Answer:
w'=(1/2)w
Explanation:
In order to calculate the angular velocity of the second wheel, you use the following formula:
[tex]\omega=\frac{v}{r}[/tex] (1)
v: speed of the wheel 1 = speed of the wheel 2
r: radius of the wheel 1
For the second wheel you have:
r'=2r
You replace this value of r' in the following equation:
[tex]\omega'=\frac{v}{r'}=\frac{v}{2r}=\frac{1}{2}\frac{v}{r}=\frac{1}{2}\omega\\\\\omega'=\frac{1}{2}\omega[/tex]
The angular velocity of the second wheel is one half of the angular velocity of the first wheel
Calculate the total current in parallel circuit using Kirchhoff’s Current Law? anyone give example and solution...
Answer:
please check attachment and follow the steps.
The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 40.0 cm.
Answer:
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
Explanation:
Given:
Mass of α particle (m) = 6.50 × 10⁻²⁷ kg
Charge of α particle (q) = 3.20 × 10⁻¹⁹ C
Potential difference ΔV = 165 V
Find:
kinetic energy (K.E)
Computation:
kinetic energy (K.E) = (ΔV)(q)
kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)
kinetic energy (K.E) = 528 (10⁻¹⁹)
kinetic energy (K.E) = 5.28 ×10⁻¹⁷
a block of wood is pulled by a horizontal string across a rough surface at a constant velocity with a force of 20N. the coefficient of kinetic friction between the surfaces is 0.3 the force of the friction is
Answer:
6 N
Explanation:
From the laws of friction
F = ¶R = 0.3 × 20 = 6 N
The force of friction opposing the block's motion is 6 N.
The given parameters;
force applied on the block, F = 20 Ncoefficient of kinetic friction = 0.3The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.
F = ma
Fₓ = μF
Substitute the given parameters to calculate the frictional force on the object.
Fₓ = 0.3 x 20
Fₓ = 6 N
Thus, the force of friction opposing the block's motion is 6 N.
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Answer: Net electrostatic force on C is 24.2×[tex]10^{-2}[/tex] Newtons.
Explanation: Coulomb's Law is used to determine Electrostatic Force. Its formula is:
F = k.[tex]\frac{q_{0}.q_{1}}{r^{2}}[/tex]
where:
k is electrostatic constant (k = 8.987×[tex]10^{9}[/tex] Nm²/C²);
q is the charge of the object in Coulumb;
r is the distance between charges;
The net force is the sum of all the forces acting on C, so:
Force B on C:
They are both positive, so there is a relpusive force acting between them on the y-axis.
[tex]F_{BC} = 8,987.10^{9}.\frac{4.35.10^{-3}.9.67.10^{-4}}{(6.14.10^{2})^{2}}[/tex]
[tex]F_{BC} = 10.03.10^{-2}[/tex] N
Force D on C:
There is an atractive force between them on the x-axis.
[tex]F_{CD} = 8.987.10^{9}.\frac{9.67.10^{-4}.1.92.10^{-3}}{(1.42.10^{3})^{2}}[/tex]
[tex]F_{CD} = 13.64.10^{-4}[/tex] N
Force A on C:
First, find the distance between objects:
The distance is a diagonal line that divides the rectangle into a right triangle. Distance is square of the hypotenuse .
