Calculate the momentum of a 6 kg ball thrown at 20 m/s by a 3 newton
force. *
Answer:
momentum = mass × velocity = 6× 20 =120 kg.ms-1
Explanation:
not sure if this is right
all of the following elements will form ions by losing electrons except
aluminum
iron
sodium
oxygen
.) What will be the unit of time in that system in
which unit of length is m, unit of mass is kg and
unit of force is kg wt?
(a) [1/V(9.8) second (b) (9.8)2 second
(c) V(9.8) second
(d) 9.8 second
Answer:
yes it is a great question. Thank you .
Consider the following three concentric systems two thick shells and a solid sphere all conductors The radii in the increasing order are a b c d and e The small sphere is given an excess charge of 3 C and the smaller shell is given an excess charge of 7 C The larger shell is electrically neutral The system quickly comes to electrostatic equilibrium state a Note that there are 5 conducting surfaces What are the electric charges values and signs on the each of them Are these charges distributed uniformly
Answer:
Explanation:
From the given question, the small sphere was provided with an excess charge of +3 C, while the smaller shell was given an excess of -7 C, it should be -7 C and not 7 C.
So, in light of that, to determine the electric charges values & signs on each of them, we have:
on a = +3 C
on b = -7 C
on c = -7 C
on d = +3 C
on e = -7 C
Olive and her friend Wellington are playing down by Captain Don's docks when they find an old chain. The old chain has only three links. By measuring with an old fish scale which is a permanent feature of the dock area, they determine that the total mass of the chain is 3.12 kg (the scale reads in newtons, but Olive knows how to calculate the mass of the chain from its weight). While playing with the scale and the chain (the chain is hanging vertically from the end of the scale, and Olive is holding on to the top of the scale with both hands, either moving the entire system upwards or downwards), Olive notices that if she is accelerating the chain either upwards or downwards, the scale no longer accurately reads the weight of the chain. When the Scale Reading is Larger than the Weight of the Chain At one point in their experiments with the chain and the scale, Wellington observes that the scale reads 47.00 N . Part A When the scale reads 47.00 N , what is the tension in the chain at the point where the lowest two links connect
Answer:
T₁ = 15.66 N
Explanation:
From the given information:
Using the free body system in the chain.
T - mg = ma
47 - 3.12 × 9.8 = 3.12a
47 - 30.576 = 3.12a
16.424 = 3.12a
a = 16.424/3.12
a = 5.26 m/s²
Now, by the free body diagram of the lowest link; the tension (T₁ ) in the chain is:
T₁ - (3.12/3) × 9.8 = (3.12/3) × 5.26
T₁ - 1.04 × 9.8 = 1.04 × 5.26
T₁ - 10.192 = 5.4704
T₁ = 5.4704 + 10.192
T₁ = 15.6624
T₁ = 15.66 N
0.55 kg mouse moving E at 60m s or a 900 kg elephant moving E at 0.03m Which has the most momentum?
Answer:
the mouse
Explanation:
the mouse has a momentum of 33 m kg/s
while the elephant has a momentum of 27 m kg/s
i found this out using p=mv
During a home run, the batter only needs to run around all 4 bases if he wants to, since the ball cleared the outfield fence.
True
False
Answer: False
Explanation:
A spring with a spring constant of 80 N/m is at rest and is attached to a block. If the block (attached to the end of the spring) is moved from 0.6 m to 0.4 m and then let go, what will be the spring force exerted on the block? Rightward is positive.
Answer:
–I6 N
Explanation:
From the question given above, the following data were obtained:
Spring constant (K) = 80 N/m
Initial position = 0.6 m
Final position = 0.4 m
Force (F) =?
Next, we shall determine the compression of the spring. This can be obtained as follow:
Initial position = 0.6 m
Final position = 0.4 m
Compression (x) =?
Compression = Final position – Initial position
Compression (x) = 0.4 – 0.6
Compression (x) = – 0.2 m
Finally, we shall determine the force. This can be obtained as follow:
Spring constant (K) = 80 N/m
Compression (x) = – 0.2 m
Force (F) =?
F = Kx
F = 80 × –0.2
F = –I6 N
can someone please answer this for me ❤️
Answer:
I don't understand the question
Explanation:
sorry I cant help because I am just a first former
What scientific observation did Edwin Hubble use to determine distances between galaxies?
Answer: the expanding universe
Explanation:
Hope that helps!
There are around one billion light years across galaxies on ordinary. The observable cosmos has around 100 billion galaxies.
