Answer:
An object's density is the ratio of mass to volume of an object. The mass is how much it resists acceleration when a force is applied to it and generally means how much of an object or substance there is. Volume describes how much space an object takes up.
12. If the 50kg rock fell from the 20m high cliff, then what would its velocity be just before it
hits the ground?
Answer:
The velocity is 2.5 before hitting the ground
Explanation:
Mark me as brainliest
2021 edge
What force prevents a machine from having 100% efficiency? A. Magnetic B. Electric O C. Friction O D. Gravity
Answer:
I believe it would be C ? my apologies if I'm wrong.
a father and his son are racing down a mountain. if they're traveling at the same velocity who has more kinetic energy
Answer:
I think it's the son
Explanation:
bcs he might love it
2. Starting from rest, a racecar has a displacement of 201 m south while accelerating at
16 m/s?. Calculate the time it takes the car to travel the 201 m.
Help solve please!
Answer:
Explanation:
Assuming you mean a = 16 m/s²
s = ½at²
t = √(2s/a)
t = √(2(201)/16)
t = 5.012484...
t = 5.0 s
help please this is due in an hour!!!
[tex]▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ [/tex]
Charge on one electron :
[tex]1.602 \times 10 {}^{ - 19} \: \: coulombs[/tex]So, charge on ball having 97 electrons is equal to :
[tex]97 \times 1.602 \times 10 {}^{ - 19} [/tex][tex]155.394 \times 10 {}^{ - 19} [/tex][tex]1.554 \times 10 {}^{ - 17} \: \: coulombs[/tex]two balls are dropped from rest and allowed to fall for 1 s and the other for 3 s, compare the distances they have fallen
In two paragraph of lines each, describe how you
would simplify, for your
grade 10 learners the
following topics
1.1 Electricity and magnetism.
The question above wants to assess your ability to simplify complex matters. In that case, I can't answer this question for you, but I'll show you how to answer it.
First, you should research electricity and magnetism. This research will make you understand the subject and all its elements. You can do this research in textbooks and digital platforms aimed at children and teenagers since these media usually present subjects in a more simplified way.
After doing this research, you will be able to write a text about electricity and magnetism as follows:
Present the main focus of this subject.Present the elements that make up this subject.Present how this subject applies in the real world.Present the importance of this subject in human life and society.Remember that you must use simple language, without the use of technical terms, but with common terms.
More information:
https://brainly.com/question/14870576?referrer=searchResults
A block of mass 2.52 kg is kept at rest
as it compresses a horizontal massless spring
(k 116 N/m) by 9.73 cm. As the block
is released, it travels 0.537 m on a rough
horizontal surface before stopping.
The acceleration of gravity is 9.8 m/s.
Calculate the coefficient of kinetic friction
between surface and block.
By Newton's second law, the net vertical force on the block is
∑ F = N - mg = 0
so that the normal force exerted by the surface has magnitude
N = mg = (2.52 kg) (9.80 m/s²) = 24.7 N
Then as the block slides over the surface, it feels a frictional force of
f = µ (24.7 N)
where µ is the coefficient of kinetic friction.
As the block is pushed by the spring to its equilibrium position, friction performs
µ (-24.7 N) (0.0973 m) = -2.40µ J
of work (which is negative because it opposes the block's motion).
In compressing the spring by 9.73 cm = 0.0973 m, we store
1/2 (116 N/m) (0.0973 m)² = 0.549 J
of energy. This energy is released and partially converted to kinetic energy, while the rest is lost to friction.
By the work-energy theorem, the total work performed on the block as the spring pushes it towards the equilibrium position is equal to the change in its kinetic energy:
W = ∆K
0.549 J - 2.40µ J = 1/2 (2.52 kg) v ² - 0
where v is the speed of the block at the equilibrium position. Solving for v, we get
v = 0.891 √(0.549 - 2.40µ) m/s
After the block is released, the only force acting on it as it slides freely is friction. It comes to a stop after 0.537 m, so that friction performs
µ (-24.7 N) (0.537 m) = -13.3µ J
of work.
Using the work-energy theorem again, we have
W = ∆K
-13.3µ J = 0 - 1/2 (2.52 kg) v ²
Substitute the velocity we found in terms of µ, and solve for µ :
-13.3µ J = -1/2 (2.52 kg) (0.891 √(0.549 - 2.40µ) m/s)²
===> µ = 0.0350
A rock is held steady over a cliff and dropped. 1 seconds later, another rock is thrown straight down at a speed of 11.3 m/s, and hits the first rock. How far have the rocks dropped before they collide? How long is the first rock in the air before it gets hit by the second rock?
What is the displacement of the rocks when they collide? __________ (Hint: be careful of sign -- the rocks drop.)
The first rock is in the air for __________ before it is hit.