[tex]r^{2} = (6.14.10^2)^{2} + (1.42.10^{3})^{2}[/tex]
[tex]r^{2} = 37.72.10^{4}[/tex]
and hypotenuse: r = [tex]6.14.10^2[/tex]m
There is an atractive force between charges, but there are components of the force in x- and y-axis. So, because of that, force will be:
[tex]F_{CA} = F_{CA}[/tex].sinα + [tex]F_{CA}.[/tex]cosα
[tex]F_{CA} = 8.987.10^{9}.\frac{3.12.10^{-3}.9.67.10^{-4}}{37.72.10^{4}}[/tex]
[tex]F_{CA} = 7.2.10^{-2}[/tex]
The trigonometric relations is taken from the rectangle:
sinα = [tex]\frac{6.14.10^{2}}{6.14.10^{2}}[/tex]
cosα = [tex]\frac{1.42.10^{3}}{6.14.10^{2}}[/tex]
[tex]F_{CA}.[/tex]cosα = [tex]7.2.10^{-2}(\frac{1.42.10^{3}}{6.14.10^{2}})[/tex] = 0.17
[tex]F_{CA}.[/tex]sinα = [tex]7.2.10^{-2}.(\frac{6.14.10^{2}}{6.14.10^{2}} )[/tex] = 0.072
[tex]F_{CA} =[/tex] 0.17î + 0.072^j
Now, sum up all the terms in its respective axis:
X: [tex]13.64.10^{-4} + 0.17 =[/tex] 0.1714
Y: [tex]10.03.10^{-2} + 7.2.10^{-2}[/tex] = 0.1723
These forms another right triangle, whose hypotenuse is the net electrostatic force:
[tex]F_{net} = \sqrt{(0.1714)^{2} + (0.1723)^2}[/tex]
[tex]F_{net} = 24.3.10^{-2}[/tex] N
The net electrostatic force acting on C has magnitude [tex]F_{net} = 24.3.10^{-2}[/tex] N.
A child has a toy car on a horizontal platform. The car starts from rest and reaches a maximum speed in 4 s. If the mass of the car is
0.1 kg and engine has an effective pull of 0.4 N Find the acceleration of the car.
Answer:
a=4m/s²
Explanation:
F=ma
0.4=0.1a
Answer:
a=4m/s
Explanation:
F=ma
0.4=0.1a
[tex] \frac{0.4}{0.1} = \frac{0.1}{0.1} [/tex]
a =4m/ s
a crate b of mass 40kg is raised by the rope of crane from the hold of a ship. mark and name forces on the crate . find acceleration if tension is 480N
Find the acceleration, a .
Formula used:-Force = Mass × Acceleration
Solution:-We know that ,
Force = Mass × Acceleration
★ Substituting the values in the above formula,we get:
⇒ 480 = 40 × Acceleration
⇒ Acceleration, a = 480/40
⇒ Acceleration,a = 12 m/s
Thus,the acceleration of a body is 12 metres per seconds.
toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the x-y plane. The 4.00 kg puck has a velocity of 3.00 i m/s at one instant. Eight seconds later, its velocity is (8.00 i 10.00 j) m/s. Assuming the rocket engine exerts a constant force, find (a) the components of the force and (b) its magnitude.
Answer:
Fx = 2.5 N
Fy = 5 N
|F| = 5.59 N
Explanation:
Given:-
- The mass of puck, m = 4.0 kg
- The initial velocity of puck, u = 3.00 i m/s
- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s
- The time interval for the duration of force, Δt = 8 seconds
Find:-
the components of the force and (b) its magnitude.
Solution:-
- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.
- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.
[tex]F_net = \frac{m*( v - u ) }{dt}[/tex]
Where,
Fnet: The net force that acts on the puck-rocket system
- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.
- We will apply the newton's second law of motion in component forms. And determine the components of force F, as ( Fx ) and ( Fy ) as follows:
[tex]F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\[/tex]
- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:
[tex]| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590[/tex]
Answer: the magnitude of the thrust force is F = 5.59 N
Two Earth satellites, A and B, each of mass m = 980 kg , are launched into circular orbits around the Earth's center. Satellite A orbits at an altitude of 4100 km , and satellite B orbits at an altitude of 12100 km The radius of Earth RE is 6370 km.
(a) What is the ratio of the potential energy of satellite B to that of satellite A, in orbit?
(b) What is the ratio of the kinetic energy of satellite B to that of satellite A, in orbit?
(c) Which satellite has the greater total energy if each has a mass of 14.6 kg?
(d) By how much?
Answer:
Do u have a picture of the graph?
Explanation:
I can solve it with refraction
If an ant is starting from 0 radians, and travels all the way to 4.5 radians in 18 seconds, what is angular velocity? If the same ant then slows down back to 0 rad/s in 3 seconds what is its angular acceleration?
