Answer:
6.3 m/s
Explanation:
From the given information:
The displacement (x) = 15 m
time (t) = 7.0 s
initial velocity = -2.0 m/s (since it is moving in the opposite direction)
We need to determine the acceleration then find the final velocity.
By applying the kinematics equation:
[tex]x = ut + \dfrac{1}{2}at^2[/tex]
[tex]15 = (-2.0)(7.0) + \dfrac{1}{2}a(7.0)^2\\ \\ 15 = -14.0 + \dfrac{49}{2}a \\ \\ 29= 24.5a \\ \\ a= \dfrac{29}{24.5} \\ \\ a = 1.184 \ m/s^2[/tex]
Now, to determine the final velocity by using the equation:
v = u + at
v = -2 + 1.184(7.0)
v = 6.288 m/s
v ≅ 6.3 m/s
Distillation is the separation of multiple Choose... components based on their different Choose... . As the mixture is heated and the first component Choose... , its Choose... form travels through the distillation set-up and Choose... into a different container.
Answer:
Explanation:
Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container
Question 3 of 10
Which statement describes the law of conservation of energy?
A. Air resistance has no effect on the energy of a system.
B. Energy cannot be created or destroyed.
C. The total energy in a system can only increase.
D. Energy cannot change forms.
هما
SUBMIT
Answer:
B . energy cannot be created or destroyed
A horizontal force of P=100 N is just sufficient to hold the crate from sliding down the plane, and a horizontal force of P=350 N is required to just push the crate up the plane. Determine the coefficient of static friction between the plane and the crate, and find the mass of the crate.
"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.
The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)
Using Newton's second law, we set up the following equations.
• p = 100 N
∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0
∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0
• P = 350 N
∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0
∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0
(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)
Solve for n and N :
n = p sin(θ) + mg cos(θ)
N = P sin(θ) - mg cos(θ)
Substitute these into the corresponding equations containing µ, and solve for µ :
µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))
µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
Next, you would set these equal and solve for m :
(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))
...
Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with
m ≈ 36.5 kg
µ ≈ 0.256
The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.
From the given,
The force that opposes the crate by sliding is P = 100N
In X-axis, the sum of forces is zero.
ΣF = 0
Pcosθ - mgsinθ-Ff = 0
Ff = Pcosθ - mgsinθ
In Y-axis
Psinθ - mgcosθ - N = 0
N = Psinθ-mgcosθ
Frictional force, Ff = μN, μ is the coefficient of friction
Ff = μN
Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0
μ = mgsin30-Pcos30/Psin30+mgcos30 ------1
The block is sliding with the horizontal force, F = 350N
X-axis
P₂cosθ - mgsinθ-Ff = 0
Y-axis
P₂sinθ - mgcosθ - N = 0
N = P₂sinθ-mgcosθ
μ = P₂cos30-mgsin30/P₂sin30-mgcos30 -----2
Equate equations 1 and 2
mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30
4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49
41.7m² + 123m - 1.516×10⁴ = 0
-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)
83.4m² - 2207m -3.03×10⁴ = 0
m= 36.4 kg
Hence, the mass of the crate is 36.4 Kg.
Substitute the value of m in equation 1,
μ = 4.905(36.4) - 86.6 / 50 + 8.49
μ = 0.256
Thus, the coefficient of static friction is 0.256.
To learn more about friction and its types:
https://brainly.com/question/30886698
#SPJ1
The working substance of a certain Carnot engine is 1.90 of an ideal
monatomic gas. During the isothermal expansion portion of this engine's
cycle, the volume of the gas doubles, while during the adiabatic expansion
the volume increases by a factor of 5.7. The work output of the engine is
930 in each cycle.
Compute the temperatures of the two reservoirs between which this engine
operates.
Answer:
Explanation:
The energy for an isothermal expansion can be computed as:
[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)
However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:
[tex]V_b = 2V_a[/tex]
Equation (1) can be written as:
[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]
Also, in a Carnot engine, the efficiency can be computed as:
[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]
[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]
In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]
relating the above two equations together, we have:
[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]
Making the work done (W) the subject:
[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]
From equation (1):
[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]
[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]
If we consider the adiabatic expansion as well:
[tex]PV^y[/tex] = constant
i.e.
