Answer:
a)[tex]h_{max}=14536.16 m[/tex]
b)[tex]h = 15687.9 m[/tex]
c)[tex]PD=7.62\%[/tex] The estimate is low.
Explanation:
a) Using the energy conservation we have:
[tex]E_{initial}=E_{final}[/tex]
we have kinetic energy intially and gravitational potential energy at the maximum height.
[tex]\frac{1}{2}mv^{2}=mgh_{max}[/tex]
[tex]h_{max}=\frac{v^{2}}{2g}[/tex]
[tex]h_{max}=\frac{43^{2}}{2*0.0636}[/tex]
[tex]h_{max}=14536.16 m[/tex]
b) We can use the equation of the gravitational force
[tex]F=G\frac{mM}{R^{2}}[/tex] (1)
We have that:
[tex] F = ma [/tex] (2)
at the surface G will be:
[tex]G=\frac{gR^{2}}{M}[/tex]
Now the equation of an object at a distance x from the surface.
is:
[tex]F=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
[tex]m\frac{dv}{dt}=\frac{mgR^{2}}{(R+x)^{2}}[/tex]
Using that dv/dt is vdx/dt and integrating in both sides we have:
[tex]v_{0}=\sqrt{\frac{2gRh}{R+h}}[/tex]
[tex]h=\frac{v_{0}^{2}R}{2gR-v_{0}^{2}}[/tex]
[tex]h=15687.9[/tex]
c) The difference is:
So the percent difference will be:
[tex]PD=|\frac{14536.16-15687.9}{(14536.16+15687.9)/2}*100%[/tex]
[tex]PD=7.62\%[/tex]
The estimate is low.
I hope it helps you!
at the temperature at which we live, earth's core is solid or liquid?
Explanation:
The Earth has a solid inner core
Determine the minimum gauge pressure needed in the water pipe leading into a building if water is to come out of a faucet on the fourteenth floor, 43
Answer:
The gauge pressure is [tex]P = 4.2*10^{5} \ N/m^2[/tex]
Explanation:
From the question we are told that
The height of the 14th floor from the point where the water entered the building is h = 43 m
The gauge pressure is mathematically represented as
[tex]P = mgh[/tex]
Where the m is the mass of the water which is mathematically represented as
[tex]m = \frac{\rho}{V}[/tex]
Where [tex]\rho[/tex] is the density of the water which has a constant value of [tex]\rho = 1000 \ kg/m^3[/tex] and this standard value of density the volume is [tex]1 m^3[/tex] so
[tex]m = \frac{1000}{1}[/tex]
[tex]m = 1000 \ kg[/tex]
Thus
[tex]P = 1000 * 9.8 * 43[/tex]
[tex]P = 4.2*10^{5} \ N/m^2[/tex]
Consider the Earth and the Moon as a two-particle system.
Find an expression for the gravitational field g of this two-particle system as a function of the distance r from the center of the Earth. (Do not worry about points inside either the Earth or the Moon. Assume the Moon lies on the +r-axis. Give the scalar component of the gravitational field. Do not substitute numerical values; use variables only. Use the following as necessary: G, Mm, Me, r, and d for the distance from the center of Earth to the center of the Moon.)"
sorry but I don't understand
A circle has a radius of 13m Find the length of the arc intercepted by a central angle of .9 radians. Do not round any intermediate computations, and round your answer to the nearest tenth.
Answer:
11.7 m
Explanation:
The radius of the circle is 13 m.
The central angle of the arc is 0.9 radians
The length of an arc is given as:
L = r θ
where θ = central angle in radians = 0.9
=> L = 0.9 * 13 = 11.7 m
Length of the arc will be 11.7 m ≈ 10 m
What is an arc length?
Arc length refers to the distance between two points along a curve’s section.