Answer:
Angular velocity is 0.25 rad/sAngular acceleration is 0.017 rad/s²Explanation:
Given;
initial angular displacement, θ₁ = 0 radians
final angular displacement, θ₂ = 4.5 radians
Angular velocity is calculated as;
[tex]\omega = \frac{\delta \theta}{\delta t}= \frac{4.5 -0}{18} \\\\\omega = 0.25 \ rad/s[/tex]
Angular acceleration is calculated as;
[tex]\alpha = \frac{\delta \omega}{\delta t} = \frac{\omega _f - \omega_i}{t_2 -t_1}[/tex]
where;
[tex]\omega_f[/tex] is the final angular velocity = 0 rad/s
[tex]\omega _i[/tex] is the initial angular velocity = 0.25 rad/s
t₂ is the final time of the motion = 3 seconds
t₁ is the initial time of the motion = 18 seconds
[tex]\alpha =\frac{\omega _f - \omega_i}{t_2 -t_1} \\\\\alpha = \frac{0 - 0.25}{3-18} \\\\\alpha = 0.017 \ rad/s^2[/tex]
Austin is interested in working for an intelligence branch of the government. A polygraph test is part of the interview
process. Though Austin intends to be honest, he is worried that the polygraph will say he is lying when he is not.
Austin's friend, Gabe, assures Austin that polygraph tests are infallible. Is Gabe right?
Answer:
Gabe is wrong in concluding that the polygraph tests are infallible. Even though the polygraph machine is a man-made machine, there is a small margin of error factored into it.
This error could be caused as a result of the anxiety of the test taker, the heart beat rate, the blood pressure etc which would lead to the answers provided by the person to be flagged as false despite being true.
In most cases, the test result will be declared inconclusive if there was an error in the readings.
Explanation:
A WOMAN HAS A MASS OF 75.0 kg What is her weight on earth?
Answer:
735 N
Explanation:
If a woman has mass 75kg and u know that mass is constant everywhere then just apply the formula W=mg...as gravity on earth is 9.8 m/s2 , so her weight will be 735 N...hope it helps...
If a woman has a mass of 75 kilograms then her weight on the earth would be 735.75 Newtons, because the weight of the woman is the multiplication of the mass of the woman and the acceleration due to the gravity of the earth.
What is gravity?It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another,
As given in the problem if a woman has a mass of 75 kilograms we have to find the weight of the woman,
The weight of the woman = mass of the woman × acceleration due to the gravity of the earth
= 75 kilograms × 9.81
=735.75 Newtons
Thus, the weight of a woman who has a mass of 75 kilograms would be 735.75 Newtons
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If you could see stars during the day, the drawing above shows what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Taurus. Which constellation will be highest in the sky at sunset?
A. Pisces
B. Cancer
C. Gemini
D. Taurus
E. Aries
Answer:
Taurus constellation
Explanation:
Because Taurus constellation is mostly seen In the Northern Hemisphere, the bull passes through the sky from November to March, but the constellation's at its most visible in January. Taurus covers 797 square degrees.
Right ascension: 4 hours
Declination: 15 degrees
Best visible between latitudes 90 degrees and minus 65 degrees
An RLC circuit has a resistance of 200 Ω and an inductance of 15 mH. Its oscillation frequency is 7000 Hz. At time t = 0, the current is 25 mA, and there is no charge on the capacitor. After five complete cycles, the current is
Answer: 2.13 × 10^-4 A
Explanation:
Given that the RLC circuit has a resistance R = 200 Ω and an inductance L = 15 mH.
Its oscillation frequency F = 7000 Hz
The initial current I = 25 mA = 25/1000 or 25 × 10^-3 A
Since there is no charge on the capacitor, the current after complete 5 circle will be achieved by using the formula in the attached file.
Please find the attached file for the remaining explanation for the solution.
The current in the RLC circuit after five (5) complete cycles is equal to [tex]2.15 \times 10^{-4} \; Amperes[/tex]
Given the following data:
Resistance = 200 ΩInductance = 15 mH = 0.015 HOscillation frequency = 7000 HzCurrent = 25 mA = 0.025 ATo determine the current in the RLC circuit after five (5) complete cycles, we would use the following formula:
[tex]I_t = I_o e^{\frac{Rt}{2L} coswt}[/tex]Note: There is no electrical charge on the capacitor.
