Answer:
The apparent weight of the pilot is 3311 N
Explanation:
Given;
radius of the vertical circle, r = 600 m
speed of the plane, v = 150 m/s
mass of the pilot, m = 70 kg
Weight of the pilot due to his circular motion;
[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]
Real weight of the pilot;
[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]
Apparent weight - Real weight of pilot = weight due to centripetal force
[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]
Therefore, the apparent weight of the pilot is 3311 N
A 750 kg car is moving at 20.0 m/s at a height of 5.0 m above the bottom of a hill when it runs out of gas. From there, the car coasts. a. Ignoring frictional forces and air resistance, what is the car’s kinetic energy and velocity at the bottom of the hill
Answer:
Explanation:
Kinetic energy at the height = 1/2 m v²
= 1/2 x 750 x 20²
= 150000 J
Its potential energy = mgh
= 750 x 9.8 x 5
=36750 J
Total energy = 186750 J
Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy
If v be its velocity at the bottom
1/2 m v² = 186750
v = √498
= 22.31 m /s
PLEASE HELP I’LL MARK YOU BRAINLIEST!!!!
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.
Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?
Answer:
Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second
Explanation:
Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg
Velocity of the car is 2.5 meter per second
Since formula to calculate the momentum of an object is,
p = mv
Where, p = momentum of the object
m = mass of the object
v = velocity of the object
By substituting these values in the formula,
p = [tex](6.3\times 10^{-3})\times 2.5[/tex]
= [tex]1.575\times 10^{-2}[/tex] Kg meter per second
Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.
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.
Please help! Which statements correctly describe the effect of distance in determining the gravitational force and the electrical force? Check all that apply.
There are six statements on the list.
The first 2 are true, and the last 2 are true.
The 2 in the middle aren't true. They are false.
If a sample of 346 swimmers is taken from a population of 460 swimmers,
the population mean, w, is the mean of how many swimmers' times?
Answer:
It is the mean of 460 swimmers
Explanation:
In this question, we are concerned with knowing the mean of the population w
Now, according to the question at hand, we have a total population of 460 swimmers and we have taken out 346 swimmers for a study
The population mean in this case is simply the mean of the swimming times of the 460 swimmers
There is another related thing here called the sample mean. For the sample mean, we only make a reference to the mean of the 346 swimmers who were taken out from the population to conduct a separate study
So conclusively, the population mean w is simply the mean of the total 460 swimmers
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]
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.
A particle moves along a straight line with the acceleration a = (12t - 3t ^ 1/2) feet / s ^ 2, where t is in seconds. Determine your speed and position as a function of time. When t = 0, v = 0 and s = 15 feet.
Answer:
v = 6t² − 2t^³/₂
s = 2t³ − ⅘t^⁵/₂ + 15
Explanation:
a = 12t − 3t^½
Integrate to find velocity.
v = ∫ a dt
v = ∫ (12t − 3t^½) dt
v = 6t² − 2t^³/₂ + C
Use initial condition to find C.
0 = 6(0)² − 2(0)^³/₂ + C
C = 0
v = 6t² − 2t^³/₂
Integrate to find position.
s = ∫ v dt
s = ∫ (6t² − 2t^³/₂) dt
s = 2t³ − ⅘t^⁵/₂ + C
Use initial condition to find C.
15 = 2(0)³ − ⅘(0)^⁵/₂ + C
15 = C
s = 2t³ − ⅘t^⁵/₂ + 15
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
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
A box with an initial speed of 15 m/s slides along a surface where the coefficient of sliding friction is 0.45. How long does it take for the block to come to rest
Answer:
t = 3.4 s
The box will come to rest in 3.4 s
Explanation:
For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma
Frictional Force = μR = μW = μmg
Therefore,
ma = μmg
a = μg
where,
a = acceleration of box = ?
μ = coefficient of sliding friction = 0.45
g = 9.8 m/s²
Therefore,
a = (0.45)(9.8 m/s²)
a = -4.41 m/s² (negative sign due to deceleration)
Now, for the time to stop, we use first equation of motion:
Vf = Vi + at
where,
Vf = Final Speed = 0 m/s (since box stops at last)
Vi = Initial Speed = 15 m/s
t = time to stop = ?
Therefore,
0 m/s = 15 m/s + (-4.41 m/s²)t
(-15 m/s)/(-4.41 m/s²) = t
t = 3.4 s
The box will come to rest in 3.4 s
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
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 ;)
If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answer:
The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
[tex]r=\frac{L}{2 \pi}\\[/tex]
[tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]
area of the loop Is:
[tex]A_L= \pi r^2[/tex]
[tex]=100.2001\times 3.14\\\\=314.628[/tex]
change in magnetic field is:
[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]
then the induced emf is: [tex]e = A_L \times \frac{dB}{dt}[/tex]
[tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]
resistivity of the copper wire is: [tex]\rho =[/tex] 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
[tex]= 0.5 \times 10^{-3} \ m[/tex]
hence the resistance of the wire is:
[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]
[tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]
Power:
[tex]P=\frac{e^2}{R}[/tex]
[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]
The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]
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
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
Object 2 has the larger drag coefficient
Explanation:
The drag force, D, is given by the equation:
[tex]D = 0.5 c \rho A v^2[/tex]
Object 1 has twice the diameter of object 2.