Arc length = radius * theta
where
Arc length = ? to find
given :
radius = 13 m
theta ( central angle) = 0.9 radians
Arc length = 13 m * 0.9 radians
= 11.7 m ≈ 10 m
length of the arc will be 11.7 m ≈ 10 m
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5) What is the weight of a body in earth. if its weight is 5Newton
in moon?
Answer:
8.167
Explanation:
How do I find an apparent weight in N for a metal connected to a string submerged in water if a scale shows the mass 29.52 g when it is submerged ? Also how do I measure its density
The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz. The center of mass of the rod is 42.5 cm below the pivot point. What is the rotational inertia of the pendulum around its pivot point
Answer:
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
Explanation:
The angular frequency of a physical pendulum is measured by the following expression:
[tex]\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }[/tex]
Where:
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]m[/tex] - Mass of the physical pendulum, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]d[/tex] - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.
[tex]I_{O}[/tex] - Moment of inertia with respect to pivot point, measured in [tex]kg\cdot m^{2}[/tex].
In addition, frequency and angular frequency are both related by the following formula:
[tex]\omega =2\pi\cdot f[/tex]
Where:
[tex]f[/tex] - Frequency, measured in hertz.
If [tex]f = 0.658\,hz[/tex], then angular frequency of the physical pendulum is:
[tex]\omega = 2\pi \cdot (0.658\,hz)[/tex]
[tex]\omega = 4.134\,\frac{rad}{s}[/tex]
From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:
[tex]\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}[/tex]
[tex]I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}[/tex]
Given that [tex]m = 1.15\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]d = 0.425\,m[/tex] and [tex]\omega = 4.134\,\frac{rad}{s}[/tex], the moment of inertia associated with the physical pendulum is:
[tex]I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{o} = 0.280\,kg\cdot m^{2}[/tex]
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z
The question is not complete, the value of z is not given.
Assuming the value of z = 4.0m
Answer:
the magnitude of the electric field at any point having z(4.0 m) =
E = 5.65 N/C
Explanation:
given
σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²
z = 4.0 m
Recall
E =F/q (coulumb's law)
E = kQ/r²
σ = Q/A
A = 4πr²
∴ The electric field at point z =
E = σ/zε₀
E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)
E = 5.65 N/C
How much heat does it take to raise the temperature of 7.0 kg of water from
25-C to 46-C? The specific heat of water is 4.18 kJ/(kg.-C).
Use Q = mcTr-T)
A. 148 kJ
B. 176 kJ
C. 610 kJ
D. 320 kJ
Answer:
non of the above
Explanation:
Quantity of heat = mass× specific heat× change in temperature
m= 7kg c= 4.18 temp= 46-25=21°
.......H= 7×4.18×21= 614.46kJ
Answer:610 KJ
Explanation:A P E X answers
Lightbulbs are typically rated by their power dissipation when operated at a given voltage. Which of the following lightbulbs has the largest resistance when operated at the voltage for which it's rated?
A. 0.8 W, 1.5 V
B. 6 W 3 V
C. 4 W, 4.5 V
D. 8 W, 6 V
Answer:
The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V
Explanation:
The equation for electric power is
power P = IV
also, I = V/R,
substituting into the equation, we have
[tex]P = \frac{V^{2} }{R}[/tex]
[tex]R = \frac{V^{2} }{P}[/tex]
a) [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω
b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω
c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω
d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω
from the calculations, one can see that the lightbulb with te greates resistance is
C. 4 W, 4.5 V
A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?
Explanation:
KE = q
½ mv² = mCΔT
ΔT = v² / (2C)
ΔT = (200 m/s)² / (2 × 236 J/kg/°C)
ΔT = 84.7°C
This question involves the concepts of the law of conservation of energy.
The temperature change of the bullet is "84.38°C".
What is the Law of Conservation of Energy?According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.