[tex]P_bV_b^y = P_cV_c^y[/tex]
From ideal gas PV = nRT
we can have:
[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]
[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]
From the question, let us recall aw we are being informed that:
If the volumes changes by a factor = 5.7
Then, it implies that:
[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]
∴
[tex]T_H = T_L (5.7)^{y-1}[/tex]
In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]
As such:
[tex]T_H = T_L (5.7)^{1.6-1}[/tex]
[tex]T_H = T_L (5.7)^{0.67}[/tex]
Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]
[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]
From in the question:
W = 930 J and the moles = 1.90
using 8.314 as constant
Then:
[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]
[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]
[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]
[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]
From [tex]T_H = T_L (5.7)^{0.67}[/tex]
[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]
[tex]\mathbf{T_H \simeq 125K}[/tex]
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
A 10.0kg of desk initial is pushed along a frictionless surface by a constant horizontal of force magnitude 12N Find the speed of the desk after it has moved through a horizontal distance of 5.0m
kylydljty many true dvx*&;'*+$_5+
A tank is full of water. Find the work (in J) required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1,000 kg/m3 as the density of water. Round your answer to the nearest whole number.)
The electric potential ( relative to infinity ) due to a single point charge Q is 400 V at a point that is 0.6 m to the right of Q. The electric potential (relative to infinity) at a point that is 0.90 m to the left of 0 is:_____.
A. + 400 V.
B. -400 V.
C. + 200 V.
Answer:
The potential at a distance of 0.9 m is 266.67 V.
Explanation:
Charge = Q
Potential is 400 V at a distance 0.6 m .
Let the potential is V at a distance 0.9 m.
Use the formula of potential.
[tex]V = \frac{Kq}{r}\\\\\frac{V}{400}=\frac{0.6}{0.9}\\\\V = 266.67 V[/tex]
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building
Answer:
The length of his shadow is decreasing at a rate of 1.13 m/s
Explanation:
The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.
Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.
By similar triangles,
H/D = h/L
L = hD/H
Also, when the man is 4 m from the building, the length of his shadow is L = D - 4
So, D - 4 = hD/H
H(D - 4) = hD
H = hD/(D - 4)
Since h = 2 m and D = 12 m,
H = 2 m × 12 m/(12 m - 4 m)
H = 24 m²/8 m
H = 3 m
Since L = hD/H
and h and H are constant, differentiating L with respect to time, we have
dL/dt = d(hD/H)/dt
dL/dt = h(dD/dt)/H
Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)
Since h = 2 m and H = 3 m,
dL/dt = h(dD/dt)/H
dL/dt = 2 m(-1.7 m/s)/3 m
dL/dt = -3.4/3 m/s
dL/dt = -1.13 m/s
So, the length of his shadow is decreasing at a rate of 1.13 m/s
The relation of mass m, angular velocity o and radius of the circular path r of an object with the centripetal force is-
a. F = m²wr
b. F = mwr²
c. F = mw²r
d. F = mwr.
Answer:
Correct option not indicated
Explanation:
There are few mistakes in the question. The angular velocity ought to have been denoted with "ω" and not "o" (as also suggested in the options).
The formula to calculate a centripetal force (F) is
F = mv²/r
Where m is mass, v is velocity and r is radius
where
While the formula to calculate a centrifugal force (F) is
F = mω²r
where m is mass, ω is angular velocity and r is radius of the circular path.
From the above, it can be denoted that the relationship been referred to in the question is that of a centrifugal force and not centripetal force, thus the correct option should be C.
NOTE: Centripetal force is the force required to keep an object moving in a circular path/motion and acts inward towards the centre of rotation while centrifugal force is the force felt by an object in circular motion which acts outward away from the centre of rotation.
Explain the following defects of a simple electric cell:
a.Polarization,
ß. Local action.
Answer:
Explanation:
The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.
a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.
b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.
A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute. If the package descends at a constant vertical acceleration of 6.89 m/s2 the initial plane horizontal speed was 86.9 m/s, how far away from the survivor will it hit the waves
Answer: 19.15 meters on the waves away from the survivor.
Explanation:
two resistors with resistance values 4.5 ohms and 2.3 ohms are connected in series or parallel across a potential difference of 30V to a light bulb find the current flowing through the light bulb in both cases
Answer:
Look at work
Explanation:
Series:
I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.
Parallel:
V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.
In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.
With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.
Part A
What is the most significant conclusion that Thomson was able to draw from his measurements?
He found a different value of q/m for different cathode materials.
He found the same value of q/m for different cathode materials.
From measurements of q/m he was able to calculate the charge of an electron.
From measurements of q/m he was able to calculate the mass of an electron.
