Answer: The volume of an irregularly shaped object is 0.50 ml
Explanation:
To calculate the volume, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Density of object = [tex]6.0g/ml[/tex]
mass of object = 3.0 g
Volume of object = ?
Putting in the values we get:
[tex]6.0g/ml=\frac{3.0g}{\text{Volume of substance}}[/tex]
[tex]{\text{Volume of substance}}=0.50ml[/tex]
Thus the volume of an irregularly shaped object is 0.50 ml
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex] (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]
The strength of the magntetic field is 4.1mT
A commercial diffraction grating has 500 lines per mm. Part A When a student shines a 480 nm laser through this grating, how many bright spots could be seen on a screen behind the grating
Answer:
The number of bright spot is m =4
Explanation:
From the question we are told that
The number of lines is [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]
The wavelength of the laser is [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]
Now the the slit is mathematically evaluated as
[tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]
Generally the diffraction grating is mathematically represented as
[tex]dsin\theta = m \lambda[/tex]
Here m is the order of fringes (bright fringes) and at maximum m [tex]\theta = 90^o[/tex]
So
[tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]
=> [tex]m = 4[/tex]
This implies that the number of bright spot is m =4
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by
Answer:
the two beams must travel paths that differ by one-half of a wavelength.
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Complete Question
A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex] on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is [tex]E = 4.5198 *10^{5} \ N/C[/tex]
Explanation:
From the question we are told that
The charge density is [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]
The position outside the surface is [tex]a = 3.00 \ mm = 0.003 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon _o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with values [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]
substituting values
[tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]
[tex]E = 4.5198 *10^{5} \ N/C[/tex]
1. The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere
Complete Question
The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).
B = N*mue*I/(2*r)
# of loops = 140
radius of the coil = 0.14m
Answer:
The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]
Explanation:
From the question we are told that
The coil current is [tex]I = 1.5 \ A[/tex]
The number of loops is [tex]N = 140[/tex]
The magnetic field due to the current is mathematically represented as
[tex]B = \mu_o * N * I[/tex]
[tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]
[tex]B = 2.639*19^{-4} \ T[/tex]
From question
(1 Tesla = [tex]10^4 \ Gauss[/tex]).
=> [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]
=> [tex]B = 2.639 \ Gauss[/tex]
In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change
Answer:
The angular speed of the ball will increase
Explanation:
the angular speed of the ball will increase because the force of hit by the players will sum up in opposite direction to increase the angular speed
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
Answer:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
Explanation:
The magnitude of force applied by each charge on one another can be given by Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
where,
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now, in the final state the charges on both spheres are halved. Therefore,
q₁' = q₁/2
q₂' = q₂/2
Hence, the new force will be:
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
using equation 1:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
The magnitude of the electrostatic force will be F' = F/4
The magnitude of the electrostatic force:Here we used Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
Here
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now
q₁' = q₁/2
q₂' = q₂/2
So, the new force should be
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
So,
F' = F/4
Learn more about force here: https://brainly.com/question/14282312
A 16.0-m uniform ladder weighing 520 N rests against a frictionless wall. The ladder makes a 65.0° angle with the horizontal.
(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.20 m along the ladder from the bottom. Horizontal Force Magnitude = Direction = Vertical Force Magnitude = Direction =
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
Answer:
we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
Coefficient of static friction = 0.26
Explanation:
Given that:
length of the ladder = 16.0 m
weight of the ladder = 520 N
angle θ = 65.0°
(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :
Force = 850 N
distance of the climber from the base of the ladder = 4.20 m
The diagrammatic illustration representing what the given information entails can be seen from the attached file below.
Let consider the Ladder being at point A with the horizontal layer of the ground.
From the whole system; the condition for the equilibrium at the point A can be computed as :
[tex]N_2 (16 \ Sin\ 65) = 850(4.2 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(1.7749 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) = 1508.665+1758.016[/tex]
[tex]N_2 (14.50) = 3266.681[/tex]
[tex]N_2 =\dfrac{ 3266.681}{14.50}[/tex]
[tex]N_2 =225.28 \ N[/tex]
[tex]N_1 = mg+F\\[/tex]
where ;
w =mg
[tex]N_1 = 520+850[/tex]
[tex]N_1 = 1370 \ N[/tex]
Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.