[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]
where,
ΔT = change in temperature of the bullet = ?C = specific heat capacity of silver = 237 J/kg°Cv = speed of bullet = 200 m/sTherefore,
[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]
ΔT = 84.38°C
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Two small identical speakers are connected (in phase) to the same source. The speakers are 3 m apart and at ear level. An observer stands at X, 4 m in front of one speaker. If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
a. 1 m
b. 2 m
c. 3 m
d. 4 m
e. 5 m
Answer:
b. 2 m
Explanation:
Given that:
the identical speakers are connected in phases ;
Let assume ; we have speaker A and speaker B which are = 3 meter apart
An observer stands at X = 4m in front of one speaker.
If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:
From above; the distance between speaker A and speaker B can be expressed as:
[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]
The path length difference will now be:
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive interference for that path length will be half the wavelength; which is
= [tex]\dfrac{1}{2}*4 \ m[/tex]
= 2 m
The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.
Given data:
The distance between the speakers is, d = 3 m.
The distance between the observer and speaker is, s = 4 m.
The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then the distance between speaker A and speaker B can be expressed as:
[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]
And the path length difference is,
= 5 m - 4 m
= 1 m
Since , we are to determine the least intense sound; the destructive
interference for that path length will be half the wavelength; which is
[tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]
= 2 m
Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.
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A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration?
b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.
Answer:
8.7 m/s^2
82.15°
Explanation:
Given:-
- The initial speed of the car, vi = 25 m/s
- The radius of track, r = 129 m
- Car makes a circular " quarter turn "
- The constant tangential acceleration, at = 1.2 m/s^2
Solution:-
- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:
[tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]
- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:
at = r*α
α = ( 1.2 / 129 )
α = 0.00930 rad/s^2
- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).
- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:
[tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]
- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:
[tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]
- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:
[tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]
- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:
[tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]
- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:
[tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex]
Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .
200 J of heat is added to two gases, each in a sealed container. Gas 1 is in a rigid container that does not change volume. Gas 2 expands as it is heated, pushing out a piston that lifts a small weight. Which gas has the greater increase in its thermal energy?Which gas has the greater increase in its thermal energy?Gas 1Gas 2Both gases have the same increase in thermal energy.
Answer:
Gas 1
Explanation:
The reason for this is that for gases attached to both gases or containers, with a heat of 200 J, the change in volume is only observed in gas 2, whereas the volume of gas 1 is the same as that of gas. Therefore, the internal energy (heat) or thermal energy of the system is not utilized for Gas 1 and hence the absorption and transfer of energy is the same, whereas Gas 2 is propagated by the use of additional heat of heat. Thus there is a large increase in the thermal energy of Gas1.When a certain capacitor carries charge of magnitude Q on each of its plates, it stores energy Ep. In order to store twice as much energy, how much charge should it have on its plates
2Q
Explanation:
When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;
E = [tex]\frac{1}{2}qV[/tex] ------------(i)
Where;
q = charge between the plates of the capacitor
V = potential difference between the plates of the capacitor
From the question;
q = Q
E = Ep
Therefore, equation (i) becomes;
Ep = [tex]\frac{1}{2} QV[/tex] ----------------(ii)
Make V subject of the formula in equation (ii)
V = [tex]\frac{2E_{p}}{Q}[/tex]
Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;
2Ep = [tex]\frac{1}{2}qV[/tex]
Substitute the value of V into the equation above;
2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])
Solve for q;
[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]
[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]
[tex]q = 2Q[/tex]
Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q
Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a phosphor screen. The magnetic fields were created by coils of wire on either side of the tube carrying large currents. In one such TV set, the phosphor screen is 51.2 cm wide, and is 11.1 cm away from the center of the magnetic deflection coils (that is, the center of the region of magnetic field). The electron beam is first accelerated through a 22,000 V potential difference before it enters the magnetic field region, which is 1.00 cm wide. The field is approximately uniform and perpendicular to the velocity of the electrons. If the field were turned off, the electrons would hit the center of the screen. What magnitude of magnetic field (in mT) is needed to deflect the electrons so that they hit the far edge of the screen
Answer:
B = 0.046T
Explanation:
given
size of the screen = 51.2cm
distance from center = 11.1cm
region of magnetic field = 1.00cm
V= 22000V= 22kV
An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?
