Answer:
A) B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West
Explanation:
A) For this exercise we must use the expression of Faraday's law for a moving body
fem = [tex]- \frac{d \phi }{dt}[/tex]
fem = [tex]- \frac{d (B l y}{dt}= - B l v[/tex]- d (B l y) / dt = - B lv
B = [tex]- \frac{fem}{l \ v}[/tex]
we calculate
B = - 7.9 10⁻⁴ /(0.73 20)
B = 5.4 10⁻⁵ T
B) to determine which side of the bar is positive, we must use the right hand rule
the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.
Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West
Which two options are forms of kinetic energy?
A. Thermal energy
OO
O B. Nuclear energy
c. Sound energy
O D. Gravitational energy
O E. Chemical energy
Answer:
Themal energy is the best answer from you
Which type of land usage would increase as the population increases?
Answer:
If the population were to INCREASE land use would also INCREASE, if the human population DECREASED so would land use because there wouldn’t be as many people to using it. hope i helped
Answer:
D. Urban use
Explanation:
It shows here on this graph that overtime urban use is rising as the population is increasing! The other answers that are on this graph does not show it increasing!
Most modern vehicles have padded dashboards. This reduces collision injuries by...
1: reducing the driver's change in momentum
2:increasing the driver's change in momentum
3:increasing the force of impact
4:increasing the time of impact
Answer:
Option 4. is correct
Explanation:
A dashboard refers to a control panel that lies in front of the driver of the vehicle. Padded dashboards are designed to reduce injuries including face injuries and chest injuries to the driver as well as the front passenger on collisions.
Most modern vehicles have padded dashboards. This reduces collision injuries by increasing the time of impact.
Therefore,
Option 4. is correct
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 ×102 T, what is the number of turns per meter for this solenoid?
Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;
[tex]B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m[/tex]
Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? Suppose you are standing at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward. You have a compass that is free to swivel in any direction. Which way does your compass point? It would point up. It would point east. It would point down. It would point west.
Answer:
It would point up.
Explanation:
Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.
Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.
So, the compass would point up.
Which type of biological molecule would contain fats?
A) Amino Acids.
C) Nucleic Acids.
B) Lipids.
D) Carbohydrates.
B
Explanation:
lipids contains fat
hope it helps
Fifty grams of ice at 0◦ C is placed in a thermos bottle containing one hundred grams of water
at 6◦ C. How many grams of ice will melt? The heat of fusion of water is 333 kJ/kg and the
specific heat is 4190 J/kg · K.Immersive Reader
Answer:
7.55 g
Explanation:
Given that:
Heat of fusion = 333kj/kg
Heat capacity, c = 4190 j/kg /k
The Number of grams of ice that will melt can be represented as y:
Number of grams of ice that will melt * heat of fusion = specific heat capacity * temperature change
y * 333 * 10^3 J = (4190) * (6 - 0)
333000y = 25140
y = 25140 / 333000
y = 0.0754954 kg
y = 0.0754954 * 100
y = 7.549 g
Hence, Number of grams of ice that will melt = 7.55 g
You are planning a deep space exploration and want to take your cat with you. If your spacecraft achieves a maximum speed of 87% of the speed of light, how much additional energy is required to bring your cat (during the acceleration only)? Express this quantity in units of Petajoules (1015).Assume you start from zero velocity and your magical cat does not require additional food or litter. The cat weighs 8.5 lbf. Use LaTeX: 3\times10^8 3 × 10 8 m/s for the speed of light.
Answer:
E = 1.7 10² PJ
Explanation:
Let's use the special relativity relations, specifically the energy of a body is
E = γ mc²
γ = [tex]\sqrt{ 1 - \frac{v^2}{c^2} }[/tex]
where m is the rest mass
for that case they tell us that the speed of the body is 87% of the speed of light
= 0.87
let's calculate
γ = [tex]\sqrt{1 - 0.87^2}[/tex]
γ = 0.49305
let's reduce the mass of the jack to SI units
W = 8.5 lb (4.448 N / 1lb) = 37.808 N
W = mg
m = W / g
m = 37.808 / 9.8
m = 3.86 kg
let's look for energy
E = 3.86 (3 10⁸ )² 0.49305
E = 1.7 10¹⁷ J
let's reduce to take PJ
E = 1.7 10¹⁷ J ([tex]\frac{ 1 PJ}{10^{15} J}[/tex] )
E = 1.7 10² PJ
3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
(10 Points)
0.05 kg.m/s
8000 kg.m/s
80,000 kg.m/s
20 kg.m/s
Answer:
momentum=mass×velocity
momentum =400kg×20m/s=8000kg.m/s
why do fishermen like fishing during a full moon?