Part B
What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.
Part C
Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.
What is the speed v0 of the electrons in this case?
Express your answer in terms of E0 and B0.
Answer:
a) He found the same value of q/m for different cathode materials.
b) y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex] , c) v = [tex]\frac{E_o}{B_o}[/tex]
Explanation:
In Thomson's experiments he was able to measure the deflection of the light beam under the effect of the magnetic field and with these results find the e / m relationship, which in all cases is the same, therefore the most important conclusion is that the value e E / m is constant for all materials.
b) In the part of the plates the electrons are accelerated by the electric field,
F = ma
- e E = m a
a = - (e/m) E₀
the distance traveled is
X axis
x = v₀ t
the separation of the plates is x = d
t = vo / d
Y axis
y = v_{oy} t + ½ to t²
y = ½ a t²
y = [tex]- \frac{e}{m}\ \frac{E_o v_o^2 }{2d^2}[/tex]
c) In this case there is a magnetic field B₀ and the electrons have no deflection
F = - e E + e v x B
if there is no deviation F = 0
e E = e v B
v = [tex]\frac{E_o}{B_o}[/tex]
When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is 1.0 megohm (106 Ω). How much current passes through the insulation of the motor if the test voltage is 1000 V?
Answer:
0.001 A
Explanation:
Applying,
V = IR.............. Equation 1
Where V = Voltage of the motor, I = current, R = resistance
make I the subject of the equation
I = V/R.............. Equation 2
From the question,
Given: V = 1000 V, R = 1 MΩ = 10⁶ Ω
Substitute these values into equation 2
I = 1000/10⁶
I = 10⁻³ A
I = 0.001 A
A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Answer:
a. 8p
Explanation:
We are given that
Radius of hollow sphere , R1=R
Density of hollow sphere=[tex]\rho[/tex]
After compress
Radius of hollow sphere, R2=R/2
We have to find density of the compressed sphere.
We know that
[tex]Density=\frac{mass}{volume}[/tex]
[tex]Mass=Density\times volume=Constant[/tex]
Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]
Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]
Using the formula
[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]
[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]
[tex]\rho_2=8\rho[/tex]
Hence, the density of the compressed sphere=[tex]8\rho[/tex]
Option a is correct.
A piston-cylinder device contains 5 kg of refrigerant-134a at 0.7 MPa and 60°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24°C. If the surroundings are at 100 kPa and-24°C, determine:
(a) the exergy of the refrigerant at the initial and the final states and
(b) the exergy destroyed during this process.
Answer:
Yes sure, keep it going, and never give up because your dreams are so important
A) The exergy of the refrigerant at the initial and final states are :
Initial state = - 135.5285 kJ Final state = -51.96 kJB) The exergy destroyed during this process is : - 1048.4397 kJ
Given data :
Mass ( M ) = 5 kg
P1 = 0.7 Mpa = P2
T1 = 60°C = 333 k
To = 24°C = 297 k
P2 = 100 kPa
A) Determine the exergy at initial and final states
At initial state :
U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k
exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )
= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)
≈ - 135.5285 kJ
At final state :
U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k
exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )
= -51.96 kJ
B) Determine the exergy destroyed
exergy destroyed = To * M ( S2 - S1 )
= 297 * 5 ( 0.31958 - 1.0256 )
= - 1048.4397 KJ
Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ
Learn more about exergy : https://brainly.com/question/25534266
Find the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
Preocupada com o aumento da tarifa na conta de luz, uma pessoa resolve economizar diminuindo o tempo de banho de 20 para 15 minutos. Seu chuveiro possui as seguintes especificações: 4200 W e 220V. Sabendo que o kWh custa R$0,30, a economia feita em 10 dias foi de aproximadamente
Answer:
The mount saved is $ 0.105.
Explanation:
Concerned about the increase in the electricity bill, a person decides to save by reducing bathing time from 20 to 15 minutes. Your shower has the following specifications: 4200 W and 220V. Knowing that the kWh costs R$0.30, the savings made in 10 days were approximately.
The electrical energy is given by
E = P x t
where, P is the electrical power and t is the time.