Answer:
a
[tex]\theta = 0.0022 rad[/tex]
b
[tex]I = 0.000304 I_o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
The distance of the slit separation is [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]
Generally the condition for two slit interference is
[tex]dsin \theta = m \lambda[/tex]
Where m is the order which is given from the question as m = 2
=> [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
substituting values
[tex]\theta = 0.0022 rad[/tex]
Now on the second question
The distance of separation of the slit is
[tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]
The intensity at the the angular position in part "a" is mathematically evaluated as
[tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]
Where [tex]\beta[/tex] is mathematically evaluated as
[tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]
[tex]\beta = 0.06581[/tex]
So the intensity is
[tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]
[tex]I = 0.000304 I_o[/tex]
An experimenter finds that standing waves on a string fixed at both ends occur at 24 Hz and 32 Hz , but at no frequencies in between. Part A What is the fundamental frequency
Answer:
8 Hz
Explanation:
Given that
Standing wave at one end is 24 Hz
Standing wave at the other end is 32 Hz.
Then the frequency of the standing wave mode of a string having a length, l, is usually given as
f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc
Also, another formula is given as
f(m) = m.f(1), where f(1) is the fundamental frequency..
Thus, we could say that
f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)
And as such,
f(1) = 32 - 24
f(1) = 8 Hz
Then, the fundamental frequency needed is 8 Hz
Problem 2: A 21-W horizontal beam of light of wavelength 430 nm, travelling at speed c, passes through a rectangular opening of width 0.0048 m and height 0.011 m. The light then strikes a screen at a distance 0.36 m behind the opening.
a) E = 4.62E-19 Joules
b) N = 4.543E+19 # of photons emitted
c) If the beam of light emitted by the source has a constant circular cross section whose radius is twice the height of the opening the beam is approaching, find the flow density of photons as the number of photons passing through a square meter of cross-sectional area per second.
d) Calculate the number of photons that pass through the rectangular opening per second.
e) The quantity NO that you found in part (d) gives the rate at which photons enter the region between the opening and the screen. Assuming no reflection from the screen, find the number of photons in that region at any time.
f) How would the number of photons in the region between the opening and the screen change, if the photons traveled more slowly? Assume no change in any other quantity, including the speed of light.
Answer:
a. E = 4.62 × 10⁻¹⁹J
b. n = 4.54 × 10¹⁹photons
c. 2.99 × 10²²photons/m²
d. 1.58 ×10¹⁸photons/seconds
e. n = 1.896 × 10 ⁹ photons
f. the number of photons will be more if it travels slowly
Explanation:
The electron density in copper is 8.49 × 1028 electrons/m3. When a 1.00 A current is present in a copper wire with a 0.40 cm2 cross-section, the electron drift velocity, in m/s, with direction defined relative to the current density, is
Answer:
Drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].
Explanation:
Formula for Drift Velocity is:
[tex]u = \dfrac{I}{nAq}[/tex]
Where I is the current
n is the number of electrons in 1 [tex]m^3[/tex] or the electron density
A is the area of cross section and
q is the charge of one electron.
We are given the following:
n = [tex]8.49 \times 10^{28}\ electrons/m^3[/tex]
I = 1 A
A = 0.40 [tex]cm^2[/tex] = 40 [tex]\times 10^{-4}[/tex] [tex]m^{2}[/tex]
We know that q = [tex]1.6\times10^{-19} C[/tex]
Putting all the values to find drift velocity:
[tex]u = \dfrac{1}{8.49 \times 10 ^{28} \times 40 \times 10^{-4}\times 1.6 \times 10^{-19}}\\u = \dfrac{1}{543.36 \times 10 ^{5} }\\u = 1.84 \times 10^{-8}\ m/s[/tex]
So, drift velocity in m/s is [tex]1.84 \times 10^{-8}\ m/s[/tex].
