Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly
Answer:
Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force, air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
The speed of the arrow after passing through the target is 30.1 meters per second.
Explanation:
The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:
[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]
Where:
[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.
[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.
[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.
The final speed of the arrow is now cleared:
[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]
[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]
If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:
[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]
[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]
The speed of the arrow after passing through the target is 30.1 meters per second.
Stress is a factor that contributes to heart disease risk.true or false
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?
Answer:
The sun will be located near the Gemini constellation at sunset
Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. If the charges are deflected in opposite directions, what are the possible relative charges and directions? (Select all that apply.)
Answer:
*If the particles are deflected in opposite directions, it implies that their charges must be opposite
*the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Explanation:
When a charged particle enters a magnetic field, it is subjected to a force given by
F = q v x B
where bold letters indicate vectors
this expression can be written in the form of a module
F = qv B sin θ
and the direction of the force is given by the right-hand rule.
In our case the magnetic field is perpendicular to the speed, therefore the angle is 90º and the sin 90 = 1
If the particles are deflected in opposite directions, it implies that their charges must be opposite, one positive and the other negative.
Furthermore, the force is perpendicular to the speed, therefore it describes a circular movement, one in the clockwise direction and the other in the counterclockwise direction.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?
Answer:
a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Explanation:
a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:
First arrow
[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
Now, the system is expanded and simplified:
[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]
[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]
Where:
[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.
[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
The initial vertical speed of the arrow is:
[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]
Where:
[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.
[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:
[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]
[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{2} - y_{1} = 56.712\,m[/tex]
Second arrow
[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]
[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{3} - y_{1} = 342.816\,m[/tex]
The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.
b) The total energy of each system is determined hereafter:
First arrow
The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:
[tex]E = U + K_{x}[/tex]
The expression is now expanded:
[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]
Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.
[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:
[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]
[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]
If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 201.720\,J[/tex]
Second arrow:
The total mechanical energy is equal to the potential gravitational energy. That is:
[tex]E = m\cdot g \cdot y_{max}[/tex]
[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]
[tex]E = 201.720\,J[/tex]
Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
On Apollo missions to the Moon, the command module orbited at an altitude of 160 km above the lunar surface. How long did it take for the command module to complete one orbit?
Answer:
T = 2.06h
Explanation:
In order to calculate the time that the Apollo takes to complete an orbit around the moon, you use the following formula, which is one of the Kepler's law:
[tex]T=\frac{2\pi r^{3/2}}{\sqrt{GM_m}}[/tex] (1)
T: time for a complete orbit = ?
r: radius of the orbit
G: Cavendish's constant = 6.674*10^-11 m^3.kg^-1.s^-2
Mm: mass of the moon = 7.34*10^22 kg
The radius of the orbit is equal to the radius of the moon plus the distance from the surface to the Apollo:
[tex]r=R_m+160km\\\\[/tex]
Rm: radius of the moon = 1737.1 km
[tex]r=1737.1km+160km=1897.1km=1897.1*10^3 m[/tex]
Then, you replace all values of the parameters in the equation (1):
[tex]T=\frac{2\pi (1897.1*10^3m)^{3/2}}{\sqrt{(6.674*10^{-11}m^3/kgs^2)(7.34*10^22kg)}}\\\\T=7417.78s[/tex]
In hours you obtain:
[tex]T=7417.78s*\frac{1h}{3600s}=2.06h[/tex]
The time that the Apollo takes to complete an orbit around the moon is 2.06h
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus
Answer:
D). Uranus.
Explanation:
Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.
As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.
HELP ILL MARK BRAINLIEST PLS!!!!
A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 0.600 g is supported by a needle, the tip of which is a circle 0.240 mm in radius, what pressure is exerted on the record in N/m2?
