Answer:
The inner planets are much smaller than the outer planets and because of this have relatively low gravity and were not able to attract large amounts of gas to their atmospheres.
What travels by vibrating particles? Mechincal Waves or ElecrtoMagnetic Waves.
Answer:mechanical waves.
Explanation:
Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.
If your friend pushes a lawnmower four times as far as you do while exerting only half the force, which one of you does more work
Answer:
Him
Explanation:
He will push less but more and he will use more energy
what does a plant need to take in from environment to live?
nitrogen gas
argon
carbon dioxide
oxygen
Answer:
Carbon Dioxide.
Explanation:
General relativity can be used to calculate what?
Answer:
Equation for general relativity
Explanation:
a TV set is pushed a distance of 2 m with a force of 20 N how much work is done on the set
Answer:
40 J
Explanation:
help me for 11 point pls
Answer:
B
Explanation:
Hope this helps. :)
How much kinetic energy does leaves moving in the street have?
Answer:
as much as the wind is blowing has
why watchman does no work but gets tired ?
Answer: the body is trained to wake up when the sun rises and get the energy from the sun and sleep and it is night on when the sun is not there. So when a watchman works at night there is no sun first of all to give him any kind of energy and secondly the body clock resists him waking up thus making him tired.
Explanation:
A projectile is fired into the air. Three characteristics of its subsequent motion are
1) the horizontal component of velocity
II) the vertical component of velocity
III) the acceleration
Neglecting air resistance, which characteristic(s) change(s) whilst the projectile is in the air.
(1 Point)
I only
Il only
II and III only
I and II only
Answer:
The answer is C) II and III only
Explanation:
Two lamps rated 60W; 240V and 100W, 240Vrespectively are connected in series to a 240V power source. Calculate;
a) The resistance of each lamp.
b) The effective resistance of the circuit.
c) The current passing the lamps. pls answer correctly
Answer:
See the answers below.
Explanation:
The total power of the circuit is equal to the sum of the powers of each lamp.
[tex]P=60+100\\P=160 [W][/tex]
Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.
[tex]P=V*I[/tex]
where:
P = power [W]
V = voltage [V]
I = current [amp]
[tex]I = P/V\\I=160/240\\I=0.67 [amp][/tex]
So this is the answer for c) I = 0.67 [amp]
We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.
a)
[tex]V=I*R[/tex]
where:
V = voltage [V]
I = current [amp]
R = resistance [ohms]
Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.
[tex]P=V*I\\and\\V = I*R\\P = (I*R)*I\\P=I^{2}*R[/tex]
[tex]60 = (0.67)^{2}*R\\R_{60}=133.66[ohm] \\and\\100=(0.67)^{2} *R\\R_{100}=100/(0.66^{2} )\\R_{100}=225 [ohm][/tex]
b)
The effective resistance of a series circuit is equal to the sum of the resistors connected in series.
[tex]R = 133.66 + 225\\R = 358.67 [ohms][/tex]
a metallic cube whose each side is 10 cm is subjected to a shearing force of 100 kg. The top force is displaced through 0.25 cm with respect to the bottom. calculate the shearing stress strain and modulus
Answer:
9.8×104Nm−2,0.025,3.92×106Nm−2
Solution :
Here, L=10cm=10×10−2m
F=100kgf=100×9.8N
ΔL=0.25cm=0.25×10−2m,
Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025
=3.92×106Nm−2
Explanation:
Question 1 (0.5 points)
Electric circuits provide a path for ___ to flow.
protons
matter
electrons
energy
Electric circuits provide a path for electrons.
The passage of electrons is facilitated by electric circuits.
An electric circuit resembles a roadway consisting of wires over which electrons are transferred.
The force that causes the electrons to travel comes from a battery or another power source. To allow electrons to travel from the source of power to the gadget and back, wires must be run from the source of electricity to the gadget and back.So, Option "C" is the correct answer to the following question.
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If the object distance in the following picture were 8.0 m and the image distance were 2.0 m, what is the focal length?
Answer options: 0.63m or 1.6m
Answer:
where is the picture. .......
Answer:
1.6m
Explanation:
To solve, we must use the equation: 1/f = 1/so + 1/si
f= focal point
so=object's distance
si=image's distance
We put in the numbers we have and we get: 1/f = 1/8.0 + 1/2.0.
1/f = 0.125 + 0.5 (changed the fractions to decimals)
1/f = 0.63 (0.625 rounded to the hundredths)
Now we solve for f:
1 = 0.63f
f = 1/0.63
f = 1.6
The focal length is 1.6m.
Hope this helped!