If the frequency of a periodic wave is cut in half while the speed remains the same, what happens to the wavelength
Answer:
The wavelength becomes twice the original wavelength
Explanation:
Recall that for regular waves, the relationship between wavelength, velocity (i.e speed) and frequency is given by
v = fλ
where,
v = velocity,
f = frequency
λ = wavelength
Before a change was made to the frequency, we have: v₁ = f₁ λ₁
After a change was made to the frequency, we have: v₂ = f₂ λ₂
We are told that the speed remains the same, so
v₁ = v₂
f₁ λ₁ = f₂ λ₂ (rearranging this)
f₁ / f₂ = λ₂/λ₁ --------(1)
we are given that the frequency is cut in half.
f₂ = (1/2) f₁ (rearranging this)
f₁/f₂ = 2 -------------(2)
if we substitute equation (2) into equation (1):
f₁ / f₂ = λ₂/λ₁
2 = λ₂/λ₁
λ₂ = 2λ₁
Hence we can see that the wavelength after the change becomes twice (i.e doubles) the initial wavelength.
A pickup truck moves at 25 m/s toward the east. Ahmed is standing in the back and throws a baseball in what to him is the southwest direction at 28 m/s (with respect to the truck). A person at rest on the ground would see the ball moving how fast in what direction? HTML EditorKeyboard Shortcuts
Answer:
Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South
Explanation:
given data
truck moves = 25 m/s toward the east.
throws a baseball = 28 m/s southwest
solution
first we take here Speed of truck w.r.to ground i.e. V(p/g) = 25 m/sec toward the east so we can say
V(p/g) = (25 i) m/sec ........................1
and
Speed of baseball w.r.t. pickup i.e. V(b/p) = 28 m/sec toward the South West and we know that south west direction is in third quadrant
and here both component (x and y) are negative
So that we can say it
V(b/p) = -28 × cos(45) i - 28 × sin(45) j = -19.8 i - 19.8 j
and
now we use here relative motion velocity for ball w.r.t ground
V(b/g) = V(b/p) + V(p/g ) ..........................2
put here value and we get
V(b/g) = (-19.8 i - 19.8 j) + 25 i = 5.2 i - 19.8 j
so
Magnitude of that velocity
| V(b/g) | = [tex]\sqrt{(5.2^2 + 19.8^2)}[/tex]
| V(b/g) | = 20.47 m/sec
so that Direction will be here
Direction = arctan (19.8 ÷ 5.2)
Direction = 75.3° South of East
so that
Speed = 20.47 m/sec at 75.3 deg South of East
and 2 significant
Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South
In a uniform electric field, the magnitude of torque is given by:-
Answer:
Electric dipole
Explanation:
the dipole axis makes an angle with the electric field. depending on direction (clockwise/aniclockwise) you get the torque
Hope this helps
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is
Answer:
1:4
Explanation:
The formula for calculating kinetic energy is:
[tex]KE=\dfrac{1}{2}mv^2[/tex]
If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!
The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].
Kinetic EnergyThe kinetic energy of an object depends on the mass and velocity with which it moves.
While under free-fall, the mass of an object does not affect the velocity with which it falls.
So, the velocities of both the balls are the same.
Let the mass of ball A is 'm'
So, the mass of ball B is '4m'
The kinetic energy of ball A is given by;
[tex]KE_{A}=\frac{1}{2} mv^2[/tex]
The kinetic energy of ball B is given by;
[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]
Therefore, the ratio of kinetic energies of A and B is,
[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]
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The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air.
By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius. (change in V/V)
Answer:
Increase of volume (F) = 8.01%
Explanation:
Given:
Density of fish = 1,080 kg/m³
Density of water = 1,000 kg/m³
density of air = 1.28 kg/m³
Find:
Increase of volume (F)
Computation:
1,080 kg/m³ + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³
1,080 + 1.28 F =1,000 F + 1,000
80 = 998.72 F
F = 0.0801 (Approx)
F = 8.01% (Approx)
To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 900 loops of wire wound on a rod 6 cm long with radius 1 cm?