When he is using the shower for 20 minutes, the energy consumed is
E = 4200 x 20 x 60 = 5040,000 J = 1.4 kWh
When he is using the shower for 15 minutes, the energy consumed is
E' = 4200 x 15 x 60 = 3780000 J = 1.05 kWh
The difference in energy is
E'' = E - E' = 1.4 - 1.05 = 0.35 kWh
The money saved is
= 4 0.3 x 0.35 = $ 0.105
In a certain cyclotron a proton moves in a circle of radius 0.530 m. The magnitude of the magnetic field is 1.30 T. (a) What is the oscillator frequency
Answer:
[tex]f=1.98\times 10^7\ Hz[/tex]
Explanation:
Given that,
The radius of circle, r = 0.53 m
The magnitude of the magnetic field, B = 1.3 T
We need to find the oscillator frequency. It is given by :
[tex]f=\dfrac{qB}{2\pi m}[/tex]
Put all the values,
[tex]f=\dfrac{1.6\times 10^{-19}\times 1.3}{2\pi \times 1.67\times 10^{-27}}\\\\f=1.98\times 10^7\ Hz[/tex]
So, the oscillator frequency is [tex]1.98\times 10^7\ Hz[/tex].
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positive direction, what is the truck's resultant displacement
Answer:
140m east
Explanation:
If East is positive then lets rephrase the problem into integers
A truck moves +70 m, then moves -120m, and finally moves +90m.
So totally Displacement = +70-120+90= +140m
Since east is positive, the trucks resultant displacement is 140 m east of origin
The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?
Answer:
a) the required angle in both radian and degree is 1.25 rad and 71.6°
b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
Explanation:
Given the data in the question;
a)
we know that The expression for the angle subtended by an arc of circle at the center of the circle is,
θ = Length / radius
given that Length is 5000 miles and radius is 4000 miles
we substitute
θ = 5000 miles / 4000 miles
θ = 1.25 rad
Radian to Degree
θ = 1.25 rad × ( 180° / π rad )
θ = 71.6°
Therefore, required angle in both radian and degree is 1.25 rad and 71.6°
b)
The flight from Kampala to Singapore take 9 hours.
the plane's angular speed relative to the earth = ?
we know that, the relation between angular velocity and angular displacement is;
ω = θ / t
given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec
we substitute
ω = 1.25 rad / 32400 sec
ω = 3.86 × 10⁻⁵ rad/sec
Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
what is the force of a body which have mass of 7 kg
Answer:
Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N
Answer:
10 m/s2 or 70 newtons.
Explanation:
............................
............
A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would result in an induced emf in the loop?
A. Moving the loop outside of the magnetic field region.
B. Change the diameter of the loop.
C. Change the magnitude of the magnetic field.
D. Spin the loop such that its axis does not consistently line up with the magnetic field direction.
Answer:
All the given options will result in an induced emf in the loop.
Explanation:
The induced emf in a conductor is directly proportional to the rate of change of flux.
[tex]emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta[/tex]
where;
A is the area of the loop
B is the strength of the magnetic field
θ is the angle between the loop and the magnetic field
Considering option A, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.
Considering option B, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.
Option C has a similar effect with option A, thus both will result in an induced emf.
Finally, considering option D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will change the angle between the loop and the magnetic field. This effect will also result in an induced emf.
Therefore, all the given options will result in an induced emf in the loop.
now suppose that we have attached not just two springs in series, but N springs. Write an equation that expresses the effective spring constant of the combination using the spring constant of the original spring k and the number of springs N
Answer:
[tex]k_{eq} = \frac{k}{N}[/tex]
Explanation:
For this exercise let's use hooke's law
F = - k x
where x is the displacement from the equilibrium position.
x = [tex]- \frac{F}{k}[/tex]
if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs
x_ {total} = ∑ x_i
we substitute
x_ {total} = ∑ -F / ki
F / k_ {eq} = -F [tex]\sum \frac{1}{k_i}[/tex]
[tex]\frac{1}{k_{eq}} = \frac{1}{k_i}[/tex] 1 / k_ {eq} = ∑ 1 / k_i
if all the springs are the same
k_i = k
[tex]\frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\[/tex]
[tex]\frac{1}{k_{eq} } = \frac{N}{k}[/tex]
[tex]k_{eq} = \frac{k}{N}[/tex]
Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Compare the time it takes each of them to reach the lake below.
a. Alice reaches the surface of the lake first
b. Tom reaches the surface of the lake first
c. Alice and Tom will reach the surface of the lake at the same time.
Part AFind the x- and y-components of the vector d⃗ = (4.0 km , 29 ∘ left of +y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.d⃗ = km Part BFind the x- and y-components of the vector v⃗ = (2.0 cm/s , −x-direction).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.v⃗ = cm/s Part CFind the x- and y-components of the vector a⃗ = (13 m/s2 , 36 ∘ left of −y-axis).Express your answer using two significant figures. Enter the x and y components of the vector separated by a comma.a⃗ x = m/s2
Solution :
Part A .