What is scientific observation?In research, observation is vital. Scientists gather and record data through observation, which allows them to create and subsequently test concepts and hypotheses. Scientists can observe in many different ways, including using their own senses or instruments like telescopes, thermometers, satellites, or stethoscopes.
In addition, Hubble proved a basic fact now known as Hubble's Law: galaxies distant from us are retreating faster than those closer to us. The Big Bang Theory's central tenet is the notion of an expanding cosmos. The first understanding of the beginnings of our cosmos came from Hubble's observations.
There are around one billion light years between galaxies on normal. The observable cosmos has around 100 billion galaxies.
More about the scientific observation link is given below.
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Explain how magnets have an attractive force and repulsive force
2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R= 15mm in each case; assume that the density of steel is 7.8 g/cm3; and assume that the density of aluminum is 2.7 g/cm3.a.A solid steel sphere slidingdown the ramp without friction.b.A solid steel sphere rollingdown the ramp without slipping.c.A spherical steel shell with shell thickness 1.0 mm rollingdown the ramp without slipping.d.A solid aluminum sphere rollingdown the ramp without slipping.
Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Explanation:
Given that;
height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum = 2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ = [tex]\frac{1}{2}[/tex] mv²
v² = 2gh₁
v = √(2gh₁)
from the second equation; s = ut + [tex]\frac{1}{2}[/tex] at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )
we plug in our values
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{5}[/tex]mR²) ω²
v = √( [tex]\frac{10}{7}[/tex]gh₁ )
so we substitute √( [tex]\frac{10}{7}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( [tex]\frac{10}{7}[/tex]gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c)
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{3}[/tex]mR²) ω²
v = √( [tex]\frac{6}{5}[/tex]gh₁ )
so we substitute √( [tex]\frac{6}{5}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √( [tex]\frac{6}{5}[/tex]gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] [tex]I_{}[/tex]ω²
mgh₁ = [tex]\frac{1}{2}[/tex] mv² + [tex]\frac{1}{2}[/tex] ([tex]\frac{2}{5}[/tex]mR²) ω²
v = √( [tex]\frac{10}{7}[/tex]gh₁ )
so we substitute √( [tex]\frac{10}{7}[/tex]gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( [tex]\frac{10}{7}[/tex]gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
We have that for the Question it can be said that
[tex]A solid steel sphere sliding down the ramp without friction = 1.55m[/tex][tex]A solid steel sphere rolling down the ramp without slipping = 1.309m[/tex]A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping =[tex]1.2m[/tex]A solid aluminum sphere rolling down the ramp without slipping = [tex]1.309m[/tex]From the question we are told
Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R= 15mm in each case; assume that the density of steel is 7.8 g/cm3; and assume that the density of aluminum is 2.7 g/cm3.
Generally the equation for sliding without friction is mathematically given as
[tex]V = \sqrt{4Hh}[/tex]
the equation for sliding without slipping is mathematically given as
[tex]X = \sqrt{\frac{4Hh}{1+I/mR^2}}[/tex]
A) A solid steel sphere sliding down the ramp without friction.[tex]V = \sqrt{4*1.5*0.4}\\\\= 1.55m[/tex]
B) .A solid steel sphere rollingdown the ramp without slipping.[tex]I = 2/5 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m[/tex]
C) A spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping.
[tex]I = 2/3 mR^2\\\\X = \sqrt{\frac{4*1.5*0.4}{1+2/3}}\\\\= 1.2m[/tex]
D) A solid aluminum sphere rolling down the ramp without slipping.
[tex]X = \sqrt{\frac{4*1.5*0.4}{1+2/5}}\\\\= 1.309m[/tex]
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HELP ASAP PLS ILL GIVE YOU BRAINLIEST
Select the correct answer.
Which civilization was the first to record a solar eclipse?
A.
Greek
B.
Chinese
C.
Aztec
D.
Polynesian
Answer:
chinese
Explanation:
Answer:
B chinese
hope this helps you
What is the subject Physic about
Explanation:
Physics is the branch of science concerned with the study of the properties and interactions of space, time, matter and energy.