After five (5) complete cycles, the formula becomes:
[tex]I_5 = I_o e^{\frac{-R}{2L} \frac{5}{F} }[/tex]Substituting the given parameters into the formula, we have;
[tex]I_5 = 0.025 e^{\frac{-200}{2\; \times \;0.015} \times \frac{5}{7000} }\\\\I_5 = 0.025 e^{\frac{-200}{0.03} \times \frac{5}{7000} }\\\\I_5 = 0.025 e^{\frac{-1000}{210} }\\\\I_5 = 0.025 e^{-4.7619}\\\\I_5 = 0.025 \times 0.0086\\\\I_5 = 0.00215 \\\\I_5 = 2.15 \times 10^{-3} \; Amps[/tex]
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An object is placed at 10.2 cm in front of a diverging lens with a focal length of -10.6 cm. What is the magnification
Answer:
M= -0.51
Explanation:
After i calculated my v to be -5.2cm from the formula 1/f=1/v+1/u
Then m=v/u which is -0.51
The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen
Answer:
The speed is [tex]v =10.27 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The voltage is [tex]V = 30 kV = 30*10^{3} V[/tex]
The initial velocity of the electron is [tex]u = 0 \ m/s[/tex]
Generally according to the law of energy conservation
Electric potential Energy = Kinetic energy of the electron
So
[tex]PE = KE[/tex]
Where
[tex]KE = \frac{1}{2} * m* v^2[/tex]
Here m is the mass of the electron with a value of [tex]m = 9.11 *10^{-31} \ kg[/tex]
and
[tex]PE = e * V[/tex]
Here e is the charge on the electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]e * V = \frac{1}{2} * m * v^2[/tex]
=> [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]
[tex]v =10.27 *10^{7} \ m/s[/tex]
A train station bell gives off a fundamental tone of 500 Hz as the train approaches the station at a speed of 20 m/s. If the speed of sound in air is 334 m/s, what will be the apparent frequency of the bell to an observer riding the train
Answer: 529.9 Hz
Explanation:
Here we need to use the Doppler equation, so we have:
f' = f*(v + v0)/(v - vs)
Here, f is the frequency = 500Hz
v is the velocity of the wave, = 334m/s
v0 is the velocity of the observer = 20m/s
vs is the velocity of the source = 0m/s
Then we have:
f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz
A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.
Answer:
21,920 Pascals
Explanation:
P = ρgh
P = (1096 kg/m³) (10 m/s²) (2 m)
P = 21,920 Pa
Which of the following statements is accurate? A) Compressions and rarefactions occur throughout a transverse wave. B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave. C) Sound waves passing through the air will do so as transverse waves, which vibrate vertically and still retain their horizontal positions. D) Amplitude of longitudinal waves is measured at right angles to the direction of the travel of the wave and represents the maximum distance the molecule has moved from its normal position.
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave
Answer:
B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.
Explanation: hope this helps ;)
Complete the following statement: When a net torque is applied to a rigid object, it always produces a:______.
a. constant acceleration.
b. rotational equilibrium.
c. constant angular velocity.
d. constant angular momentum.
e. change in angular velocity.
When a net torque is applied to a rigid object, it always produces a change in angular velocity. (e.)
The word which best completes the given sentence, "When a net torque is applied to a rigid object, it always produces a_____" is:
E. Change in angular velocityAccording to the given question, we are asked to ask what would be produced when a net torque is applied to a rigid object based on the vector quantities involved.
As a result of this, we can see that there is a change in angular velocity when there is a net torque applied as the rigid object would change the rate at which it rotates.
Therefore, the correct answer is option E
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Copper wire of diameter 0.289 cm is used to connect a set of appliances at 120 V, which draw 1850 W of power total. The resistivity of copper is 1.68×10−8Ω⋅m.
A. What power is wasted in 26.0 m of this wire?
B. What is your answer if wire of diameter 0.417 cm is used?
Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
What does the vertical polarization axis of polarized sunglasses indicate about the direction of polarization of light bouncing off a horizontal surface, such as a wet road or lake surface
Answer:
it is desired that the lenses stop this ray, its polarization must be vertical
Explanation:
To answer this exercise, let's analyze the rays of light reflected on a horizontal surface, when the incident light that we consider non-polarized is reflected on a surface, the electric field of light moves the electrons on the surface horizontally and this re-emits the radiation same shape, that is horizontal.
The other vertical direction the atoms have a lot of movement restricted by the attraction on the surface, so for the reflected ray this polarization is attenuated, this does not stop the transmitted ray where the two polarizations are transmitted.