Answer:
4.27s
Explanation:
If "t" represents the time traveled from the time rock 2 is dropped until the collision, then the time traveled for rock 1 = t + 1. And, since rock #1 is dropped making its initial velocity = 0, then:
The distance rock 1 travels is
x = (0)(t + 1) + 1/2(-9.8)(t + 1)2 = -4.9(t2 + 2t + 1) = -4.9t2 - 9.8t - 4.9
The distance rock 2 travels
x = -11.3t + 1/2(-9.8)t2 = -11.3t - 4.9t2
For the distances must be equal when the rocks collide:
-4.9t2 - 9.8t - 4.9 = -11.3t - 4.9t2
-9.8t - 4.9 = -11.3t
-4.9 = -1.5t
t = 3.267 s
Now, the distance they traveled can be found by plugging the 3.267 s back into either equation:
x = -11.3(3.267) - 4.9(3.267)2 = -89.2 m or 89.2 m below where they began
The time the first rock was in the air is t + 1 = 3.267 + 1 = 4.267 s = 4.27 s
Let
First rocks time be x
Second rocks time be x+1
initial velocity=u=11.3m/s
Distance of both rocks be s1 and s2
[tex]\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2[/tex]
Now
[tex]\\ \sf\longmapsto s1=11.3x+5x^2[/tex]
[tex]\\ \sf\longmapsto s2=11.3(x+1)+5(x+1)^2[/tex]
As both collide then
[tex]\\ \sf\longmapsto s1=s2[/tex]
[tex]\\ \sf\longmapsto 11.3x+5x^2=11.3x+11.3+5(x+1)^2[/tex]
[tex]\\ \sf\longmapsto 5x^2=11.3+5x^2+10x+1[/tex]
[tex]\\ \sf\longmapsto 10x+12.3=0[/tex]
[tex]\\ \sf\longmapsto x=1.23s[/tex]
Displacement
[tex]\\ \sf\longmapsto 11.3(1.23)=13.8m[/tex]
Done
Say hi :P...................................................................
HELLOOOOOOOOOOOOOOOOOOOOO
find the current and voltage across each resistor in the given circuit.
Answer:
R1, R4: 1.3433 VR2: 0 VR3: 6.3134 VExplanation:
The resistor R2 is shorted out, so is not in the circuit. The remaining resistors are in a series circuit with a total resistance of ...
R1 +R3 +R4 = 1k +4.7k +1k = 6.7k
The current is the ratio of the applied voltage to this total resistance:
I = V/R = (9 v)/(6.7k) ≈ 1.3433 mA
The voltage across each resistor is the product of this current value and the resistance:
R1 voltage = (1 kΩ)(1.3433 mA) = 1.3433 volts
R2 voltage = 0
R3 voltage = (4.7 kΩ)(1.3433 mA) = 6.3134 volts
R4 voltage = (1 kΩ)(1.3433 mA) = 1.3433 volts
please can anybody tell me what are these lab equipments
Answer:
5 is the tripoid stand
Thanks have a bangtastic day
Answer:
hii there
5 ) tripod
6 ) test tube holder
Explanation:
Hope it helps
Have a nice day : )
What is the earths layers?
Everytime a train passes Ms. Lindsey's house, one specific item on her dresser rattles. Use physics to explain why this occurs.
Answer: gravity
Explanation: the moving and sifting around in the train
Which of the following is the best hypothesis
Answer:
Melting and frezzing are physical changes
plz help owo
What characteristic is shared by sunspots, solar flares, and coronal mass ejections? Select the two correct responses.
They are caused by outward fluctuations of the sun's magnetic field.
They appear as dark spots on the sun's surface.
They tend to occur during active periods of the solar cycle.
They appear as bright spots on the sun's surface.
Answer:
They are caused by outward fluctuations of the sun's magnetic field.
They tend to occur during active periods of the solar cycle.
Explanation:
While sun spots will appear darker than the surrounding surface, flares will tend to appear brighter. CME's are associated with massive flares.
The two correct responses are as follows:
They are caused by outward fluctuations of the sun's magnetic field.They tend to occur during active periods of the solar cycle.The following information should be considered:
At the time when sunspots should appear darker as compared to the surrounding surface, so flares should be appear brighter. CME's are along with massive flares.Due to this, the above answers are considered.Learn more: brainly.com/question/16911495
For question 51, do you use a=d/t
No. That won't help. It's not even true.
PLEASEEEEEE HELPPPPPPP MEEEEEEEE!!!!!!!!!!!!!
Answer:
[tex]2 \\ because \\ 2 - 2 = 0 \\ 5 - 5 = 0[/tex]
Helppppppppppppppp please and explain!
A 500 kg airplane wing is hanging on a rope from the ceiling. What is the tension force being supplied by the rope?
a.5000
b.2500
c.250
d.500
Answer:
a. 5000 N
Explanation:
If gravity is 10 m/s² and the wing is motionless.