Answer:
if you are very excited to know why ball doesn't go up
For Water, determine the missing property and determine the state. 1) P = 300 kPA, v = 0.5 m3/kg. Find T 2) P = 28 MPa, T = 200oC. Find v 3) P = 1 MPa, T = 405 oC. Find v 4) T = 100oC, x = 60%. Find v
Answer:
1) LIQUID, 2) LIQUID, 3) GASEOU, 4) GASEOUS, v = 0.76 m³ / kg
Explanation:
In this exercise it is asked to determine the state of the water, that is, if it is solid, liquid or gas. For this we must use a phase diagram of water which is a graph of Pressure versus Temperature. Let's describe the water diagram
* Below 0ºC and normal pressure the water is in a solid state
* Below this temperature and at low pressure it becomes a gaseous state
* Above 0.01ºC and normal pressure is in liquid state
* Above 0.01ºC and low pressure is in a gaseous state
there are two important points
* The triple point at t + 0.01ºC and P = 0.006 atm where the three states coexist
* The critical point T = 374ºC and P = 218 atm where water decomposes into hydrogen and oxygen
Specific volume is related to density
v = 1 / ρ
ρ = 1 / v
the density of water is approximately 1000 kg / m³ in the liquid state at t = 4ºC and decreases with increasing pressure 960 kg / m³ at T = 100ºC (but without changing to the gaseous state
With the above considerations we can answer the questions
1) P = 300 kPa = 3 105 Pa
v = 0.5 m³ / kg
atmospheric pressure is Patm = 1.01 105 Pa
P = 3 105 Pa (1 atm / 1.01 105 Pa) = 3 atm
From the phase diagram described, the water can be in two liquid or gaseous states, depending on the temperature, as indicated by the speed of the same state, the water must be LIQUID
the parameter v that you indicate is the
with the other parameter we can calculate the density
rho = 1 / 0.5
rho = 2.0 kg / m³
in a thermodynamic system the three basic properties are: pressure, volume and temperature,
we can calculate the body temperature
The body must be at a temperature between 0 <T <100ºC
2) P = 28 Mpa = 28 106 Pa
P = 28 106 Mpa (1 atm / 1.01 105 Pa) = 280 atm
T = 200ºC
When examining the diagram it can be seen that the water is in the range of the LIQUID state
3) P = 1MPa
P = 1 106 Pa (1atm / 1.01 105Pa) = 10 atm
T = 405ºC
In this case, the only accessible state is the GASEOUS
4) T = 100ºC
x = 60%
Examining the phase diagram at this temperature depending on the pressure the possible states are Vapor and liquid, for pressures below 1 atm the state is GASEOUS
for the gaseous state we can use the ideal gas equation
PV = nR T
let us perform the calculation for a mole of gas n = 1, the ideal gas constant is R = 8,206 10-2 atm / mol K
V = nRT / p
V = 1 8,206 10-2 100/1
V = 8,206 m³
v = V / m
The mass of the water is 18 Kg / mol, which indicates that 60% is in the gaseous state, so the mass in this state is
m = 0.60 18
m = 10.8 kg
v = 8.206 / 10.8
v = 0.76 m³ / kg
How far can you get away from your little
brother with the squirt gun filled with
paint if you can travel at 3 m/s and you
have 15s before he sees you?
Know
Find
Equation
Solve
spray him in the eyes and you have until he washes it put
State the four factors the magnetic force moving in magnetic field depends on
Explanation:
It depends upon the magnitude of the charge, the velocity of the particle and its strength .
If the force of static friction on a crate is 67 N and the weight of the create is 289 N, what is the coefficient of static friction?
a
19363
b
0.23
c
4.31
d
222
Answer:
B) μ = 0.23
Explanation:
The coefficient of friction is equal to μ = F / N where μ (mu) is the coefficient of friction, F is the friction force, and N is the normal force (the force of an object being applied onto the earth by gravity).
F, the friction force, is given as 67 N
Since the weight of the crate is 289 N, that means the normal force is 289 N.
This means that the coefficient of friction is μ = F / N = 67 N / 289 N = 0.23183391 = 0.23
Therefore, B is the correct answer
If 32g of kerosene of densities of 0.80gcm-3 are mixed with 8g of water, what is the densities of the resulting
mixtures? Take the densities of water to be 1.0g/cm3.
(3mks)
Answer:
volume = 8 cm^3 + 32 g / 0.8 g/ cm^3 = 48 cm^3
mass = 32 + 8 = 40 g
40 g / 48 cm^3
Explanation:
The density of a mixture can be calculated from the individual densities. The average density of the mixture of 32 g kerosene and 8 g of water is 0.9 g/cm³.
What is density?Density of a substance is the measure of its mass per unit volume. Thus, mathematically it is the ratio of mass to the volume of the substance. Density depends on the mass, volume, temperature,bond type and pressure.