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, d = [tex]\text{4 km 29}[/tex] degree left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -4 x sin (29°) = -1.939 km
[tex]y[/tex] component is = 4 x cos (29°) = 3.498 km
Part B
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{v = 2 cm/s}[/tex] , [tex]\text{-x direction}[/tex]
So the [tex]x[/tex] component is = -2 cm/s
[tex]y[/tex] component is = 0
Part C
Given : The [tex]x[/tex] and [tex]y[/tex] components of the vector, [tex]\text{a = 13 m/s, 36 degree}[/tex] left of [tex]y[/tex]-axis.
So the [tex]x[/tex] component is = -13 x sin (36°) = -7.6412 [tex]m/S^2[/tex]
[tex]y[/tex] component is = -13 x cos (36°) = -10.517 [tex]m/S^2[/tex]
The x- and y-components of the vectors is mathematically given as as follows for each Part respectively
x= -1.939 km, y= 3.498 km
x= -2 cm/s, 0
y=, x= -7.6412m/s^2, -10.517m/s^2
What are the x- and y-components of the vectors?
Question Parameters:
Generally, we follow a basic principle where
x component= Fsin\theta
y component= Fcos\theta
Therefore
For A
x component is
x= -4 x sin (29°)
x= -1.939 km
y component is
y= 4 x cos (29°)
y= 3.498 km
For B
x component is
x= -2 cm/s
y component is
y= 0
For C
x component is
x= -13 x sin (36°)
x= -7.6412m/s^2
y component is
y= -13 x cos (36°)
y= -10.517m/s^2
Read more about Cartession co ordinate
https://brainly.com/question/9410676
a circuit shown below is Wheastone Bridge used to determine the valve of unknown resistor X by comparison with three resistors M,N,P whose resistances can be varied. For each setting, the resistances of each resistor is precisely known. With switches k1and k2 closed, these resistors are varied until the current in the galvanometer G is zero; the bridge is then said to be balanced. (a) if the galvanometer G shows zero deflection when M=850.0, N=15.00 and P=33.48, what is the unknown resistance X?
Answer:
X = 0.6
Explanation:
The resistance of the unknown resistor can be found by using the formula of the Wheatstone bridge:
[tex]\frac{M}{N}=\frac{P}{X}\\\\\frac{850}{15} = \frac{33.48}{X}\\\\X = \frac{(33.48)(15)}{850}[/tex]
X = 0.6
Hence, the unknown value of resistance is found to be 0.6 units.
A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.
Answer:
a)[tex]V=1.067\: m/s[/tex]
b)[tex]v=434.65\: m/s [/tex]
Explanation:
a)
Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.
[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]
Where:
M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)Then, let's find the initial speed of the bullet-block system.
[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]
[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]
[tex]V=1.067\: m/s[/tex]
b)
Using the conservation of momentum we can find the velocity of the bullet.
[tex]mv=MV[/tex]
[tex]v=\frac{MV}{m}[/tex]
[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]
[tex]v=434.65\: m/s [/tex]
I hope it helps you!
A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?
Answer:
3.44 rad
Explanation:
The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk
Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is ΔK = 21 J/s × 3.3 s = 69.3 J
So, ΔK = 1/2I(ω² - ω₀²)
Since ω₀ = 0 rad/s
ΔK = 1/2I(ω² - 0)
ΔK = 1/2Iω²
ΔK = 1/2(MR²/2)ω²
ΔK = MR²ω²/4
ω² = (4ΔK/MR²)
ω = √(4ΔK/MR²)
ω = 2√(ΔK/MR²)
Substituting the values of the variables into the equation, we have
ω = 2√(ΔK/MR²)
ω = 2√(69.3 J/( 4 kg × (4 m)²))
ω = 2√(69.3 J/[ 4 kg × 16 m²])
ω = 2√(69.3 J/64 kgm²)
ω = 2√(1.083 J/kgm²)
ω = 2 × 1.041 rad/s
ω = 2.082 rad/s
The angular displacement θ is gotten from
θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²
Substituting the values of the variables into the equation, we have
θ = ω₀t + 1/2αt²
θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²
θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²
θ = 1/2 × 6.87159 rad
θ = 3.436 rad
θ ≅ 3.44 rad