Total polarizations only for one angle, but in general as we approach dominant polarization it horizontal. Specifically the angle for full polarization is
n = tan teaP
Now we can analyze what polarization the lenses have, if the ray that comes is polarized horizontally and it is desired that the lenses stop this ray, its polarization must be vertical
g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, the mass is released from rest at x = 10.0 cm. ( That is, the spring is stretched by 10.0 cm.) (a) Determine the frequency of the oscillations. (b) Determine the maximum speed of the mass. Where dos the maximum speed occur? (c) Determine the maximum acceleration of the mass. Where does the maximum acceleration occur? (d) Determine the total energy of teh oscillating system. (e) Express the displacement as a function of time.
Answer:
(a) f = 0.58Hz
(b) vmax = 0.364m/s
(c) amax = 1.32m/s^2
(d) E = 0.1J
(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)
Explanation:
(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex] (1)
k: spring constant = 20.0N/m
m: mass = 1.5kg
you replace the values of m and k for getting f:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz[/tex]
The frequency of the oscillation is 0.58Hz
(b) The maximum speed is given by the following relation:
[tex]v_{max}=\omega A=2\pi f A[/tex] (2)
A: amplitude of the oscillations = 10.0cm = 0.10m
[tex]v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}[/tex]
The maximum speed of the mass is 0.364 m/s.
The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.
(c) The maximum acceleration is given by the following formula:
[tex]a_{max}=\omega^2A=(2\pi f)^2 A[/tex]
[tex]a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}[/tex]
The maximum acceleration is 1.32 m/s^2
The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.
(d) The total energy of the system is calculated with the maximum potential elastic energy:
[tex]E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J[/tex]
The total energy is 0.1J
(e) The displacement as a function of time is:
[tex]x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)[/tex]
An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.38 m/s, and its maximum acceleration is 6.83 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum
Answer:
t = 0.31s
Explanation:
In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:
[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]
A: amplitude
v_max = 1.38m/s
a_max = 6.83m/s^2
w: angular frequency
From the previous equations you can obtain the angular frequency w.
You divide vmax and amax, and solve for w:
[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]
Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.
You calculate the period by using the information about the angular frequency:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]
Then the required time is:
[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]
Find acceleration. Will give brainliest!
Answer:
16200 km/s
270 km/min
4.5 km/h
Explanation:
Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)
Simply plug in our known variables and solve:
a = (45.0 - 0)/10
a = 45.0/10
a = 4.5 km/h
Answer:
[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]
Explanation:
[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]
[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]
[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]
[tex]\displaystyle \mathrm{a = 4.5}[/tex]
[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]
Two resistors, R1=3.85 Ω and R2=6.47 Ω , are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current ????1 through R1 and the potential difference V2 across R2 .
Answer:
(a) 2.33 A
(b) 15.075 V
Explanation:
From the question,
The total resistance (Rt) = R1+R2 = 3.85+6.47
R(t) = 10.32 ohms.
Applying ohm's law,
V = IR(t)..........equation 1
Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.
Note: Since both resistors are connected in series, the current flowing through them is the same.
Therefore,
I = V/R(t)............. Equation 2
Given: V = 24 V, R(t) = 10.32 ohms
Substitute these values into equation 2
I = 24/10.32
I = 2.33 A.
Hence the current through R1 = 2.33 A.
V2 = IR2.............. Equation 3
V2 = 2.33(6.47)
V2 = 15.075 V
The current is 2.33 A
And, the potential difference is 15.075 V
Calculation of the current and the potential difference:Since Two resistors, R1=3.85 Ω and R2=6.47 Ω
So,
The total resistance (Rt) is
= R1+R2
= 3.85+6.47
= 10.32 ohms.
Now here we applied ohm law
V = IR(t)..........equation 1
Here V = Emf of the battery,
I = current
R(t) = combined resistance
Now
I = V/R(t)............. Equation 2
Here
Given: V = 24 V, R(t) = 10.32 ohms
So, I should be
I = 24/10.32
I = 2.33 A.
