Answer:
Resistance = 1.5 ohms
Explanation:
Given:
Potential difference = 3v
Current flow = 2 A
Find:
Resistance
Computation:
Resistance = Potential difference / Current flow
Resistance = 3 / 2
Resistance = 1.5 ohms
It requires 3480 J to move a 675 N object how far?
PLEASE HELP The United States spends over $20 billion a year on space exploration through NASA. Do you think that this has been worth the cost? In three to five sentences, provide two specific examples of things we have learned from space exploration, and explain how these examples influence your opinion.(4 points)
Answer: I think $20 billion a year it’s worth the cost. The reasoning behind that is because we can conduct research on various things that could help out humanity. Therefore we can conclude that’s spending billions of dollars every year is worth it.
Explanation:
____ is a chemical messenger released by virus infected cells
Answer:
Cytokines is the answer
Explanation:
it is another word for the chemical messanger
Why are protists difficult classify?
Answer:
Protists are difficult to characterize because of the great diversity of the kingdom. These organisms vary in body form, nutrition, and reproduction. They may be unicellular, colonial, or multicellular.
Answer:
There are many varying characteristics and exceptions to each type of protist.
They have been previously categorized based on what they are not
Recent studies show that protists have not descended from one common ancestor.
Explanation:
its right i took test :)
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
mark me brainliest
A 430 kg motorcycle starts from rest and accelerates to a speed of 12 m/s.
Calculate the net work done on the motorcycle.
a. 42 kJ
b. 31 kJ
c. 38 kJ
d. 35 kJ
e. none of these
Answer:
Vi = 0
Vf = 12 m/s
ΔV = Vf - Vi
ΔV = 12 m/s
Change in kinetic energy
ΔKE = Kf - Ki
ΔKE = 1/2 mv^2 - 0
ΔKE = 1/2 * 430 kg * (12 m/s)^2
ΔKE = 30,960
ΔKE = 30,960 joules
Work = ΔK
Work = 30,960 J
A motorcycle traveling due east at a constant speed covers 75 kilometers in
1.5 hours. What is its velocity in km/h?
Answer:
The velocity of the motorcycle is 50km/hr due east.
A projectile is fired with an initial velocity of 120.0 m/s at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second, then angle θ measures approximately
Answer:
algm sabe tô precisando muito
1. Alexandra and Rachel are on a train that sounds a whistle at a constant frequency as
it leaves the train station. Compared to the sound emitted by the whistle, the sound that
the passengers standing on the platform hear has a frequency that is
a. lower, because the sound-wave fronts reach the platform at a frequency
lower than the frequency at which they are produced
b. lower, because the sound waves travel more slowly in the still air above the
platform than in the rushing air near the train
c. higher, because the sound-wave fronts reach the platform at a frequency
higher than the frequency at which they are produced
d. higher, because the sound waves travel faster in the still air above the
platform than in the rushing air near the train
Answer: the answer would be C trust me i took the test if its not that its b
hope that helps
Explanation: i took the test
answer:
a) lower because the sound-wave fronts reach the platform at a frequency lower than the frequency at which they are produced
explanation :3
If the train is leaving the train station, then the people who are standing on the platform would hear a sound with a lower frequency since the train is moving further away. ^^
What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)
total lunar eclipse
total solar eclipse
partial lunar eclipse
partial solar eclipse
Answer: I'm not sure, but I think it would be a total lunar eclipse
When the moon passes through the penumbra of Earth’s shadow it is referred to as partial lunar eclipse. The correct option is C.
What is partial lunar eclipse?A partial lunar eclipse occurs when the moon is not completely immersed in the umbra of the earth's shadow.
During a partial solar eclipse, the Moon, Sun, and Earth do not align perfectly straight, and the Moon casts only the penumbra of its shadow on Earth. From our vantage point, it appears that the Moon has eaten the Sun.
A shadow's penumbra is the lighter outer edge. Partial solar eclipses are caused by the Moon's penumbra, while penumbral lunar eclipses are caused by the Earth's penumbra. The penumbra is a type of lighter shadow.
