Answer:
i Believe the correct answer is "An open circuit being changed into a closed circuit"
Explanation:
an object has a mass of 2000kg what is its weight on earth
Answer:
19600 N
Explanation:
weight = mass x gravity
We know that gravity = 9.8 m/s^2 and mass = 2000 kg.
w = m x g
w = 2000 kg x 9.8 m/s^2
w = 19600 N
The weight of the object is 19600 N (newtons).
Answer:
the answer i 2000kg
Explanation:
First to answer gets brainliest
Answer:
Sodium (K)
Explanation:
An object undergoes constant acceleration after starting from rest and then travels 5m in the first second. Determine how far it will go in the next second
The speed will be 10 m/s after the 1st and 20 m/s after the 2nd for an average of 15 m/s. So it will travel 15 m during that 2nd second
Mark as brainlist
The object, which undergoes constant acceleration after starting from rest, will go in the next second 15 m.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given parameters:
initial velocity of object: u = 0.
time = 1 second.
distance travelled: d= 5 m.
So, acceleration of the object: a = 2d/t² = (2×5)/1² m/s² = 10 m/s².
Hence, it will go in the next second = 1/2×a(2²-1²) m
= 1/2×10×3 m.
= 15 m.
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At what height does a 3500-kg truck have a potential energy of 90,000 J gravitational potential energy relative to the ground?
Answer:
MGH=energy
3500*9.8*h=90000
h=90000/34300
h=2.62m
1)A rocket expels gas at high speed for a short period of time. We are going to treat the rocket as being far away from any gravitational objects.a)Draw a momentum chart for the rocket expelling gas in space.Take the initial time before expelling gas and the final time after the rocket has finished expelling gas. The rocket has an initial constant speed.Put the rocket and the expelled gas on separate rows.b)Use your chart to explainhow the speed of the rocket changes. c)Does the rocket have to keep expelling gas to stay at a constant speed
We have that
a) We Draw a graph to follow the equation
[tex]v=\frac{m_0u}{m_0-dt}[/tex]
b) The speed of the rocket changes because Momentum is conserved because with the law of conservation in mind mass is reduced thereby causing increase in speed
c) The rocket does NOT have to keep expelling gas to stay at a constant speed because On the account of maintaining speed the Rocket need not expel gas as that will only increase speed and not maintain it
a)
Let
Mass =m
Time t
Generally, the equation for mass ejection constant is mathematically given by
[tex]\phi=\frac{d_m}{d_t}[/tex]
Therefore
[tex]m=m_0-\phi t[/tex]
where
[tex]m_0=initial\ mass[/tex]
Apply the law of conservation of momentum which states that
Conservation of momentum, states that momentum can neither be lost nor gained in an isolated system
[tex]m_o\mu=mv[/tex]
[tex]v=\frac{m_0u}{m_0-dt}[/tex]
b)
Momentum is conserved because with the law of conservation in mind mass is reduced thereby causing increase in speed
c)
On the account of maintaining speed the Rocket need not expel gas as that will only increase speed and not maintain it.Therefore the answer is NO
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Which statement applies only to magnetic force instead of both electric and magnetic forces? O A. It acts between a north pole and a south pole. O B. It can push objects apart. O C. It can pull objects together. D. It acts between objects that do not touch.
Answer:
the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.
Answer:
A
Explanation:
I did the test on ap3x
Ang larong Latin at Sisiw ay________________________.
larong pinoy
Explanation:
ito ay larong Pinoy
compare and contrast speed and velocity.
Speed is the time rate of an object moving from one place to another, while velocity is the rate and direction of the object's movement. They are very similar but they don't mean the same thing.
Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3 feet. Group B runs a similar experiment 3 times and calculates that the best estimate of the distance slid by green blocks is 25 + 4 feet. Using what you learned in the above video, find the t' parameter for the comparison of the results of Groups A and B.
In comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
The given parameters;
in group A, distance traveled by the red block, s = 15 ± 3 feet
in group B, distance traveled by the green block, s = 25 ± 4 feet
To find:
the t' parameter for comparison of the results of Groups A and B.Since both groups performed the experiment at equal times, we assume the time for both motion = t
Also, assume the initial velocity of both blocks = 0
For group A, we set-up the equation of motion as follows;
[tex]s = v_0t + \frac{1}{2} at^2 \\\\15+ 3 = 0 + 0.5\times a_1t^2\\\\18 = 0.5a_1t^2\\\\t^2 = \frac{18}{0.5a_1} \\\\t^2 = \frac{36}{a_1}[/tex]
For group B, we set-up the equation of motion as follows;
[tex]25 + 4 = v_0t + \frac{1}{2} a_2t^2\\\\29 = 0.5\times a_2t^2\\\\t^2 = \frac{29}{0.5a_2} = \frac{58}{a_2}[/tex]
Solve the first equation and the second equation together;
[tex]\frac{36}{a_1} = \frac{58}{a_2} \\\\\frac{a_2}{a_1} = \frac{58}{36} \\\\\frac{a_2}{a_1} = 1.61[/tex]
The ratio of error margin of both experiments;
[tex]\frac{4}{3} = 1.33[/tex]
The resulting parameter for comparison;
[tex]parameter, t' = 1.33 \times 1.61 = 2.14 \approx 2.0[/tex]
Thus, in comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
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How does energy move in relation to the medium in a transverse wave?
Answer:
Only the energy of the wave travels through the medium. In a transverse wave, particles of the medium vibrate up and down perpendicular to the direction of the wave. ... In a surface wave, particles of the medium vibrate both up and down and back and forth, so they end up moving in a circle.
Calculate the magnitude and direction of the resultant of the following forces
Answer:
add them
100+150 = 250N same direction
that's the resultant
same direction
At last year's Homecoming Pep Rally, Trudy U. Skool (attempting to generate a little excitement) slid down a 42.0 degree incline from the sports dome to the courtyard below. The coefficient of friction between Trudy's jeans and the incline was 0.650. Determine Trudy's acceleration along the incline.
Answer:
a = - 1.8 m / s²
Explanation:
To solve this exercise let's use Newton's second law, let's start by defining a reference system with the x axis parallel to the plane, in the adjoint we can see a diagram of the forces
let's break down the weight
sin 42 = Wₓ / W
cos 42 = W_y / W
Wₓ = W sin 42
W_y = W cos 42
Y axis
N- W_y = 0
N = W_y
N = mg cos 42
X axis
fr -Wₓ = m a
a = (fr - mg sin 42) / m
the friction force has the formula
fr = μ N
fr = μ mg cos 42
we substitute
a = (μ g cos 42– g sin 42)
a = g (μ cos 42 - sin 42)
let's calculate
a = 9.8 (0.650 cos 42 - sin 42)
a = - 1.8 m / s²
The negative sign indicates that the acceleration is downward, so the boy is descending.
1. Hydrogen: For an electron in the lowest energy it can orbit around proton, they have a separation of 5.3 *10-11 m. If you have a 4.5*10-19J photon (bit of light) hit the electron, it will transfer all of its energy to the proton electron interaction and the electron will start orbiting at a larger radius. Assuming all the energy went into the potential energy, what is the new distance between the electron and proton.
Answer:
rₙ = 1,325 10⁻⁹ m
Explanation:
To solve this problem we use the bohr atomic model
Eₙ = -13.606 /n² [eV]
the brackets indicate that the units are in electron volts.
let's reduce the photon energy to eV
E = 4.5 10-19J (1 eV / 1.6 10⁻¹⁹ eV) = 2.8125 eV
This energy is in the visible range, so the transition must occur in this range, this is for the Balmer series whose initial number is n₀ = 2
for an atomic transition on two levels
ΔE = Eₙ - E₀ = [tex]\frac{-13.606}{n^2} + \frac{13.606}{2^2}[/tex]
2.8125 = [tex]\frac{-13.606}{n^2} + 3.4015[/tex]
[tex]\frac{13.606}{n^2}[/tex] = 3.4015 - 2.8125 = 0.589
n² = 13.606 / 0.589
n² = 23.1
n = 4.8
as n must be an integer
n = 5
taking the quantum number as far as the electron goes, we substitute in the equation for the radius
rn = n² a₀
where ao is the radius of the lowest level a₀ = 5.3 10⁻¹¹ m
rₙ = 5 2 5.3 10⁻¹¹
rₙ = 132.5 10⁻¹¹ m
rₙ = 1,325 10⁻⁹ m
A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 7.0 m/s2; after 3.8 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 73.3 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car
Answer:
t = 13.3 s
Explanation:
The distance traveled by both cars once they are in the main speedway, assuming that the acceleration of the refueling car is constant, is given by the following kinematic equation:[tex]x = v_{o}*t + \frac{1}{2} * a * t^{2} (1)[/tex]
The refueling car (which we will call car 1) in the moment that enters to the main speedway, has achieved a speed that can be found from the definition of acceleration, rearranging terms, as follows:[tex]v_{f1} = a* t = 7.0m/s2*3.8s = 26.6 m/s (2)[/tex]
So, since vf1 = v₀ in (1), we get:[tex]x_{1} = v_{f1}*t + \frac{1}{2} * a * t^{2} (3)[/tex]
Now, for the other car (which we will call car 2), due to is moving at a constant speed, a=0, so we can write the following equation for x₂:[tex]x_{2} = v_{f2}*t = 73.3m/s*t (4)[/tex]
When the entering car catches up the other car, both distances will be equal each other, so x₁ = x₂, as follows:[tex]26.6m/s*t + \frac{1}{2} * 7.0m/s2* t^{2} = 73.3m/s*t[/tex] Rearranging, simplifying and solving for t:[tex]t =\frac{2*(73.3m/s-26.6m/s}{7.0m/s2} = 13.3 s (5)[/tex]Which three statements are true of all matter?
