Answer:
Explanation:
Since the compass uses a magnetic field, if anything else magnetic is near it, the compass will start acting up. Making it unreliable so keep magnets away!
convert 56km/h to m/s.
Explanation:
15.556 metres per second
What word chemical equation describes this chemical reaction?
Answer : sodium + chlorine → sodium chloride
A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?
Answer:
h = 2755102 m = 2755.102 km
Explanation:
According to the given condition:
Potential Energy = Energy Consumed by Bulb
[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]
where,
h = height = ?
P = Power of bulb = 75 W
t = time = (2 h)(3600 s/1 h) = 7200 s
m = mass of bulb = 20 g = 0.02 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]
h = 2755102 m = 2755.102 km
g A student slides her 80.0-kg desk across the level floor of her dormitory room a distance 3.00 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
The desk is in equilbrium, so Newton's second law gives
∑ F (horizontal) = p - f = 0
∑ F (vertical) = n - mg = 0
==> n = mg
==> p = f = µn = µmg = 0.400 (80.0 kg) g = 313.6 N
The student pushes the desk 3.00 m, so she performs
W = (313.6 N) (3.00 m) = 940.8 Nm ≈ 941 J
of work.
A satellite is launched to orbit the Earth at an altitude of 2.90 x10^7 m for use in the Global Positioning System (GPS). Take the mass of the Earth to be 5.97 x 10^24 kg and its radius 6.38 x10^6 m.
Required:
What is the orbital period of this GPS satellite?
Answer:
[tex]T=66262.4s[/tex]
Explanation:
From the question we are told that:
Altitude [tex]A=2.90 *10^7[/tex]
Mass [tex]m=5.97 * 10^{24} kg[/tex]
Radius [tex]r=6.38 *10^6 m.[/tex]
Generally the equation for Satellite Speed is mathematically given by
[tex]V=(\frac{GM}{d} )^{0.5}[/tex]
[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]
[tex]V=3354.83m/s[/tex]
Therefore
Period T is Given as
[tex]T=\frac{2 \pi *a}{V}[/tex]
[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]
[tex]T=66262.4s[/tex]
Starting from rest, a wheel undergoes constant angular acceleration for a period of time T. At which of the following times does the average angular acceleration equal the instantaneous angular acceleration?
a. 0.50 T
b. 0.67 T
c. 0.71 T
d. all of the above
The value found for the universal gravitational constant, G, will vary depending on the materials used for the balls of a Cavendish balance. Question 11 options: True False
Answer:
false
Explanation:
took the test
A scientist who studies fossils of ancient life forms .O ornithologist O Paleontologist O Ichthyologist O Marine Biologist .
Hurry !! First one to answer gets points !
Answer:
paleontologist
Explanation:
Paleontologists are scientists that investigate the fossils of extinct life forms. Thus, the correct option is B.
What is Fossil?A fossil is defined as the preserved trace, imprint, or proof of a once-living entity from a past geological era. Exoskeletons, bones, shells, impressions of animals or microbes in stone, objects preserved in amber, hair, petrified wood, and genetic traces are a few examples. The collection of all the fossils is called as the fossil record.
An organism from a past geologic era that has been preserved in the Earth's crust is referred to as a fossil. Paleontologists are scientists that investigate the fossils of extinct life forms. The intricate system of fossil records is the main source of information about the evolution of life on Earth.
Therefore, the correct option is B.
Learn more about Fossils, here:
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An airplane increases its speed at the average rate of 15 m/s2. How much time does it take to increase its speed from 100 m/s to 160 m/s
Answer:
4 s
Explanation:
Acceleration (a) = 15 m/s²Initial velocity (u) = 100 m/sFinal velocity (v) = 160 m/sWe are asked to calculate time taken (t).
By using the first equation of motion,
[tex]\longrightarrow[/tex] v = u + at
[tex]\longrightarrow[/tex] 160 = 100 + 15t
[tex]\longrightarrow[/tex] 160 - 100 = 15t
[tex]\longrightarrow[/tex] 60 = 15t
[tex]\longrightarrow[/tex] 60 ÷ 15 = t
[tex]\longrightarrow[/tex] 4 s = t
In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.517 ss for the value of the period.