The Penumbra is a half-shadow region that occurs when an object only partially covers a light source.
Thus, the correct option is C.
For more details regarding lunar eclipse, visit:
https://brainly.com/question/28679608
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a boy standing by a lake sees a fish in the pond and tries to thrust a spear into it he will success or not explain with reason
He will not probably success to thrust a spear into the fish in the pond because when light travels from water to air , it bends due to refractive property of light.
What is refraction of light?Refraction is the bending of a wave as it travels through different media. The two materials' different densities are what lead to the bending.
Refraction is defined as "the change in a wave's direction as it passes through a medium."
Although light refraction is one of the most frequently seen phenomena, refraction can also occur with sound and water waves. We can use optical tools like lenses, prisms, and magnifying glasses thanks to refraction. We can focus light on our retina because of the refraction of light, which is another benefit.
When light travels from water to air , it bends due to refractive property of light. So, he e will not probably success to thrust a spear into the fish.
Learn more about refraction here:
https://brainly.com/question/14760207
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In one type of mass spectrometer, ions having the same velocity move through a uniform magnetic field. The spectrometer is being used to distinguish 12C and 14C ions. The 12C ions move in a circle of diameter 45.4 cm. Use these atomic mass values: 12C, 12.0 u; 14C, 14.0 u.
Answer:
r = 0.5297 m
Explanation:
In this exercise we use Newton's second law where the force is magnetic
F = ma
centripetal acceleration
a = v² / r
F = q v x B = q v B sin θ
where the angle between the velocity and the magnetic field is 90º, therefore the sin 90 = 1
we substitute
q v B= m v² / r
r = [tex]\frac{m v^2 }{qv B}[/tex]
the mass of each isotope is
12C
m12 = 6 m_proton + 6 m_neutrons
m12 = (6 1,673 +6 1,675) 10⁻²⁷
m12 = 20.088 10-27 kg
14C
m14 = 6 m_proton + 8 m_neutron
m14 = (6 1,673 + 8 1,675) 10-27
m14 = 23,438 10⁻²⁷ kg
in the exercise they indicate that the velocity of the two particles is the same, therefore with the initial data we can calculate the parameters that do not change in the experiment.
[tex]\frac{v}{qB} = \frac{r}{m_{12}}[/tex]
v / qB = 0.454 / 20.088 10⁻²⁷
v / qb = 2.26 10²⁵
this quantity remains constant, let's use the other data to calculate the radius
r = 23.438 10⁻²⁷ 2.26 10²⁵
r = 5.297 10⁻¹ m
r = 0.5297 m
5.
2075 Set A Q.No. 20 2070 Supp. Set B Q.No. 2 B What
happens to the kinetic energy of photo electrons when
intensity of light is doubled?
[2]
Answer:
The energy of each photon can be transformed into kinetic energy and as this energy does not change, the energy of both photoelectrons remains constant,
Explanation:
The photoelectric effect was explained by Einstein, who assumed that the lz is made up of particles called photons each of a given energy, therefore the photoelectric effect can be explained as a collision of particles.
From this explanation we see that the intensity is proportional to the number of existing particles, when we double the intensity we double the number of particles, but the energy of each particle does not change, therefore if we use the conservation of energy.
The energy of each photon can be transformed into kinetic energy and as this energy does not change, the energy of both photoelectrons remains constant, only the number of electrons expelled changes.
How does the REE change as a person ages?
Explain
I go to k12 too and i am writing this test now too. I am like pretty sure my answer is correct. I guess we can kinda help eachother with this assignment cuz i am sorta stuck too.
The older a person gets, the greater REE he has.I dont know i hope this helped a bit I know its not much i need help with this too.
A foot spa machine is an electronic gadget used for soaking,bathing and massaging the feet.
Answer:
Just gonna take this at free points and yes you are right but I am confused on what you wanted us to do
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V[tex]_B[/tex] = 54 km/hr
V[tex]_A[/tex] = 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω[tex]_B[/tex] = V[tex]_B[/tex] / α
so, we substitute
ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100
ω[tex]_B[/tex] = 15 / 100
ω[tex]_B[/tex] = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
The current supplied to an air conditioner unit is 4.00 amps. The air conditioner is wired using a 10-gauge (diameter 2.588 mm) wire. The charge density is n=8.48×1028electronsm3n=8.48×1028electronsm3 . Find the magnitude of (a) current density and (b) the drift velocity.