If [tex]d_2 = d[/tex]
[tex]d_1 = 2d[/tex]
Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]
Area of object 1:
[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]
Since all other parameters are still the same except the drag coefficient:
For object 1:
[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]
For object 2:
[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]
Since the drag force for the two objects are the same:
[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]
Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient
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]
In a circus act, a uniform board (length 3.00 m, mass 25.0 kg ) is suspended from a bungie-type rope at one end, and the other end rests on a concrete pillar. When a clown (mass 79.0 kg ) steps out halfway onto the board, the board tilts so the rope end is 30∘ from the horizontal and the rope stays vertical. Calculate the force exerted by the rope on the board with the clown on it.
Answer:
Force of Rope = 122.5 N
Force of Rope = 480.2N
Explanation:
given data
length = 3.00 m
mass = 25.0 kg
clown mass = 79.0 kg
angle = 30°
solution
we get here Force of Rope on with and without Clown that is
case (1) Without Clown
pivot would be on the concrete pillar so Force of Rope will be
Force of Rope × 3m = (25kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 122.5 N
and
case (2) With Clown
so here pivot is still on concrete pillar and clown is standing on the board middle and above the centre of mass so Force of Rope will be
Force of Rope × 3m = (25kg+73kg)×(9.8ms²)×(1.5m)
solve it and we get
Force of Rope = 480.2N
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
In a sinusoidally driven series RLC circuit, the inductive resistance is XL = 100 Ω, the capacitive reactance is XC = 200 Ω, and the resistance is R = 50 Ω. The current and applied emf would be in phase if
Answer:
The current and the applied emf can be in phase if either of the two changes are made.
1) The inductance of the inductor is doubled, with everything else remaining constant.
2) The capacitance of the capacitor is doubled, with everything else remaining constant.
Explanation:
The current and applied emf for this type of circuit would be in phase when there is no phase difference between the two quantities. That is, Φ = 0°.
The phase difference between current and applied emf is given as
Φ = tan⁻¹ [(XL - Xc)/R]
XL = Impedance due to the inductor
Xc = Impedance due to the capacitor
R = Resistance of the resistor.
For Φ to be 0°, tan⁻¹ [(XL - Xc)/R] = 0
But only tan⁻¹ 0 = 0 rad
So, for the phase difference to be 0,
[(XL - Xc)/R] = 0
Meaning
XL = Xc
But for this question,
XL = 100 Ω, Xc = 200 Ω
For them to be equal, we have to find a way to increase the impedance of the inductor or reduce the impedance of the capacitor.
The impedance are given as
XL = 2πfL
Xc = (1/2πfC)
f = Frequency
L = Inductance of the inductor
C = capacitance of the capacitor
The impedance of the inductor can be increased from 100 Ω to 200 Ω by doubling the inductance of the inductor.
And the impedance of the capacitor can be reduced from 200 Ω to 100 Ω by also doubling the capacitance of the capacitor.
So, these are either of the two ways to make the current and applied emf to be in phase.
Hope this Helps!!!
A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.
Answer:
the internal energy of the mixture at final state = 238kJ/kg
Explanation:
Given
V= 0.6m³
m=5kg
R=0.287kJ/kg.K
T=320 K
from ideal gas equation
PV = nRT
where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.
Recall, mole = mass/molar mass
attached is calculation of the question.
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car
Answer:
9,375
Explanation:
Data provided
The mass of the car m = 1500 Kg.
The diameter of the circular track D = 200 m.
For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-
The radius of the circular path is
[tex]R = \frac{D}{2}[/tex]
[tex]= \frac{200}{2}[/tex]
= 100 m
after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula
[tex]Force F = \frac{mv^2}{R}[/tex]
[tex]= \frac{1500\times (25)^2}{100}[/tex]
= 9,375
Therefore for computing the magnitude of the centripetal force we simply applied the above formula.
The lower the value of the coefficient of friction, the____the resistance to sliding
Answer: lower
There are a number of factors that can affect the coefficient of friction, including surface conditions.
Values of the coefficient of sliding friction can be a good reference for specific combinations of materials. The frictional force and normal reaction are directly proportion but an increase or decrease in coefficient of friction will cause an increase or decrease in the resistance of sliding respectively
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.
Learn more here: https://brainly.com/question/18247518
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
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is
The lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.
What is diffraction?The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So, The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.
Learn more about diffraction here:
https://brainly.com/question/11176463
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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]