Trial 1 Spring constant is 117N/m, period of oscillations .37s, mass of the block is .400kg .
Trial 2 oscillation period is .52s
Answer:
[tex]M_2=0.79kg[/tex]
Explanation:
From the question we are told that:
Period [tex]T=0.517s[/tex]
Trial 1
Spring constant [tex]\mu=117N/m[/tex]
Period [tex]T_1=0.37[/tex]
Mass [tex]m=0.400kg[/tex]
Trial 2
Period [tex]T_2=0.52[/tex]
Generally the equation for Spring Constant is mathematically given by
\mu=\frac{4 \pi^2 M}{T^2}
Since
[tex]\mu _1=\mu_2[/tex]
Therefore
[tex]\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}[/tex]
[tex]M_2=M_1*(\frac{T_2}{T_1})^2[/tex]
[tex]M_2=0.400*(\frac{0.52}{0.37}})^2[/tex]
[tex]M_2=0.79kg[/tex]
what is the time taken by moving body with acceleration 0.1m/s2 if the initial or finak velocities are 20m/s and 30m/s respectively?
Answer:
t= 100s
Explanation:
use v=v0+at
plug in givens and solve for t
30=20+0.1*t
t= 100s
A block of mass 10kg is suspendet at a diameter of 20cm from the centre of a uniform bar im long, what force is required to balance it at its centre of gravity by applying the fore at the other end of the bar?
Answer:
4 kg of force
Explanation:
Force = (mass x distance to fulcrum) / length of fulcrum to end
Subsitute values
F = (10 x 20)/50
F =4
Three forces of magnitude 10N, 5N and 4N act on an object in the directions North, West and East respectively. Find the magnitude and directions of their resultant
Answer:
19N to the south
Explanation:
F =10N + 5N + 4N
a tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?
Answer: 2.4 km/hr
Explanation:
Distance = 600m
Time= 15 minutes = 15 x 60 second/minute = 900 seconds
Speed = [tex]\frac{Distance}{Time}[/tex] = [tex]\frac{600}{900}[/tex] = [tex]\frac{2}{3}[/tex] m/sec
⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)
why does self inductance acts as electrical inertia?
Answer:
self-indulgence of coil is the property by virtue of wich is tends to maintain magnatic flux link with it and opposed any any change in the flux inducing current in it
A 36.0 kg child slides down a long slide in a playground. She starts from rest at a height h1 of 24.00 m. When she is partway down the slide, at a height h2 of 11.00 m, she is moving at a speed of 7.80 m/s.
Calculate the mechanical energy lost due to friction (as heat, etc.).
Answer:
E = 3495.96 J
Explanation:
From the law of conservation of energy:
Total Mechanical Energy at h1 = Total Mechanical Energy at h2
Kinetic energy at h1 + potential energy at h1 = Kinetic energy at h2 + potential energy at h2 + Mechanical Energy Lost due to Friction
[tex]K.E_{h1}+P.E_{h1} = K.E_{h2}+P.E_{h2} + E\\\\\frac{1}{2}mv_1^2\ J + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 + E\\\\\frac{1}{2}(36\ kg)(0\ m/s)_1^2\ J + (36\ kg)(9.81\ m/s^2)(24\ m)_1 = \frac{1}{2}(36\ kg)(7.8\ m/s)_2^2 + (36\ kg)(9.81\ m/s^2)(11\ m)_2 + E\\\\0\ J + 8475.84\ J = 1095.12\ J + 3884.76\ J + E\\E = 8475.84\ J - 1095.12\ J - 3884.76\ J\\[/tex]
E = 3495.96 J
A stopped organ pipe of length L has a fundamental frequency of 220 Hz. For which of the following organ pipes will there be a resonance if a tuning fork of frequency 660 Hz is sounded next to the pipe?
a. a stopped organ pipe of length L
b. a stopped organ pipe of length 2L
c. an open organ pipe of length L;
d. an open organ pipe of length 2L.