Answer:
a. 7.6 * 10^5 A/m^2
b. 5.6 * 10-^5 m/s
Explanation:
lol
Assume your computer has a power rating of 50.0 W, and you use your computer for 7.0 hours a day for a normal school day. If the electric company charges $0.10 per kWh, how much does it cost you to use your computer each school day?
35$
$0.035
$0.005
$0.35
A 41.0-kg crate, starting from rest, is pulled across level floor with a constant horizontal force of 135 N. For the first 15.0 m the floor is essentially frictionless, whereas for the next 12.0 m the coefficient of kinetic friction is 0.320. (a) Calculate the work done by all the forces acting on the crate, during the entire 27.0 m path. (b) Calculate the total work done by all the forces. (c) Calculate the final speed of the crate after being pulled these 27.0 m.
Answer:
Explanation:
From the information given;
mass of the crate m = 41 kg
constant horizontal force = 135 N
where;
[tex]s_1 = 15.0 \ m \\ \\ s_2 = 12.0 \ m[/tex]
coefficient of kinetic friction [tex]u_k[/tex] = 0.28
a)
To start with the work done by the applied force [tex](W_f)[/tex]
[tex]W_F = F\times (s_1 +s_2) \times cos(0) \ J[/tex]
[tex]W_F = 135 \times (12 +15) \times cos(0) \ J \\ \\ W_F = (135 \times 37 )J \\ \\ W_F =4995 \ J[/tex]
Work done by friction:
[tex]W_{ff} = -\mu\_k\times m \times g \times s_2 \\ \\ W_{ff} = -0.320 \times 41 \times 9.81 \times 12 \ J \\ \\ W_{ff} = -1544.49 \ J[/tex]
Work done by gravity:
[tex]W_g = mg \times (s_1+s_2) \times cos (90)} \ J \\ \\ W_g = 0 \ j[/tex]
Work done by normal force;
[tex]W_n = N \times (s_1 + s_2) \times cos (90) \ J[/tex]
[tex]W_n = 0 \ J[/tex]
b)
total work by all forces:
[tex]W = F \times (s_1 + s_2) + \mu_k \times m \times g \times s_2 \times 180 \\ \\ W = 135 \times (15+12) \ J - 0.320 \times 41 \times 9.81 \times 12[/tex]
W = 2100.5 J
c) By applying the work-energy theorem;
total work done = ΔK.E
[tex]W = \dfrac{1}{2}\times m \times (v^2 - u^2)[/tex]
[tex]2100.5 = 0.5 \times 41 \times v^2[/tex]
[tex]v^2 = \dfrac{2100.5}{ 0.5 \times 41 }[/tex]
[tex]v^2 = 102.46 \\ \\ v = \sqrt{102.46} \\ \\ \mathbf{v = 10.1 \ m/s}[/tex]
TRUE OR FALSE
2 QUESTIONS
NEED HELP ASAP
THX :)
LOTS OF POINTS :>
Answer: Both False
Explanation:
Our Milky Way Galaxy is a spiral galaxy. Some spiral galaxies are what we call "barred spirals" because the central bulge looks elongated
Irregualuar glaxyices are all over the place
a place where two bones come together is known as an
Answer:
a place where two bones come together is known as a join
Answer:
Hey mate.....
Explanation:
This is ur answer....
JointsJoints – A place in the body where bones come together. Non-Moveable Joints (sometimes called fixed or fibrous) – A place in the body where two or more bones come together but do not move.
Hope it helps!
mark me brainliest plz....