Now the potential difference is
V2 = IR2.............. Equation 3
V2 = 2.33(6.47)
V2 = 15.075 V
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Based on the stability classification, if, over the airport, rising air reaches the point of saturation below 3000 m, it is likely that:_______
Answer:
It is likely that vertically developed cumuliform clouds would begin to form
Explanation:
This is because since the air is conditionally unstable below 3000mand air were to be forced to rise to a point of saturation within this particular layer of the atmosphere
What is its diameter when the temperature is raised to 100 degrees Celsius? (b) What temperature change is required to increase its volume
The question is incomplete, the complete question is;
A spherical steel ball bearing has a diameter of 2.540cm at 25.00∘C.
(a) What is its diameter when it's temperature is raised to
100∘C?
(b) What temperature change is required to increase its volume by
1.000% ?
Answer:
a) 2.542 cm
b) 303.03°C
Explanation:
Given;
Diameter of the ball= 2.540cm
Initial temperature= 25.0°C
Final temperature= 100.0°C
Percentage increase in volume = 1.000%
Temperature coefficient of expansion for steel =11.0×10^−6/∘C
d2= d1[1 + α(T2-T1)]
d2= 2.540[1 + 11.0×10^−6(100-25)]
d2= 2.540[1 + 8.25×10^-4]
d2= 2.542 cm
From;
%V ×1/100 = V ×3α ×∆T/ V
Substituting values;
1.000 ×1/100= 3× 11.0×10^−6 × ∆T
∆T= 0.01/3× 11.0×10^−6
∆T= 303.03°C
A negative charge of -0.550 μC exerts an upward 0.900-N force on an unknown charge that is located 0.300 m directly below the first charge.
Required:
a. What is the value of the unknown charge (magnitude and sign)?
b. What is the magnitude of the force that the unknown charge exerts on the -0.590 μC charge?
c. What is the direction of this force?
Answer:
a. q2 = 16.4μC, positive charge
b. F = 0.900N
c. downward
Explanation:
a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:
[tex]F_e=k\frac{q_1q_2}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
r: distance between the charges = 0.300m
q1: charge 1 = -0.550 μC = 0.550*10^-6C
q2: charge 2 = ?
Fe: electric force = 0.900N
The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.
You solve the equation (1) for the second charge ans replace the values of the other parameters:
[tex]q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C[/tex]
The values of the second charge is 1.64 μC
b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.
The force exerted on the first charge is 0.900N
c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.
Two 40 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb
Answer:
The power dissipated by each bulb is [tex]P = 10.0 \ W[/tex]
Explanation:
From the question we are told that
The power rating of both bulbs is [tex]P = 40 \ W[/tex]
The voltage rating of both bulb is [tex]V = 120 \ V[/tex]
The both bulbs are connected a voltage of [tex]V_C = 120 V[/tex]
The amount of power rating of each bulb is mathematically represented as
[tex]P = \frac{V^2}{R }[/tex]
=> [tex]R = \frac{V^2}{P}[/tex]
substituting values
[tex]R = \frac{ (120)^2}{40}[/tex]
[tex]R = 360 \Omega[/tex]
Now given that the bulbs are connected is series, the equivalent resistance is evaluated as
[tex]R_{eq } = R +R[/tex]
substituting values
[tex]R_{eq } = 360 + 360[/tex]
[tex]R_{eq } =720 \ \Omega[/tex]
The current flowing through the bulbs is mathematically evaluated as
[tex]I =\frac{V_C}{R_{eq}}[/tex]
substituting values
[tex]I =\frac{120}{720}[/tex]
[tex]I = 0.1667 \ A[/tex]
Now the power dissipated by both bulbs is mathematically represented as
[tex]P = I ^2 * R[/tex]
substituting values
[tex]P = 0.1668^2 * 360[/tex]
[tex]P = 10.0 \ W[/tex]
The power that should be dissipated by each bulb is P = 10.0 W.
Calculation of the power:
Since
The power rating of both bulbs is P = 40 W.
The voltage rating of both bulbs is V = 120 V.
And, both bulks that should be connected a voltage of Vc = 120V
Now the amount of power that should be rated of each bulb should be
P = V^2/R
So, R = V^2/P
= 120^2/40
= 360Ω
The equivalent resistance should be
I = Vc/Req
= 120/720
= 0.1667 A
Now the power is = 0.1668^2 * 360
= 10.0 W
Learn more about power here: https://brainly.com/question/14089891
An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V