Answer:
Option (a), (d) are correct.
Explanation:
Frequency, f = 220 Hz
Resonant frequency = 660 Hz
The next frequency of stopped organ pipe is
2f, 3 f, 4 f ....
= 2 x 220 , 3 x 220 , 4 x 220 ....
= 440 Hz, 660 Hz, 880 Hz
So, the option (a) is correct.
The next resonant frequency of open organ pipe is
3 f, 5 f,...
= 3 x 220, 5 x 220 , ..
= 660 Hz, 1100 Hz,...
So, option (d) is correct.
A closely wound, circular coil with radius 2.70 cm has 800 turns. What must the current in the coil be if the magnetic field at the center of the coil is 0.0750 T
Answer:
Approximately 4.029 A
Explanation:
We can use the formula that the B field of a few loops all with current in same direction is permeability of free space (mu)* current * Number or loops divided by 2*radius. You are given B field, radius(convert into meters), number of loops and mu is 4pi * 10^-7. Solve for current and you get 4.029 Amperes.
3. You have a variable-voltage power supply and a capacitor in the form of two metal disks of radius 0.6 m, held a distance of 1 mm apart. What is the largest voltage you can apply to the capacitor without the air becoming highly conductive
Answer:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
Explanation:
The breakdown of air occurs at a maximum voltage of 3kV/mm.
At this level of voltage the air between the plates become highly ionised and breakdown occurs. Since, the distance held between the plates is 1mm , it can withstand a maximum voltage of 3 kV.
After this voltage the air will become conductive in nature and will form ions in the air between the plates and ultimately breakdown will take place with a flash.
what are the dynamic properties of a nucleus
Two plane coaxial disks are separated by a distance 0.6 m. The lower disk is solid with a diameter 0.80 m and a temperature 300 K. The upper disk , at temperature 900 K, has the same outer diameter but is ring-shaped with an inner diameter 0.40 m. Assuming the disks to be blackbodies, calculate the net rate of radiative heat exchange from disk 1 to disk 2, in kW.
Answer:
if anyone is reading my coment your oarents will became very fine soon
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field strength in volts per meter
Answer:
7.60× 10^6 V/m
Explanation:
electric field strength can be determined as ratio of potential drop and distance, I.e
E=V/d
Where E= electric field
V= potential drop= 74.0 mV= 0.07 V
d= distance= 9.20 nm = 9.2×10^-9 m
Substitute the values
E= 0.07/ 9.2×10^-9
= 7.60× 10^6 V/m
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.240 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A. Find:
a. the force on each side of the loop
b. the torque acting on the loop.
Answer:
Explanation:
a )
Magnetic field inside solenoid B = μ₀ NI ,
μ₀ = 4π x 10⁻⁷ ; N is no of turns per meter length in solenoid and I is current B= 4π x 10⁻⁷ x 30 x 10² x 15
= .0565 T .
Force on each side of square loop = B i L
B is external magnetic field , i is current in loop and L is length of side
Force on each side of square loop = .0565 x .24 x 2 x 10⁻²
= 2.7 x 10⁻⁴ N .
b )
Torque on the loop = F x d
F is force on one side , d is distance between two sides , that is side of the square loop
= 2.7 x 10⁻⁴ x 2 x 10⁻² N.m
= 5.4 x 10⁻⁶ N.m .
recognizing forms of energy
Answer:
hi the question isn't obvious and need a photo I guess
A 3.10 mol sample of an ideal diatomic gas expands adiabatically from a volume of 0.1550 m3 to 0.742 m3 . Initially the pressure was 1.00 atm.(a) Determine the initial and final temperatures.initial Kfinal K(b) Determine the change in internal energy. J(c) Determine the heat lost by the gas. J(d) Determine the work done on the gas. J
Answer:
a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy is -18279.78 J
c) heat lost by the gas is zero or 0
d) the work done on the gas is -18279.78 J
Explanation:
Given the data in the question;
P[tex]_i[/tex] = 1 atm = 101325 pascal
P[tex]_f[/tex] = ?