Follow me! :)
(5 Points)
a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)
Answer:
1. [tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Explanation:
1. From Boyles' law;
[tex]P_{1}[/tex][tex]V_{1}[/tex] = [tex]P_{2}[/tex][tex]V_{2}[/tex]
[tex]P_{1}[/tex] = 1 x [tex]10^{5}[/tex] Pa
[tex]V_{1}[/tex] = 9.6 [tex]m^{3}[/tex]
[tex]V_{2}[/tex] = 12 [tex]m^{3}[/tex]
Thus,
1 x [tex]10^{5}[/tex] x 9.6 = [tex]P_{2}[/tex] x 12
[tex]P_{2}[/tex] = [tex]\frac{100000 x 9.6}{12}[/tex]
= 80000
[tex]P_{2}[/tex] = 8 x [tex]10^{4}[/tex] Pa
2. Pressure, P = ρhg
where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/[tex]s^{2}[/tex]).
Thus,
P = 1020 x 15 x 9.8
= 149940
P = 1.5 x [tex]10^{5}[/tex] N/[tex]m^{2}[/tex]
Solve the below problems being sure to provide the correct significant figures.
1) 1000 ÷ 4.886 = __________
2) 240 ÷ 12.3 = __________
3) 80 x 4.6 = __________
4) 4.527 x 30 = __________
5) 86 x 63.855 x 8000 = __________
6) 700 x 91.186 = __________
7) 7.1 x 348 = __________
8) 50 ÷ 29.1 = __________
9) 98.773 x 24.891 x 409 = __________
10) 0.065 x 3 x 3007 = __________
Answer:
1) 204.6663938
2) 19.51219512
3) 368
4) 135.81
5) 43932240
6) 63830.2
7) 2470.8
8) 1.718213058
9) 1005550.526
10) 586.365
Most of the questions you asked were in repeating decimal form.
Explanation:
discuss two reasons why people find transition between school and university
Answer:
Is that your answer
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
[4] A tortoise and a hare cover the same distance in a race. The hare goes very fast but stops frequently while the tortoise has a steady pace and finish first
Answer:
I know that story where the hare sleeps
Water enter the horizontal, circular cross-sectional, sudden-contraction nozzle sketched below at section (1) with a uniformly distributed velocity of 30 ft/s and a pressure of 80 psi. The water exits from the nozzle into the atmosphere at section (2) where the uniformly distributed velocity is 100 ft/s. Determinethe axial component of the anchoring force required to hold the contraction in place.
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the axial component of the anchoring force required to hold the contraction in place is 365.6 lb
Explanation:
Given the data in the question and as illustrated in the image below;
first we calculate the area at section 1
A₁ = (πD²)/4
we substitute
A₁ = (π(3 in)²)/4
A₁ = 7.06858 in²
we know that; 1 ft = 12 in
A₁ = ( 7.06858 / (12²) ) ft²
A₁ = ( 7.06858 / 144 ) ft²
A₁ = 0.0491 ft²
now, we write the elation for area at section 2
A₂ = πd²/4
here, d is the diameter at section 2
next, we use the conservation of mass equation between the two section;
m" = pV₁A₁ = pV₂A₂
we calculate the mass flow rate;
m" = pV₁A₁
= (1.94[tex]\frac{slug}{ft^2}[/tex]) × 30[tex]\frac{ft}{s}[/tex] × 0.0491 ft²
= 2.8576 slug/s
Now, Apply the linear momentum along the horizontal direction for the control volume between 1 - 2
-pV₁A₁V₁ = pV₂A₂V₂ = P₁A₁ - F[tex]_A[/tex] - P₂A₂
m"( V₂ - V₁ ) = P₁A₁ - F[tex]_A[/tex] - P₂A₂
F[tex]_A[/tex] = P₁A₁ - P₂A₂ - m"( V₂ - V₁ )
we substitute
F[tex]_A[/tex] = ((80×[tex]\frac{144 in^2}{1 ft^2}[/tex])×0.0491 ft²) - (0×(πd²/4)) - 2.8576( 100 - 30 )ft/s
F[tex]_A[/tex] = 565.632 - 0 - 200.032
F[tex]_A[/tex] = 565.632 - 200.032
F[tex]_A[/tex] = 365.6 lb
Therefore, the axial component of the anchoring force required to hold the contraction in place is 365.6 lb