A.
It is filled with air.
B.
It takes up space.
C.
It contains aluminum.
D.
It has mass.
E.
It is made up of atoms
Answer:
B, D and E, not all matter can be filled with air
The Earth’s orbit _____.
is an ellipse
goes around the moon
is a circle
causes day and night
In the diagram above, where is light energy converted to chemical energy?
A. Sun
B. Grass
C. Grasshopper
B. Hawk
Answer:
B. Grass
Explanation:
In the diagram above, Grass is where, light energy converted to chemical energy. Hence option B is correct.
What is Visible light ?Visible light spectrum is nothing but the range of wavelength of radiation from 4000 angstrom to 7000 angstrom(Violet to Red). light is a energy packet. Every Photon having different wavelength travels with same velocity c (velocity of light). When we focus numbers of colors from visible spectrum to a point, that point appears as a white light. hence white light is composed of numbers of Colors in it.
Light itself is a energy when it falls on the grass, grass grows by taking energy from the light and convert that energy into biomass that is a chemical energy.
Hence option B is correct.
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Two blocks of the same mass but made of different material slide across a horizontal, rough surface and eventually come to rest. A graph of the kinetic energy of each block as a function of position along the surface . Which of the following is a true statement about the frictional force Ff that is exerted on the two blocks?
a. Fr=2F8, since the force of friction is represented as the slope for each of the two curves.
b. Fr.-12Fri, since the force of friction is represented as the inverse slope for each of the two curves.
c. Ff:=2Ffi, since the force of friction is represented as the inverse of the area bound by each curve and th horizontal axis.
d. Fe=1/2Fr., since the force of friction is represented as the area bound by each curve and the horizontal axis.
Answer:
a. [tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex], [tex]\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}[/tex]
Explanation:
From the information given;
Using the work-energy theorem
ΔKE = W = [tex]\mathbf{ F_f \times r}[/tex]
K = [tex]\mathbf{ F_f \times r}[/tex]
∴
[tex]\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})[/tex]
Since [tex]K_1 = K_2[/tex] and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)
Then;
[tex]1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})[/tex]
[tex]\mathbf{F_{f_2} = 2 F_{f1}}[/tex]
The cart is given an initial push up the ramp. After this push, as the car moves up the ramp, the direction of the acceleration of the cart is ________ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _______.
Answer:
Explanation:
The only force acting on the cart is a component of its weight parallel to ramp downwards . No other force acts parallel to the ramp .
Even when the cart is moving up after the initial push , its weight is acting downwards so acceleration is acting downwards .
When the cart is stationary at the top position , its weight is acting downwards so acceleration is downwards at that moment also . When the cart is going downwards , still its weight is acting down so acceleration is acting downwards .
After this push, as the car moves up the ramp, the direction of the acceleration of the cart is _down _______ the ramp. After the reaches its highest point, turns around, and begins moving down the ramp, the direction of the acceleration of the cart is _down ________ the ramp. At the highest point the cart reaches on the ramp, when the cart momentarily comes to rest, the magnitude of the acceleration of the cart is _downwards ______.
If a motorbike accelerates from 15m/s to 25m/s in 15 seconds how far does it travel in that time
Answer:
distance = 0.2330142 miles
= 0.2330142 mi
Explanation:
not sure if its right though....
True or false it is impossible to determine weather you are moving unless you can touch another object
Answer: false
Explanation:
Answer:
false
Explanation:
According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.
In the absence of air resistance, ___ accelerates all objects at the same rate of 9.8 m/s2
Answer:
Gravity...this was proven by NASA and was examined by Leonardo da Vinci.