(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?
the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:
[tex]N_2 (16 \ Sin\ 65) = 850(9.4 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]
[tex]N_2 (14.50) = 850(3.9726 )+ 520 (8) \times 0.4226[/tex]
[tex]N_2 (14.50) =3376.71+1758.016[/tex]
[tex]N_2 (14.50) =5134.726[/tex]
[tex]N_2 =\dfrac{5134.726}{14.50}[/tex]
[tex]N_2 =354.12 \ N[/tex]
Therefore; the coefficient of the static friction is;
[tex]\mu = \dfrac{f_s}{N_1}[/tex]
[tex]\mu = \dfrac{354.12}{1370}[/tex]
[tex]\mu[/tex] = 0.26
Coefficient of static friction = 0.26
Inexperienced physics teachers often demonstrate the use of the electroscope by touching it with a charged glass rod at a single point. More experienced teachers typically drag the length of the rod across the top of the electroscope to increase the desired effect. Why does this help
Answer:
This is because The glass rod of the electroscope is an insulator therefore only charge transferred to the ball is at the point of contact on the rod. Thus, When the charge rod is dragged across the top of the electroscope, by the experienced teacher the more charge is transferred to electroscope thereby producing a greater effect
1. In a Millikan type experiment, two horizontal plates are 2.5 cm apart. A latex sphere of
mass 1.5 x 10-15 kg remains stationary when the potential difference between the
plates is 460 V, with the upper plate positive. [2+2+2+2 = 8 marks]
a. Is the sphere charged negatively or positively?
b. What is the magnitude of the electric field intensity between the plates?
C. Calculate the magnitude of the charge on the latex sphere.
d. How many excess or deficit electrons does the sphere have?
Answer:
Explanation:
a. Is the sphere charged negatively or positively?
The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.
Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.
b. What is the magnitude of the electric field intensity between the plates?
The magnitude of the electric field intensity between the plates is 18400v/m.
C. Calculate the magnitude of the charge on the latex sphere.
The magnitude of the charge on the latex sphere hae been solved and attached
d. How many excess or deficit electrons does the sphere have?
There are 5 excess electrons that the sphere has.
Check the attachment for further explanation.
What is the increase in the internal energy of the system, if it absorbs 500 cal and at the same time does 85J work on its surroundings? (1 cal = 4.18J)
Answer:
2005 J
Explanation:
500 cal × (4.18 J/cal) = 2090 J
ΔU = q + w
ΔU = 2090 J − 85 J
ΔU = 2005 J
a car slows down from - 27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m .what is its acceleration?
Answer:
3.09 m/s²
Explanation:
Given:
Δx = -105 m
v₀ = -27.7 m/s
v = -10.9 m/s
Find: a
v² = v₀² + 2aΔx
(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)
a = 3.09 m/s²
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.
Complete question is;
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown
Answer:
It is proved that the final speed is truly 31.1 m/s
Explanation:
From energy - conservation principle;
E_i = Initial potential energy + Initial Kinetic Energy
Or
E_i = U_i + K_i
Similarly, for final energy
E_f = U_f + K_f
So, expressing the formulas for potential and kinetic energies, we now have;
E_i = (m × g × y_i) + (½ × m × v_i²)
Similarly,
E_f = (m × g × y_f) + (½ × m × v_f²)
We are given;
y_i = 25 m
y_f = 0 m
v_i = 20 m/s
v_f = 31.1 m/s
So, plugging in relevant values;
E_i = m((9.8 × 25) + (½ × 20²))
E_i = 485m
Similarly;
E_f = m((9.8 × 0) + (½ × v_f²)
E_f ≈ ½m•v_f²
From energy conservation principle, E_i = E_f.
Thus;
485m = ½m•v_f²
m will cancel out to give;
½v_f² = 485
v_f² = 485 × 2
v_f² = 970
v_f = √970
v_f ≈ 31.1 m/s
A projectile is launched with V0 = 7.6 m/s and initial angle = 1.27 radians above the horizontal. What is the initial horizontal component of the projectile velocity in miles per hour?
Answer:
The horizontal component is [tex]v_h = 1.7096 \ m/s[/tex]
Explanation:
A diagram illustrating the projection is shown on the first uploaded image (from IB Maths Resources from British international school Phuket )
From the question we are told that
The initial velocity is [tex]v_o = 7.6 \ m/s[/tex]
The angle of projection is [tex]\theta = 1.27 \ rad = 72.77^o[/tex]
The horizontal component of this projectile velocity is mathematically represented as
[tex]v_h = v_o * cos (\theta )[/tex]
substituting values
[tex]v_h = 7.6 * cos (72.77 )[/tex]
[tex]v_h = 1.7096 \ m/s[/tex]
James drove his car down a mountain road at an average rate of 30 mph and retunred over the same road at an average rate of 20 mph. How far did he drive down the road before he turned around and drove back if his trip took 5 hours?