Answer:
The full moon attracts more fish towards the surface of the water.
Explanation:
It is a proven fact that fish are attracted to shiny/glowing items. for example, a metal fishing lure. Or, an angler fish light-bulb.
The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the current in the filament? Select the best answer from the choices provided. View Available Hint(s) Select the best answer from the choices provided. The current is turning on and off. The magnitude of the current is varying with time. The current switches direction, going one way and then the other.
Answer:
the correct answer is: The magnitude of the current varies with time.
Explanation:
The movement of a filament by a current is due to the electric force when electrons accumulate on one side attract positively charged nuclei.
In this case the voters in a period of time are on one side, let's call it forward and in another period of time it accumulates on the other side, let's call it back, this explains the very small movement of the filament, therefore we have a current that varies over time.
The other two options are not correct because turning off the current the filament goes to the central position and does not go back, the other explanation the current goes in one direction or another, the voters are always on the same side therefore the filament only goes in a sense.
Consequently the correct answer is: The magnitude of the current varies with time.
Sound energy travels through matter in waves. Do sound waves travel faster through air or water? Explain your answer.
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
Where do animals such as snakes and frogs, which do not make their own body heat, usually get their heat?
fuel
the sun
hot lava
friction
plssssssss answer correctly
The answer to this question is the Sun.
As im writing this, 12,826 people vote the Sun.
animals such as snakes and frogs, which do not make their own body heat. They usually get their heat from the sun. Hence option B is correct.
What is Cold-blooded animals ?A body temperature that is only slightly higher than the ambient temperature. This condition differentiates cold-blooded, or homoiothermic, animals from fishes, amphibians, reptiles, and invertebrates (birds and mammals). Due of their reliance on external warmth for metabolic activity, terrestrial cold-blooded species are restricted to locations with temperatures ranging from 5-10° to 35-40° C (41-50° to 95-104° F).
Cold-blooded creatures cannot create their own body heat, but they may control it by modifying their surroundings. Alligators and other reptiles frequently lay in the sun to warm up. They, on the other hand, cool off by swimming, going into a burrow in the earth, or moving inside the sade of a rock.
Hence option B is correct.
To know more about Cold-blooded animals :
https://brainly.com/question/30914929
#SPJ2.
when an object slides over a smooth horizontal surface, how does the force of friction depend on the surface area of blocks that's are in contact with the table?
Answer: with a greater surface area, there will be a greater force of friction
Explanation:
A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power
Answer:
The power exerted by the mountain lion is 1,472.35 W.
Explanation:
Given;
mass of mountain, m₁ = 21 kg
mass of the cub, m₂ = 3 kg
height jumped by the mountain lion, h = 2 m
time taken for the mountain lion to jump, t = 1 s
Determine the weight of the lions on the top rock;
W = F = (m₁ + m₂)g
F = (21 + 3) x 9.8
F = (24) x 9.8
F = 235.2 N
Determine the final velocity of the mountain rock as it jumped to the top;
v² = u² + 2gh
where;
u is the initial velocity = 0
h is the height jumped = 2 m
v² = 0 + 2 x 9.8 x 2
v² = 39.2
v = √39.2
v = 6.26 m/s
The power exerted by the mountain lion is calculated as;
P = Fv
P = 235.2 x 6.26
P = 1,472.35 W
Therefore, the power exerted by the mountain lion is 1,472.35 W.
tlo
A Cu rod whose length at 80 degree is 10.0 metres is heated at 50 degree find the new length take x-for cu as 0.000 to 17 k - 1
Answer:
9.949 m
Explanation:
From the question,
L' = L+LαΔT................. Equation 1
Where L' = New length, L = Original length, α = linear expansivity of Cu, ΔT = change in temperature
Given: L = 10 m, α = 0.00017 K⁻¹, ΔT = 50-80
L' = 10+10(0.00017)(50-80)
L' = 10-0.051
L' = 9.949 m
Hence the new legth of Cu is 9.949 m
what happens to the loudness of the sound as the amplitude increases
it increases and gets louder
and it amplifies the noise
When an object is moving on a surface with a lot of texture, how does this affect the amount of friction created?
A. The friction is stopped.
B. Less friction is created.
C. More friction is created.
D. The amount of friction is the same
Answer:
b) less friction is created
Answer:
B
Explanation:
A sphere of uniform density with mass 15 kg and radius 0.6 m is spinning, making one complete revolution every 0.3 s. The center of mass of the sphere has a speed of 3 m/s. (a) What is the rotational kinetic energy of the sphere
Answer:
[tex]473.56\ \text{J}[/tex]
Explanation:
T = Time taken to complete one revolution = 0.3 s
m = Mass of sphere = 15 kg
r = Radius of sphere = 0.6 m
I = Moment of inertia of sphere = [tex]\dfrac{2}{5}mr^2[/tex]
Angular speed of the sphere is
[tex]\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.3}\\\Rightarrow \omega=20.94\ \text{rad/s}[/tex]
Rotational kinetic energy is given by
[tex]K_r=\dfrac{1}{2}I\omega^2=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2\\\Rightarrow K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 15\times 0.6^2\times 20.94^2\\\Rightarrow K_r=473.56\ \text{J}[/tex]
The rotational kinetic energy of the sphere is [tex]473.56\ \text{J}[/tex].