Answer:
Gravity
Explanation:
In the absence of air resistance, Gravity accelerates all objects at the same rate of 9.8 m/s2
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
The stars, Rigel and Betelgeuse, are both found in the constellation Orion. Rigel is a blue supergiant, and Betelgeuse is a red supergiant. Which of the following correctly compares the temperatures of Rigel and Betelgeuse?
Answer:
batrix
Explanation:
A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 75.0 mi/h can safely round it regardless of the condition of their tires
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;
[tex]\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0[/tex]
Therefore, the banking angle of the road is 24.2⁰
The angle of banking is 24 degrees.
What is the angle of banking?As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.
The angle of banking depends on the speed of the vehicle and the radius of the curve.
θ = v^2/rg
speed = 75.0 mi/h or 33.5 m/s
r = 255 m
g = 9.8 ms-1
θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1
θ = tan-1(1122.3/2499)
θ = 24 degrees
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someone help me with this exercise ?
1. if a body with a mass of 350kg is subjected to a fare of 90n what will be its mass
?
Mass remains mass no matter what you do to it.
An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta leading away from the heart. Since blood within the heart is essentially stationary, this pressure difference can be inferred from a measurement of the speed of blood flow in the aorta. Take the speed of sound in stationary blood to be c.
a. Sound sent by a transmitter placed directly inline with the aorta will be reflected back to a receiver and show a frequency shift with each heartbeat. If the maximum speed of blood in the aorta is v, what frequency will the receiver detect? Note that you cannot simply use the textbook Doppler Shift formula because the detector is the same device as the source, receiving sound after reflection.
b. Show that in the limit of low blood velocity (v <
f= 2fo v/c
Answer:
a) f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex] , b) Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo[tex]\frac{c+v}{c}[/tex]
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’ [tex]\frac{c}{ c-v}[/tex]
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀ [tex]\frac{c}{c-v}[/tex]
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀ [tex]\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }[/tex]
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
[tex]( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}[/tex]
f ’’ = fo [tex]( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )[/tex]
f ’’ = fo [tex]( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )[/tex]
leave the linear term
f ’’ = f₀ + f₀ 2[tex]\frac{v}{c}[/tex]
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀ [tex]\frac{v}{c}[/tex]
When a space shuttle takes off, the chemical reactions of the fuel give the shuttle the kinetic energy to leave Earth's atmosphere as shown in the figure below. The kinetic energy of the space shuttle is less than the potential energy of the fuel used. Which statement best explains this idea?
A.) The potential energy is used to overcome Earth’s gravity.
B.) The potential energy is also converted to light, thermal energy, and sound energy.
C.) The potential energy must be consumed to make the fuel burn.
D.) The potential energy is destroyed by the warmth of the reaction.
Answer:a
Explanation:
Because its has to use tihs potential energy to overcome the atmosphere so the shuttle will not go back down
The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2 , determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answer:
158.32 K = -114.83 °C
Explanation:
Since P = P' where P = power absorbed and P' = power radiated
P = αAQ where α = absorptivity = 0.3, A = area of spacecraft and Q = rate of incident solar radiation = 950 W/m²
Also, P' = εσAT⁴ where ε = emissivity of spacecraft = 0.8, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴, A = area of spacecraft and T = surface temperature of spacecraft.
So, P = P'
αAQ = εσAT⁴
T⁴ = αQ/εσ
T = ⁴√(αQ/εσ)
Substituting the values of the variables into the equation, we have
T = ⁴√(0.3 × 950 W/m²/(0.8 × 5.67 × 10⁻⁸ W/m²K⁴))
T = ⁴√(285 W/m²/(45.36 × 10⁻⁸ W/m²K⁴))
T = ⁴√(6.2831 × 10⁸ K⁴))
T = 1.5832 × 10² K
T = 158.32 K
In Celsius T(C) = 158.32 - 273.15 = -114.83 °C
The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. What is the angle?
Answer:
θ = 90º
Explanation:
The velocity is given by
v = [tex]\frac{dr}{dt}[/tex]
calculate
v = 3 i ^ + √2 j ^ + 2t k ^
acceleration is defined by
a = dv / dt
a = 2 k ^
one way to find the angle is with the dot product
v. a = | v | | a | cos θ
cos θ= v.a / | v | | a |
Let's look for the value of each term
v. a = 4 t
| v | = [tex]\sqrt{3^2 + 2 + (2t)^2 }[/tex] = [tex]\sqrt{ 11 + 4t^2}[/tex]
| a | = 2
they ask us for the angle for time t = 0
v. a = 0
| v | = √11 = 3.317
we substitute
cos θ = 0 /√11
cos θ = 0
therefore the angles must be θ = 90º