/tex]
or
[tex]\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}[/tex]
Now for the maxima [tex]\frac{dP}{dI}=0[/tex]
thus,
[tex]0=\frac{-90I^2+810)}{(I^2+I+9)^2}[/tex]
or
[tex]-90I^2+810=0[/tex]
or
[tex]I^2=\frac{810}{90}[/tex]
or
[tex]I^2=9[/tex]
or
I = 3
thus, for the value of intensity I = 3, the P is maximum
at I = 3
[tex]P=\frac{90\times3}{3^2+3+9}[/tex]
or
[tex]P=\frac{270}{21}[/tex]
or
[tex]P=12.85[/tex]
A gas is compressed by an adiabatic process that decreases its volume by a factor of 2.In this process, the pressurea) increases by a factor of more than 2.b) increases by a factor of 2.c) does not change.d) increases by a factor of less than 2.
Answer:
a) Increase by a factor of more than 2
Explanation:
The compression process is an adiabatic one. The opposite of this kind of process is the isothermal process, where the heat flow makes the temperature to be constant.
Let's consider this kind of process by means of an equation of state; for example, the ideal gas law:
[tex]P=\frac{RT}{v}[/tex]
Be [tex]v_{0}[/tex] the initial volume. And [tex]v_{f}=2v_{0}[/tex] the final volume.
If we consider the ideal gas law, it is evident that if the temperature remains constant (isothermal process), the pressure increases by a factor of 2; but in an adiabatic process the temperature of a gas tends to increase its temperature, so the pressure will be a higher than the resultant for the isothermal process.
A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant rate of 15.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T, which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.A) Find the magnitude of the emf E induced in the loop after exactly 8.00 s has passed since the circumference of the loop started to decrease.B) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field. (Clockwise or Counterclockwise?)
Answer:
Part a)
[tex]EMF = 5.6 \times 10^{-3} V[/tex]
Part b)
Since the radius is decreasing so induced current will increase the flux through the coil
So it would be clockwise in direction
Explanation:
As we know that magnetic flux linked with the coil is given as
[tex]\phi = \pi r^2 B[/tex]
now the rate of change in flux is given as
[tex]\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B[/tex]
now we know that circumference is decreasing at rate of 15 cm/s
so here we know the length of circumference as
[tex]C = 2\pi r[/tex]
So rate of change in circumference is
[tex]\frac{dC}{dt} = 2\pi \frac{dr}{dt}[/tex]
[tex]\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}[/tex]
final length of circumference at t = 8 s
[tex]C = 167 - (15)(8) = 47[/tex]
Part a)
Now the induced EMF is given as
[tex]EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)[/tex]
[tex]EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)[/tex]
[tex]EMF = 5.6 \times 10^{-3} V[/tex]
Part b)
Since the radius is decreasing so induced current will increase the flux through the coil
So it would be clockwise in direction
The problem is solved using Faraday's Law of Electromagnetic Induction: the induced emf equals to the change in magnetic flux divided by change in time. Then, Lenz's law is used to determine the direction of the induced current.
Explanation:The problem you're dealing with is an application of Faraday's Law of Electromagnetic Induction. This law states that the electromotive force (or emf) induced in a circuit is equivalent to the rate of change of magnetic flux through that circuit.
For Part A, you know that the rate of change of the circumference is constant and you can convert that to a rate of change of radius (since circumference = 2πr). Using these formulas, you can determine the rate of change of area as πr(dr/dt), where dr/dt is the rate of change of radius. The induced emf is proportional to the rate of change of magnetic flux, which is the product of magnetic field strength and area. Thus, induced emf equals to (change in flux/change in time), or -Bπr(dr/dt).
For part B, you apply Lenz's law, which states that the direction of induced current is such that the magnetic field due to it opposes the change in the initial magnetic field. Since the field is getting smaller (since the loop is contracting), then the current will act in such a way as to try to keep the size of the field the same. This means going in the counterclockwise direction.
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Suppose that on a hot (33.0°C) and sticky (80% humidity) afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab (volume of 180.0 m³) was 1.00 atm before the storm and 0.800 atm after the storm, to what volume would the laboratory try to expand in order to make up for the large pressure difference outside?