F = ma = 500(10) = 5000 N
You're out for run. Your initial velocity is 1.5 m/s. Suddenly a crazy dog starts chasing you and you accelerate to a velocity of 3 m/s. It takes you 10 seconds to find a safe place to hide from the Chihuahua. What was your displacement?
Show your work for displacement.
Include units.
The displacement of your motion during the entire motion is 22.5 m
The given parameters;
your initial velocity, u = 1.5 m/syour final velocity , v = 3 m/stime of motion, t = 10 sThe displacement of your motion during the entire motion is calculated from your average velocity and time of motion.
This magnitude of this displacement is calculated as follows;
[tex]s = (\frac{u+ v}{2} )t\\\\s = (\frac{1.5 + 3}{2} ) \times 10\\\\s = 22.5 \ m[/tex]
Thus, the displacement of your motion during the entire motion is 22.5 m.
Learn more here: https://brainly.com/question/17345815
Which statement describes Kepler’s third law of orbital motion?
Answer:
The square of orbital period is proportional to the cube of the semi-major axis.
Explanation:
I just took the quick check
Which stament describes Newton’s law of universal gravitation
Mass has little effect on gravity between objects
Gravity pushes objects away from earths center
Gravity does not act between earth and the moon
Every object in the universe attracts every other object
I’ll give brainliest
We can only tell the earth is rotating almost 960 miles an hour if we
a
look at other people around us as a reference point.
b
look at mountains far away from us as a reference point.
c
use a reference that is not on the earth, like the sun or stars.
Answer:
C
It is C i have to type this because i need 20 characters.
If a material were being designed to demonstrate no thermal expansions, how would the energy plot look like?
The characteristics of thermal expansion allow finding that the response for a material without thermal expansion is
The length variation is zero In the graph the line is horizontal so there is no change in length with temperature
Thermal expansion is the macroscopic sum of the changes in the length of the bonds when the energy (temperature) changes, it can be written
ΔL = α L₀ ΔT
Where ΔL is the change in length, α the coefficient of linear expansion, L₀ the initial length and ΔT the change in body temperature
In this case, a material is being designed that the thermal expansion is very small, for this the material must be made up of several compounds where some of them present a contraction with temperature, some examples: water at low temperature, liquefied gases , ceramic tile, quartz, etc.
The thermal expansion measurement processes control the body temperature and measure the change in length, in this case the change in length must be zero, in the attachment we can see a graph of a composite material with these characteristics, an example of this type of material is Invar an alloy of nickel and iron α = 3.7 10⁻⁶ ºC⁻¹
In conclusion, using the characteristics of thermal expansion we can find that the response of material without thermal expansion is
The length variation is zero In the graph the line is horizontal so there is no change in length with temperature
Learn more here: brainly.com/question/18717902
the atmosphere of mercury and mars are very thin.what effects does the thin atmosphere on the temperature on the surface of this planet?
Answer:
The effects of the thin atmosphere are that the surface of the planet is going to be much hotter and not only that but also more radioactive than a planet with a thicker atmosphere. The thick atmosphere on other planets prevents a lot of radiation from going straight to the surface of the planet.
Explanation:
A dog is running at an initial speed of 10 m/s. He covers 50 m in 4 seconds. What is the acceleration of the dog?
a.
1.25 m/s2
b.
1.04 m/s2
c.
5 m/s2
d.
3.2 m/s2
Michael stands upright and allows his right arm to swing back and forth, naturally and without forcing, like a pendulum. He measures the time for 21 oscillations (complete cycles) to be 28.1 s. Assume his arm (length L) behaves as a simple pendulum, with an effective length of 2/3 L. Find the length of his arm, L.
Answer:
ytdgnrjjksmzjj3hfgbdhehkj jfjbcjbwbjrhrhb jdhbbjhhrhrhhjrjjr jfjnekrj
The length of his arm, L = 0.67 m.
What is time period?
A basic pendulum consists of a point mass hung by a tightly supported, weightless, and inextensible string.
T = 2π√ L/g, where L is the pendulum's length and g is the acceleration brought on by gravity, is the formula for a pendulum's period.
Given in the question the time for 21 oscillations (complete cycles) to be 28.1 s. Assume his arm (length L) behaves as a simple pendulum, with an effective length of 2/3 L.
T = 2π√ L/g
T = 2π√2l/3g
T = 28.1/21
T = 1.33
2π√2l/3g = 1.33
l = .67 m
The length of his arm, L = 0.67 m.
To learn more about time refer to the link:
brainly.com/question/13764813
#SPJ5
PLEASE HELP!!
What process is illustrated in the figure?
Answer: A
Explanation:
If an athletic field has 440 metre, how many rounds must an athlete do to cover a half marathon, I.e 21km and 200 metre
Answer:
53
Explanation:
Given:
perimeter of the field = 440 m
Half of the distance of marathon = 21 km and 200 meter
Step-by-Step Explanation:
21 km 200 meter
=
21200 Meter (1km= 1000M)
21200 ÷ 400
= 53