Assuming that it could maintain the same acceleration at higher speeds, how long would it take the car to go from 0 to 120 mph
The question is incomplete, the complete question is;
A car can go from 0 to 60 mph in 7.0 s. Assuming that it could maintain the same acceleration at higher speeds, how long would it take the car to go from 0 to 120 mph? A. 10 s B. 14 s C. 21 s D. 28 s
Answer:
B
Explanation:
We have;
vo = 0 m/s
v1 = 60 mph or 26.8 m/s
v2 = 120 mph or 53.6 m/s
t1 = 7s
To obtain a
v1 = vo + at1
but vo = 0 m/s
a = v1/t1 = 26.8 m/s/7s = 3.8 ms-2
To obtain t2 for the car;
v2 = vo + at2
but vo = 0 m/s
t2= v2/a = 53.6m/s/3.8ms-2
t2 = 14.1 s
The air is an example of a gas in a gas solution. True or false
A racecar is driving around a circular track. The car is moving with a speed of 49.7 m/s, and the track has a radius of 114 m. What is the magnitude of the centripetal acceleration of the car?
Answer:
21.67 m/s²
Explanation:
From the question given above, the following data were obtained:
Velocity (v) = 49.7 m/s
Radius (r) = 114 m
Centripetal acceleration (a) =?
The centripetal acceleration can be obtained by using the following formula:
a = v²/r
a = 49.7² / 114
a = 2470.09 / 114
a = 21.67 m/s²
Therefore, the centripetal acceleration of the car is 21.67 m/s²
A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 4.0 seconds later, approximately how high is the cliff? I NEED HELP FOR THIS QUIZ LOL
Answer:
78.4m
Explanation:
In these situations, vertical and horizontal components need to be considered separately;
So, horizontal speed is irrelevant since we are talking about vertical height, i.e. the height of the cliff;
But, the fact that the speed is horizontal is relevant because this means the vertical speed initially is 0m/s;
The downwards movement of the ball is due to gravity, which is 9.8m/s²;
The time is given as 4.0s from the top of the cliff to the bottom;
We know the acceleration, time and initial speed so we can use this kinematic equation to solve for distance:
s = ut + ¹/₂at²
s = displacement (/distance)
u = initial vertical velocity = 0
t = time (time taken for the ball to fall) = 4
a = acceleration (downwards) = 9.8
Substitute these values into the equation to get the distance:
s = 0(4) + ¹/₂(9.8)(4)²
s = 78.4m
Acceleration is a change in speed or direction over time. In what two ways does the sled accelerate as it descends?
Answer:
By force and sloppy surface.
Explanation:
By applying force on the sled and slope of the path are the two ways the sled accelerate as it descends. If there is more friction between sled and the ground then force is required to push the sled to move downward while on the other hand, if the path on which sled moves is sloppy then it will move automatically without the use of force, so these two ways can accelerate the sled.
A 250-kg moose stands in the middle of the railroad tracks in Sweden, frozen by the lights of an oncoming 10,000kg train traveling at 20m/s. Even though the engineer attempted in vain to slow the train down in time to avoid hitting the moose, the moose rides down the remaining track sitting on the train’s cowcatcher. What is the final velocity of the train and moose after the collision?
(Momentum & Impulse)
Answer:
The final velocity of the train and the moose after collision is approximately 19.51 m/s
Explanation:
The given mass of the moose, m₁ = 250 kg
The velocity of the moose, v₁ = 0
The mass of the oncoming train, m₂ = 10,000 kg
The velocity of the train, v₂ = 20 m/s
The velocity of the moose and the train after collision = v₃
By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃
Therefore, by substitution, we have;
250×0 + 10,000× 20 = (10,000 + 250) × v₃
200,000 = 10,250 × v₃
v₃ = 200,000/10,250 ≈ 19.51 m/s
The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s
The main difference between distance and displacement is tat
Answer:
Distance is the length of the path taken by an object whereas displacement is the simply the distance between where the object started and where it ended up.
Explanation:
Hope this helps! Good Luck on the rest of your assignments! :)
Answer:
Distance is quite literally the total amount of distance you have covered. Displacement is the total displacement from your origin. For example, if you were to move 5m north then 2m south, your total displacement from your origin would be 3m north.
Which of the following is the most sensible way to conserve natural resources?
O Don't use any products made from trees
O Stop using electricity entirely
Never ride in a car
O Bike or walk to school instead of taking a gasoline-powered car.
Answer:Bike or walk to school instead of taking a gasoline powered car.
Explanation: gasoline is not infinite. trees can be replanted and regrown. electricity can by generated and is not a natural recourse.
help pls its been a day
Answer:
161.28m
Explanation:
The resultant is calculated using the formula
R² = dx²+dy²
R = √dx²+dy²
From the diagram. dx = 101cos60°+85
dx = 101(0.5) + 65
dx = 50.5 + 85
dx = 135.5m
For the vertical component dy
dy = dsin theta
dy = 101 sin 60
dy = 101(0.8660)
dy = 87.47
R = √135.5²+87.47²
R = √18,360.25+7,651.0009
R = √26,011.2509
R = 161.28m
Hence the magnitude of the sum of the vectors is 161.28m
Answer:
The sum of the two vectors is 161.278 m.