1. 211 m3
2. 134 m3
3. 1800 m3
4. 7,150 m3
Answer:
V₂ = 225 m^{3}
so no exact option is match with the calculated answer.
Explanation:
According to Boyle's Law, we have
P_1 V_1 = P_2 V_2 ----------- (1)
Data Given;
P_1 = 1.0 atm
V_1 = 180 m³
P_2 = 0.80 atm
V_2 = ?
Solving equation 1 for V₂,
[tex]V_2= \frac{P_1 V_1}{P_1}[/tex]
Putting values,
[tex]V_2 = \frac{1.0*180}{0.80}[/tex]
[tex]V_2 = 225 m^{3}[/tex]
so no exact option is match with the calculated answer.
Answer:
E
Explanation:
The frequency of a wave is 1) A) measured in cycles per second. B) measured in hertz (Hz). C) the number of peaks passing by any point each second. D) equal to the speed of the wave divided by the wavelength of the wave. E) all of the above
The frequency of a wave is measured in hertz (Hz). Hence the correct answer is option E
Explanation:The frequency of a wave is measured in hertz (Hz). Hertz is the unit used to measure the number of waves passing by a point in one second. This means that option B) 'measured in hertz (Hz)' is the correct answer. Frequency is not equal to the speed of the wave divided by the wavelength of the wave, as stated in option D). While option C) 'the number of peaks passing by any point each second' is related to frequency, it does not encompass the entire definition. Therefore, option E) 'all of the above' is also incorrect.
A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object moving to the left at 1.0 m/s. What is the total kinetic energy after the collision?
Answer:
The kinetic energy of the system after the collision is 9 J.
Explanation:
It is given that,
Mass of object 1, m₁ = 3 kg
Speed of object 1, v₁ = 2 m/s
Mass of object 2, m₂ = 6 kg
Speed of object 2, v₂ = -1 m/s (it is moving in left)
Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,
[tex]E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2[/tex]
[tex]E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2[/tex]
E = 9 J
So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.
A sensor on a traffic light is most likely to produce electromagnetic waves at which of these frequencies?
Answer:
Option-(D): 10¹¹ waves per second.
Explanation:
Electromagnetic waves:The electromagnetic waves is such form of a energy transfer or wave propagation through any space with or without having any medium(particles).As, the medium or particles inside a space are able to transfer the amount of energy from the origin towards the receiver, which makes it very easy for the wave propagation through a medium.Now, the electromagnetic waves are generated from the sensor on a traffic light when the frequency,f level of the wave generation is about 10¹¹ Hertz(Hz).
Answer:
answer is A
Explanation:
Question Part Points Submissions Used If an object with mass m is dropped from rest, one model for its speed v after t seconds, taking air resistance into account, is v = mg c (1 − e−ct/m) where g is the acceleration due to gravity and c is a positive constant describing air resistance. (a) Calculate lim t→∞ v.
Answer:
[tex]\lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]
Explanation:
the velocity as a function of time is
[tex]v(t)=\frac{mg}{c}(1-e^{\frac{-ct}{m}})[/tex]
[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})[/tex]
[tex]\therefore v(t)=\frac{mg}{c}(1-\frac{1}{e^{\frac{ct}{m}}})\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-\frac{1}{\infty })\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}(1-0)\\\\\therefore \lim_{t\rightarrow \mathbb{\infty }}v(t)=\frac{mg}{c}[/tex]
The limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance considered, is the terminal velocity mg/c. The exponential term in the equation approaches zero as time becomes very large, leading to the object reaching a constant terminal velocity.
Explanation:
To calculate the limit as t approaches infinity of the speed v for an object with mass m dropped from rest with air resistance taken into account, we use the provided equation v = mg/c (1 − e^{-ct/m}), where g is the acceleration due to gravity, which averages 9.80 m/s², and c is a positive constant representing air resistance. The limit represents the object's terminal velocity, which is the constant speed an object reaches when the force of gravity is balanced by the drag force of air resistance.
As t → ∞ (increases towards infinity), the exponential term e^{-ct/m} approaches zero. Hence, the limit of the speed v becomes:
∑ lim t→∞ v = lim t→∞ mg/c (1 − e^{-ct/m}) = mg/c.
The value mg/c is known as the terminal velocity, which is the maximum speed that the object will reach as it continues to fall.
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