Your friend said that the star in this picture with the highest apparent magnitude must definitely have the highest absolute brightness as well.
Answer:
A white dwarf, also called a degenerate dwarf
Explanation:
sorry if im wrong im kind of du-m
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.
Answer:
q = 8.61 10⁻¹¹ m
charge does not depend on the distance between the two ships.
it is a very small charge value so it should be easy to create in each one
Explanation:
In this exercise we have two forces in balance: the electric force and the gravitational force
F_e -F_g = 0
F_e = F_g
Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.
Let's write Coulomb's law and gravitational attraction
[tex]k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}[/tex]
In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.
k q² = G m²
q = [tex]\sqrt{ \frac{G}{k} }[/tex] m
we substitute
q = [tex]\sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }[/tex] m
q = [tex]\sqrt{0.7419 \ 10^{-20}}[/tex] m
q = 0.861 10⁻¹⁰ m
q = 8.61 10⁻¹¹ m
This amount of charge does not depend on the distance between the two ships.
It is also proportional to the mass of the ships with the proportionality factor found.
Suppose the ships have a mass of m = 1000 kg, let's find the cargo
q = 8.61 10⁻¹¹ 10³
q = 8.61 10⁻⁸ C
this is a very small charge value so it should be easy to create in each one
Score
A 4400 Kg Track Travelling at Intial speed 52m\s can be stopeed in 42 sec. By Gently Break, also the track can be stoped
in 7.6 m\s if the driver hit the wall.
a. What is the Impulse Excerted on the Vehicle?
b. Whta is the Averge force is exeted on the Track in each stops ?
Answer:
(a) J = 10560 kg-m/s (b) 251.42 N
Explanation:
Given that,
Mass, m = 4400 kg
Initial speed, u = 5.2 m/s
Final speed, v = 7.6 m/s
Time, t = 42 s
(a) Let J be the impulse exerted on the vehicle. Impuse is equal to the change in momentum such as :
J = m(v-u)
J = 4400 (7.6-5.2)
J = 10560 kg-m/s
(b) Impulse = Force × t
[tex]F=\dfrac{J}{t}\\\\F=\dfrac{10560}{42}\\F=251.42\ N[/tex]
Hence, this is the required solution.
The teacher asked the children to turn the scales sideways and adjust the scales to zero. Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason’s push measures 80 N. What is the size of Kerry’s push? *
Answer: J
Explanation:
Jason pushes on the top of the scales and Kerry pushes on the base. They hold the scales still. Jason's push measures 80N. What is the size of Kerry's push?
he pushes zero
What is the wavelength of microwaves with a frequency of 3x10^10 Hz?
Answer:
0.01 m
Explanation:
Since the speed of light is 3.0×10^8 m/s
Use the equation,
Wavelength = speed ÷ frequency
Wavelength = 3.0×10^8 ÷ 3×10^10
Wavelength = 0.01m
Can someone please help
ummmmmmmmmmm thats confusing
Light undergoes refraction as it travels from water to glass. The refractive index of water = 1.33 and refractive index of glass = 1.5. Calculate the angle of refraction inside glass if angle of incidence in the water is 32 degrees
Answer:
28°
Explanation:
Snell's Law and equation
A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of 29.5 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 24.4 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track
Answer: [tex]2.27\ m/s^2[/tex]
Explanation:
Given
Length of the race track [tex]L=200\ m[/tex]
the radius of curvature of the track [tex]r=29.5\ m[/tex]
time taken to run on track is [tex]t=24.4\ s[/tex]
Speed of runner is
[tex]\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s[/tex]
Centripetal acceleration is
[tex]\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2[/tex]
How much Work is done when a 5 kg object is picked up a distance of 8 m.
Answer:
392J?
Explanation:
An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single constant force <220, 460, -200> N acted on the object while the object moved from location <8, -30, -4> m to location <14, -21, -7> m. Then a different single constant force <100, 260, 210> N acted on the object while the object moved from location <14, -21, -7> m to location <17, -27, -3> m. What is the speed of the object at this final location
Answer:
v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s
Explanation:
We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).
X axis.
They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.