Explanation:
Given;
vector, B = 101 m inclined at angle 60⁰
vector, A = 85 m inclined at angle 0⁰
Y-component of the vectors;
[tex]R_y = 101(sin \ 60^0) + 85(sin \ 0^0) =87.466 \ m[/tex]
X-component of the vectors;
[tex]R_x = 101(cos \ 60^0) + 85(cos \ 0^0)\\\\R_x = 50.5 + 85 = 135.5 \ m[/tex]
Sum of the two vectors;
[tex]R = A+ B\\\\R = \sqrt{R_y^2 + R_x^2} \\\\R = \sqrt{87.466^2 \ + \ 135.5^2} \\\\R = \sqrt{26010.55} \\\\R = 161.278 \ m[/tex]
Therefore, the sum of the two vectors is 161.278 m.
what is it that makes a magnet different from a piece of iron that is not mangetic?
Answer:
Under normal conditions, a magnetic material like iron doesn't behave like a magnet because the domains don't have a preferred direction of alignment. On the other hand, the domains of a magnet (or a magnetized iron) are all aligned in s specific direction.
Calculate the mass of a man jogging north along the highway at a rate of 2.65 m/s and a momentum of 225.3 kg*m/s
Answer:
85.02 kgExplanation:
The mass of the man can be found by using the formula
[tex]m = \frac{p}{v} \\ [/tex]
v is the velocity
p is the momentum
From the question we have
[tex]m = \frac{225.3}{2.65} \\ = 85.01886...[/tex]
We have the final answer as
85.02 kgHope this helps you
PLEASE HELP! FOR BRAINLIEST!
Answer:
i read the whole thing the answer that matches the most is C
hope this helped :D
Explanation:
1. The geologic time scale divides time into years and centuries, true or false
Answer:
The answer is "False"
Explanation:
The geologic time scale is the "schedule" for occasions in Earth history. It partitions time into named units of unique time called in descending order of duration "eons, eras, periods, epochs, and ages". The specification of those geologic time units depends on stratigraphy, which is the relationship and order of rock layers. The fossil structures that happen in the stones, nonetheless, give the central methods for setting up a geologic time scale, with the circumstance of the development and vanishing of far and wide species from the fossil record being used to outline the beginnings and endings of ages,, periods, and different stretches.
Geologic time is the broad time period involved by the geologic history of Earth. Formal geologic time starts toward the beginning of the Archean Eon (4.0 billion to 2.5 billion years back) and proceeds to the current day.
Calculate the kinetic energy of a 50 kg cart moving at a speed of 18.6 m/s.
Answer:
8649 J
Explanation:
KE = 1/2mv^2
1/2(50)(18.6)^2
1/2(50)(345.96) = 8649 J
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
What would a space cadet who weighs 800 N on Earth weigh on the exoplanet?
Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N
[tex]The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N[/tex]
Therefore, for the weight of the cadet on the exoplanet, W₁, we have;
[tex]W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N[/tex]
The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
3
Question 2 (1 point)
If a runner is racing at a velocity of 1.7 m/s North for 3.0 seconds, how far did they
run?
5.1 m [N]
0.57 m IN)
57 m
51 m[N]
O 1.8 m [N]
Answer:
A, 5.1 m [N]
Explanation:
1.7 times three equals 5.1, and since the runner was running north, the direction was north.
Using Velocity vs Time Graphs to Find Acceleration
A graph titled velocity versus time has horizontal axis time (seconds) and vertical axis velocity (meters per second). A line has 4 straight segments. Line segment A runs from 0 seconds 0 meters per second to 1 seconds 15 meters per second. Then segment B runs to 2 seconds 20 meters per second. Then segment C runs to 4 seconds 20 meters per second. Then segment D runs to 5 seconds 0 meters per second.
The acceleration of segment D is m/s2.
Rank segments A, B, and C from least acceleration to greatest acceleration.
Least:
Greatest:
The acceleration at segment D is -20m/s²
The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)
Acceleration is the change in velocity with respect to time.
a = v-u/t
Acceleration at segment A:
Aa = 15-0/1-0
Aa = 15m/s²
Acceleration at segment B:
Ab = 20-15/2-1
Ab = 5m/s²
Acceleration at segment C:
Aa = 0-0/4-2
Aa = 0m/s²
Acceleration at segment D:
Ac = 0-20/5-4
Ac = -20m/s²
Hence the acceleration at segment D is -20m/s²
The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)
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