V[tex]_i[/tex] = 0.1550 m³
V[tex]_f[/tex] = 0.742 m³
we know that for an adiabatic process γ = 1.4
P[tex]_i[/tex]V[tex]_i^Y[/tex] = P[tex]_f[/tex]V[tex]_f^Y[/tex]
P[tex]_f[/tex] = P[tex]_i[/tex][tex]([/tex] V[tex]_i[/tex] / V[tex]_f[/tex] [tex])^Y[/tex]
we substitute
P[tex]_f[/tex] = 1 × [tex]([/tex] 0.1550 / 0.742 [tex])^{1.4[/tex]
= [tex]([/tex] 0.2088948787 [tex])^{1.4[/tex]
= 0.11166 atm
a) the initial and final temperatures
Initial temperature
T[tex]_i[/tex] = P[tex]_i[/tex]V[tex]_i[/tex] / nR
given that n = 3.10 mol
= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )
= 15705.375 / 25.7734
T[tex]_i[/tex] = 609.4 K
Final temperature
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )
= 8394.95 / 25.7734
= 325.7 K
Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K
b) the change in internal energy
ΔE[tex]_{int[/tex] = nC[tex]_v[/tex]ΔT
here, C[tex]_v[/tex] = ( 5/2 )R
ΔE[tex]_{int[/tex] = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J
c) the heat lost by the gas
Since its an adiabatic process,
Q = 0
Therefore, heat lost by the gas is zero or 0
d) the work done on the gas
W = ΔE[tex]_{int[/tex] - Q
= -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J
a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.
b) The change in internal energy is -18279.78 J.
c) The heat lost by the gas is zero.
d) The work done on the gas is -18279.78 J.
Given data:
The moles of sample is, n = 3.10 mol.
The initial volume of sample is, [tex]V_{1}=0.1550 \;\rm m^{3}[/tex].
The final volume of sample is, [tex]V_{2}=0.742 \;\rm m^{3}[/tex].
The initial pressure of the sample is, [tex]P_{1}=1.00 \;\rm atm[/tex].
(a)
We know that the relation between the pressure and volume for an adiabatic process is as follows,
[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]
Here, [tex]\gamma[/tex] is a adiabatic index. And for air, its value is 1.41.
Solving as,
[tex]P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm[/tex]
Now, calculate the final temperature using the ideal gas equation as,
[tex]P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K[/tex]
Similarly, calculate the initial temperature as,
[tex]P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K[/tex]
Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.
(b)
The change in internal energy is given as,
ΔE = nCΔT
here, C = ( 5/2 )R
ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )
= -18279.78 J
Therefore, the change in internal energy is -18279.78 J.
c)
The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.
Q = 0
Therefore, heat lost by the gas is zero or 0
d)
The work done on the gas
W = ΔE - Q
W = -18279.78 J - 0
W = -18279.78 J
Therefore, the work done on the gas is -18279.78 J.
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which is the correct formula for calculating the age of meteor right if using half life
Answer:
n × t_1/2
Exmplanation:
The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;
Age of meteorite= n × t_1/2
where;
n = quantity of the meteorite
t_/2 = half life of the meteorite
Thus:
The correct formula is; n × t_1/2
What is significant about the primary colors of pigments?
They can be mixed together to make almost any other color.
Any two primary colors of pigments combine to make white pigment.
Each primary color of pigment absorbs all other colors.
Any two primary colors of pigments combine to make black pigment.
Answer:
They can be mixed together to make almost any other color.
Explanation:
All the three primary colors can mix to form white color.
Blue and red mix to form a black color.
Suppose the pucks start spinning after the collision, whereas they were not before. Will this affect your momentum conservation results
Answer:
No, it will not affect the results.
Explanation:
For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.
What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.