As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry
the angle fi is with respect to the in z and the angle theta with respect to the x axis.
sin φ = z / r
Cos φ = r_p / r
z = r sin φ
r_p = r cos φ
the modulus of the vector r can be found with the Pythagorean theorem
r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²
r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2
r = √126
r = 11.23 m
Let's find the angle with respect to the z axis (φfi)
φ = sin⁻¹ z / r
φ = sin⁻¹ ( [tex]\frac{-7+4}{11.23}[/tex] )
φ = 15.5º
Let's find the projection of the position vector (r_p)
r_p = r cos φ
r_p = 11.23 cos 15.5
r_p = 10.82 m
This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.
cos θ = x / r_p
θ = cos⁻¹ x / r_p
θ = cos⁻¹ ( [tex]\frac{14-8}{10.82}[/tex])
θ = 56.3º
taking the angles we can decompose the initial velocity.
sin φ = v_z / v₀
cos φ = v_p / v₀
v_z = v₀ sin φ
v_z = 5.5 sin 15.5 = 1.47 m / z
v_p = vo cos φ
v_p = 5.5 cos 15.5 = 5.30 m / s
cos θ = vₓ / v_p
sin θ = v_y / v_p
vₓ = v_p cos θ
v_y = v_p sin θ
vₓ = 5.30 cos 56.3 = 2.94 m / s
v_y = 5.30 sin 56.3 = 4.41 m / s
we already have the components of the initial velocity
v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s
let's find the acceleration on this axis (ax1) using Newton's second law
Fₓx = m aₓ₁
aₓ₁ = Fₓ / m
aₓ₁ = 220/100
aₓ₁ = 2.20 m / s²
Let's look for the velocity at the end of this interval (vx1)
Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.
vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)
let's calculate
vₓ₁² = 2.94² + 2 2.20 (14-8)
vₓ₁ = √35.04
vₓ₁ = 5.92 m / s
to the second interval
they relate it to xf
aₓ₂ = Fₓ₂ / m
aₓ₂ = 100/100
aₓ₂ = 1 m / s²
final speed
v_{xf}² = vₓ₁² + 2 aₓ₂ (x_f- x₁)
v_{xf}² = 5.92² + 2 1 (17-14)
v_{xf} =√41.05
v_{xf} = 6.41 m / s
We carry out the same calculation for each of the other axes.
Axis y
acceleration (a_{y1})
a_{y1} = F_y / m
a_{y1} = 460/100
a_[y1} = 4.60 m / s²
the velocity at the end of the interval (v_{y1})
v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)
v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)
v_{y1} = √102.25
v_{y1} = 10.11 m / s
second interval
acceleration (a_{y2})
a_{y2} = F_{y2} / m
a_{y2} = 260/100
a_{y2} = 2.60 m / s2
final speed
v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)
v_{yf}² = 10.11² + 2 2.60 (-27 + 21)
v_{yf} = √ 71.01
v_{yf} = 8.43 m / s
here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity
z axis
first interval, relate (a_{z1})
a_{z1} = F_{z1} / m
a_{z1} = -200/100
a_{z1} = -2 m / s
the negative sign indicates that the acceleration is the negative direction of the z axis
the speed at the end of the interval
v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)
v_{z1}² = 1.47² + 2 (-2) (-7 + 4)
v_{z1} = √14.16
v_{z1} = -3.76 m / s
second interval, acceleration (a_{z2})
a_{z2} = F_{z2} / m
a_{z2} = 210/100
a_{z2} = 2.10 m / s2
final speed
v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)
v_{fz}² = 3.14² - 2 2.10 (-3 + 7)
v_{fz} = √6.94
v_{fz} = 2.63 m / s
speed is v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s
d. Two point charges, q1 = +25 nC and q2 = -75 nC, are separated by a distance of 3.0 cm. Find the magnitude and direction of; i. the electric force q1 exerts on q2 [5] ii. the force that q2 exerts on q1 [4] (take k = 9.0 x 109 N.m2 /C2 )
Answer:
a) F₂₁ = 0.02 N, attracting.
b) F₁₂ = 0.02 N, attracting.
Explanation:
a)
The magnitude of the force that q₁ exerts on q₂ (F₂₁) is given by Coulomb's Law, as follows:[tex]F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)[/tex]
Since q₁ and q₂ have opposite signs, the force between them will be always attractive, i.e., from q₂ towards q₁, along the line that joins both charges.b)
The magnitude of the force on q₁ due to q₂ can be obtained applying Newton's 3rd Law, or using (1), because all parameters are the same, so F₁₂ (in magnitude) = F₂₁ = 0.02 NAs we have already said, it must be opposite to the one found in a) so it must go from q₁ towards q₂, it is an attracting force also.