A. 1B. 2C. 3D. 4E. 5
Answer:
150 miles
Explanation:
5 times 30 is 150
A commuter train passes a passenger platform at a constant speed of 40.0 m/s. The train horn is sounded at its characteristic frequency of 320 Hz.
(a) What overall change in requency is detected by a person on the platform as the train moves from approaching to receding?
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a) -75 Hz
b) 0.11 [tex]m^{-1}[/tex]
Explanation:
a) Let us first find the frequency detected by the person on the platform.
We have to find the frequency observed by the person when the train was approaching and when the train was receding.
When the train was approaching:
[tex]f_o = \frac{v}{v - v_s} f_s[/tex]
where fo = frequency observed
fs = frequency from the source = 320 Hz
v = speed of sound = 343 m/s
vs = speed of the train = 40 m/s
Therefore:
[tex]f_o = \frac{343}{343 - 40} * 320\\\\f_o = \frac{343}{303} * 320\\\\f_o = 362 Hz[/tex]
The person on the platform heard the sound at a frequency of 362 Hz when the train was approaching.
When the train was receding:
[tex]f_o = \frac{v}{v + v_s} f_s[/tex]
[tex]f_o = \frac{343}{343 + 40} * 320\\\\f_o = \frac{343}{383} * 320\\\\f_o = 287 Hz[/tex]
The person on the platform heard the sound at a frequency of 287 Hz when the train was receding.
Therefore, the frequency change is given as:
Δf = 287 - 362 = -75 Hz
b) We can find the wavelength detected by the person on the platform as the train approaches by using the formula for speed:
[tex]v = \lambda f[/tex]
where λ = wavelength
f = frequency of the train as it approaches = 362 Hz
v = speed of train = 40 m/s
Therefore, the wavelength detected is:
40 = λ * 362
λ = 40 / 362 = 0.11 [tex]m^{-1}[/tex]
mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Answer:
Explanation:
The current through the resistor is 0.83 A
.
Part b
The current through resistor is 0.53 A
.
Part c
The current through resistor is 0.30 A
Two charged concentric spheres have radii of 0.008 m and 0.018 m. The charge on the inner sphere is 3.62 10-8 C and that on the outer sphere is 1.62 10-8 C. Find the magnitude of the electric field (in N/C) at 0.012 m.
Answer:
The electric field is [tex]E = 2.2625 *10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner sphere is [tex]r_1 = 0.008\ m[/tex]
The radius of the outer sphere is [tex]r _2 = 0.018 \ m[/tex]
The charge on the inner sphere is [tex]q_1 = 3.62 *10^{-8} \ C[/tex]
The charge on the outer sphere is [tex]q_2 = 1.62 *10^{-8} \ C[/tex]
The position from the origin is [tex]d = 0.012 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{k (q_1 )}{ r^2}[/tex]
The reason for using [tex]q_1[/tex] for the calculation is due to the fact that the position considered is greater than the [tex]r_1[/tex] but less than [tex]r_2[/tex]
Here k is the Coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}[/tex]
So
[tex]E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}[/tex]
[tex]E = 2.2625 *10^{6} \ N/C[/tex]
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?
An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move to reduce his exposure to 2mRem/hr.? I1/I2 = (D2)2/(D1)2 -------> I1(D1)2 = I2(D2)2
Answer:
4.7ft
Explanation:
Pls see attached file
Which best describes the relationship between heat, internal energy, and thermal energy?
Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.
Mark this and return
Save and Exit
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Submit
Answer:
I think it is the 4th answer choice
Explanation:
Heat is thermal energy that flows in the direction of high temp to low temp, and internal energy is the "energy contained in a system", and thermal energy is a part of that.
Suppose you take two non-zero displacements represented by vectors A & B.The magnitude of A is 5 m and the magnitude of B is 6 m. Under what circumstances can you end up back at your starting point? What is the magnitude of the largest displacement you can end up from the starting point?
Suppose you take two non-zero displacements represented by vectors A & B which are perpendicular to each other. The magnitude of A is 5 m and the magnitude of B is 6 m. What is the component of vector B along the direction of vector A measured in m?
Answer:
a. When the total displacement is -(A + B)
b. A + B = 1 m or -(A + B) = -11 m
c. 0 m
Explanation:
a. Under what circumstances can you end up back at your starting point?
If we have the displacement A and displacement B. The total displacement is A + B. We would end up at the starting point if we take a displacement -(A + B) from point B
b. What is the magnitude of the largest displacement you can end up from the starting point?