The ratio of density of a substance to the density of water is called specific gravity. If the specific gravity is greater than 1 then the object will sink on water and if it is less than 1 then it will float on water.
If two liquids of densities ρ₁ and ρ₂ then after mixing their densities will be:
[tex]\rho =\frac{ \rho_{1} + \rho_{2} }{2}[/tex]
Hence the density of mixture of water (1 g/cm³) and kerosene (0.8 g/cm³) is :
= (0.8 + 1) /2
=0.9 g/cm³.
Thus the density of the resulting mixture be 0.9 g/cm³
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How long must a 0.54-mm-diameter aluminum wire be to have a 0.42 A current when connected to the terminals of a 1.5 V flashlight battery
Answer:
L = 30.85 m
Explanation:
First, we calculate the resistance of the wire by using Ohm's Law:
V = IR
where,
V = Potential Difference = 1.5 V
I = Current = 0.42 A
R = Resistance of Wire = ?
Therefore,
[tex]R = \frac{1.5\ V}{0.42\ A}\\\\R = 3.57\ Ohms[/tex]
Now, the cross-sectional area of wire will be:
[tex]Area = A = \frac{\pi d^{2}}{4}\\\\A = \frac{\pi (0.00054\ m)^{2}}{4}\\\\A = 2.29\ x\ 10^{-7}\ m^{2}[/tex]
Now, the resistance of the wire is given as:
[tex]R = \frac{\rho L}{A}\\\\L = \frac{RA}{\rho}[/tex]
where,
L = Length of Wire = ?
ρ = resistivity of aluminum = 2.65×10⁻⁸ Ohm.m
Therefore,
[tex]L = \frac{(3.57\ Ohms)(2.29\ x\ 10^{-7}\ m^{2})}{2.65\ x\ 10^{-8}\ Ohm.m}[/tex]
L = 30.85 m
A 1.3-kg book rests on a table. A downward force of 15 N is exerted on the top of the book by a hand pushing down on the
book
What is the net force on the book?
Answer:
The net force will be:
[tex] F_{net} = 142.53\: N[/tex]
Explanation:
The net force is given by:
[tex] F_{net} = W_{b}+F[/tex]
[tex] F_{net} = m_{b}g+15[/tex]
[tex] F_{net} = 142.53\: N[/tex]
I hope it helps you!
can a body have an east velocity while expressing westward acceleration
Yes it can, an object can be moving a certain direction while the ACCELERATION is in the opposite direction.
Lets say your riding a bike... if your squeezing your handle bar breaks, the acceleration of the bike would be pushing in the opposite direction of the direction the bike is moving.
Hope this helped!
5. Atoms may emit light energy when
O A electrons move to a lower energy level
OB protons move to a lower energy level
C electrons move to a higher energy level
OD protons move to a higher energy level
Hi
How are you all
Please I need help from anyone who know physics.
In these lessons the questions will be friction,vectors, forces in two dimensions.
So, Please to anyone who can help Write a comment,And I'll be in touch with him, thanks everyone
Answer:
umm it says he but I'm a she :p
I need help on how to start my essay on the 3 laws of motion
Answer:
for the first paragraph introduce the definition. for the second paragraph write about the first law, for the third paragraph write about the 2nd for the fourth paragraph write about the 3rd law. for the last paragraph do a brief summary of what you wrote and a conclusion about the laws.
i hope this helps a bit
Answer:
You can start off with the first law of motion (newtons first law), talk about what it is or what it does, give examples.
A 0.15kg baseball is pushed with a 100N force. What will its acceleration be?
What is the difference between inertia and momentum?
Is inertia a force (will give brainleist for first answer)
Answer:
Yes.
Explanation:
Answer:
I do believe it is. (more characters for character limit)
Find the moment of 300N force about B
Answer:
300
Explanation Hope I'm not wrong.
g Consider a (12.5 A) cm long metal bar moving horizontally across a vertical magnetic field at a speed of (2.40 B) m/s. The magnetic field strength is 2.45 T. If the ends of the bar are connected to a (1.20 C) ohm resistor, find the power dissipated in the resistor while the bar is moving. Calculate the answer in watts (W) and rounded to three significant figures.
Answer:
Explanation:
emf due to movement of a rod of length L in a perpendicular to magnetic field B with velocity v is given as
emf = BLv
Putting in the given values ,
E = 2.45 x 12.5 x 10⁻² x 2.4
= .735 V
This emf produces current in resistance . Power consumed by resistance
V² / R where R is resistance , V is emf induced .
Power = .735² / 1.2
= .45 W .