Answer:
[tex]P=3.25x10^{4}\frac{N}{m^2}[/tex]
Explanation:
Hello,
In this case, since pressure is defined as the force applied over a surface:
[tex]P=\frac{F}{A}[/tex]
We can associate the force with the weight of the needle computed by using the acceleration of the gravity:
[tex]F=0.600g*\frac{1kg}{1000g}*9.8\frac{m}{s^2} =5.88x10^{-3}N[/tex]
And the area of the the tip (circle) in meters:
[tex]A=\pi r^2=\pi (0.240mm)^2=\pi (0.240mm*\frac{1m}{1000mm} )^2\\\\A=1.81x10^{-7}m^2[/tex]
Thus, the pressure exerted on the record turns out:
[tex]P=\frac{5.88x10^{-3}N}{1.81x10^{-7}m^2} \\\\P=3.25x10^{4}\frac{N}{m^2}[/tex]
Which is truly a large value due to the tiny area on which the pressure is exerted.
Best regards.
An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?
Answer:
The power used is 0.96 watts.
Explanation:
Recall the formula for electric power (P) as the product of the voltage applied times the circulating current:
[tex]P=V\,\,I[/tex]
and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:
[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]
Then we can re-write the power expression as:
[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]
which in our case becomes:
[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]
describe the relation among density, temperature, and volume when the pressure is constant, and explain the blackbody radiation curve
Answer:
in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,
Explanation:
Let's start for ya dream gas
PV = nRT
Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.
The density of is defined by
ρ = m / V
As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.
This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship
E = h f
The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.
Answer:
(c) always equal to the weight of the liquid displaced.
Explanation:
Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.
Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.
Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a height of 45 cm above the bottom. What is its speed at the top of the loop
Answer:
v = 1.7 m/s
Explanation:
By applying conservation of energy principle in this situation, we know that:
Loss in Potential Energy of Car = Gain in Kinetic Energy of Car
mgΔh = (1/2)mv²
2gΔh = v²
v = √(2gΔh)
where,
v = velocity of car at top of the loop = ?
g = 9.8 m/s²
Δh = change in height = 45 cm - Diameter of Loop
Δh = 45 cm - 30 cm = 15 cm = 0.15 m
Therefore,
v = √(2)(9.8 m/s²)(0.15 m)
v = 1.7 m/s
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?
Answer:
14.1 m/s
Explanation:
From the question,
μk = a/g...................... Equation 1
Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.
make a the subject of the equation
a = μk(g).................. Equation 2
Given: μk = 0.160, g = 9.8 m/s²
Substitute into equation 2
a = 0.16(9.8)
a = 1.568 m/s²
Using,
F = ma
Where F = force, m = mass.
Make m the subject of the equation
m = F/a................... Equation 3
m = 160/1.568
m = 102.04 kg.
Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.
Assuming the initial velocity of the skier to be zero
Fd+mgμ = 1/2mv²........................Equation 4
Where v = speed of the skier after crossing the patch, d = distance/width of the patch.
v = √2(Fd+mgμ)/m)................ Equation 5
Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16
Substitute these values into equation 5
v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]
v = √197.57
v = 14.1 m/s
v = 9.86 m/s
A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element
Answer:
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.Explanation:
The formula for calculating the power expended in a circuit is P = I²R... 1
i is the current (in amperes)
R is the resistance (in ohms)
If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I
New power dissipated P₂ = (I₂)²R
P₂ = (2I)²R
P₂ = 4I²R ... 2
Dividing equation 2 by 1 will give;
P₂/P = 4I²R/I²R
P₂/P = 4
P₂ = 4P
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.
An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is
Answer:
15amps
Explanation:
V=IR
I=V/R
I = 120/8
I = 15 amps
Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal
Answer:
Halide
Explanation:
It has at least one element from the halogen group (17)
Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.
What are halides?When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.
These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.
Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.
Thus, halides are compounds that have at least one element from group 17.
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A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.
By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
What is experiment ?An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.
What is hydrochloric acid?Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.
Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.
By the experiment "By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
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The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N