Answer:
The self-inductance is [tex]L = 0.0053 \ H[/tex]
Explanation:
From the question we are told that
The number of loops is [tex]N = 900[/tex]
The length of the rod is [tex]l =6 \ cm = 0.06 \ m[/tex]
The radius of the rod is [tex]r = 1 \ cm = 0.01 \ m[/tex]
The self-inductance for the solenoid is mathematically represented as
[tex]L = \frac{\mu_o * A * N^2 }{l}[/tex]
Now the cross-sectional of the solenoid is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A =3.142 * 0.01 ^2[/tex]
[tex]A = 3.142 *10^{-4} \ m^2[/tex]
and [tex]\mu_o[/tex] is the permeability of free space with a value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values into above equation
[tex]L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}[/tex]
[tex]L = 0.0053 \ H[/tex]
The back wall of an auditorium is 30.0 m from the stage. If you are seated in the middle row, how much time elapses between a sound from the stage reaching your ear directly and the same sound reaching your ear after reflecting from the back wall?
Answer:
0.0875 secExplanation:
Using the relationship to calculate the time that elapse 2x = vt
x = distance between the source and the reflector
v is the velocity of sound in air
t = time elapsed
Given x = 30.0m and v = 343m/s
substituting the values into the formula above to get the total time elapsed t;
t = 2x/v
t = 2(30)/343
t= 60/343
t = 0.175sec
If you are seated in the middle row, the time that will elapse between the sound from the stage reaching your ear directly will be;
0.175/2 = 0.0875 secs
A 2.5-kg object falls vertically downward in a viscous medium at a constant speed of 2.5 m/s. How much work is done by the force the viscous medium exerts on the object as it falls 80 cm?
Answer:
The workdone is [tex]W_v = - 20 \ J[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]m = 2.5 \ kg[/tex]
The speed of fall is [tex]v = 2.5 \ m/s[/tex]
The depth of fall is [tex]d = 80\ cm = 0.8 \ m[/tex]
Generally according to the work energy theorem
[tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]
Now here given the that the velocity is constant i.e [tex]v_1 = v_2 = v[/tex] then
We have that
[tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]
So in terms of workdone by the potential energy of the object and that of the viscous liquid we have
[tex]W = W_v - W_p[/tex]
Where [tex]W_v[/tex] is workdone by viscous liquid
[tex]W_p[/tex] is the workdone by the object which is mathematically represented as
[tex]W_p = mgd[/tex]
So
[tex]0 = W_v + mgd[/tex]
=> [tex]W_v = - m * g * d[/tex]
substituting values
[tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]
[tex]W_v = - 20 \ J[/tex]
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to:_______
a. one-half.
b. double.
c. reduce to one-fourth.
d. quadruple.
Answer:
D. quadrupleExplanation:
The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.
Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :
D. Quadruple
"Energy"Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.
The stored energy shifts with the square of the electric charge put away within the capacitor.
In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.
Thus, the correct answer is D.
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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
Answer:
The maximum number of bright spot is [tex]n_{max} =5001[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]
The wavelength is [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]
Generally the condition for interference is
[tex]n * \lambda = d * sin \theta[/tex]
Where n is the number of fringe(bright spots) for the number of bright spots to be maximum [tex]\theta = 90[/tex]
=> [tex]sin( 90 )= 1[/tex]
So
[tex]n = \frac{d }{\lambda }[/tex]
substituting values
[tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]
[tex]n = 2500[/tex]
given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as
[tex]n_{max} = 2 * n + 1[/tex]
The 1 here represented the central bright spot
So
[tex]n_{max} = 2 * 2500 + 1[/tex]
[tex]n_{max} =5001[/tex]