The maximum displacement we can obtain is when A and B are in the same direction. So A + B = 5 m + 6 m = 11 m or -A - B = -(A + B) = -11 m.
c. When A and B are perpendicular, what is the component of B in the direction of A?
Since A is perpendicular to B, the angle between A and B is 90°
So the component of B in A,s direction is Bcos90° = B × 0 = 0 m
?
Which statement about energy transfer in a wave is true?
Particles of the medium absorb energy from the atmosphere.
Particles move from one location
Answer:
Check Explanation.
The two statements given aren't true.
Explanation:
Although the question seems incomplete, I will address the concept of energy transfer during a wave's propagation.
The particles involved in wave's propagation move back and forth perpendicularly to the way the wave is moving, but do not move (at least, no significant movement is noticeable) in the direction of the wave. The particles ‘participate’ in the wave propagation by bumping into one another and transferring energy. This is exactly why energy can be transferred, although the average position of the particles doesn’t change.
So, the particles of the medium do not absorb energy from the atmosphere and do not significantly move from one location to another.
Hope this Helps!!!
A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use g=9.8m/sec2 as the acceleration of gravity.
Answer:
θ₀ = 84.78° (OR) 5.22°
Explanation:
This situation can be treated as projectile motion. The parameters of this projectile motion are:
R = Range of Projectile = 150 m
V₀ = Launch Speed of Projectile = 90 m/s
g = 9.8 m/s²
θ₀ = Launch angle (OR) Angle of Elevation = ?
The formula for range of a projectile is given as:
R = V₀² Sin 2θ₀/g
Sin 2θ₀ = Rg/V₀²
Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²
2θ₀ = Sin⁻¹ (0.18)
θ₀ = 10.45°/2
θ₀ = 5.22°
Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:
θ₀ = 90° - 5.22°
θ₀ = 84.78°
How does an atom of rubidium-85 become a rubidium ion with a +1 charge?
Answer:
C. The atom loses 1 electron to have a total of 36.
Explanation:
Cations have a positive charge. Cations lose electrons.
The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.
A hard drive disk rotates at 7200 rpm. The disk has a diameter of 5.1 in (13 cm). What is the speed of a point 6.0 cm from the center axle
Answer:
The speed will be "3.4×10⁴ m/s²".
Explanation:
The given values are:
Angular speed,
w = 7200 rpm
i.e.,
= [tex]7200 \times \frac{2 \pi}{60}[/tex]
= [tex]753.6 \ rad/s[/tex]
Speed from the center,
r = 6.0 cm
As we know,
⇒ Linear speed, [tex]v=wr[/tex]
On putting the estimated values, we get
[tex]=753.6\times 0.06[/tex]
[tex]=45.216 \ m[/tex]
Now,
Acceleration on disk will be:
⇒ [tex]a=\frac{v^2}{r}[/tex]
[tex]=34074 \ m/s^2[/tex]
[tex]=3.4\times 10^4 \ m/s^2[/tex]
How far apart (in mm) must two point charges of 90.0 nC (typical of static electricity) be to have a force of 3.80 N between them
Answer:
The distance between the two charges is =4.4mm
A 975-kg pickup comes to rest from a speed of 87.5 km/h in a distance of 125 m. Suppose the pickup is initially traveling in the positive direction.
(a) If the brakes are the only thing making the car come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the car .
(b) Suppose instead of braking that the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the car in this case 33%
(c) What is the ratio of the force on the car from the concrete to the braking force?
Answer:
A) Force = 2303.925 N in the negative x-direction
B) F ≈ 143998.28 N
C) Ratio = 62.5
Explanation:
A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.
Let's find the deceleration using Newton's law of motion formula;
v² = u² + 2as
where;
v = final velocity,
u = initial velocity,
s = displacement
a = acceleration
We are given;
u = 87.5 km/h = 24.3056 m/s
s = 125 m
v = 0 m/s
Thus;
0 = (24.3056)² + 2a(125)
- (24.3056)²= 250a
a = - 24.3056²/250
a = - 2.363 m/s²
Now, force = mass × acceleration
We are given mass = 975 kg
Thus;
Force = 975 x (-2.363)
Force = 2303.925 N in the negative x-direction
B) formula for kinetic energy is
KE = ½mv²
KE = ½(975)(24.3056)²
= 287996.568288 J
Work done on impact = F x 2
Thus;
2F = 287996.568288
F = 287996.568288/2
F ≈ 143998.28 N
C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5