what is fire proof plastic
Explanation:
.Flame retar,dant plastic additi,ves are compounds added to plastics and other materials to inhibit, suppress or de,lay combustion. These compounds are useful in impending burning in the ign,ition phase of fire. They do not prevent char,ring or melting nor do they increase the heat resistance of a material
A truck hits a brick wall with a force of 100000 kg ms² the truck comes to a stop in 0.5 s what is the impulse
Answer:
Impulse = 50,000 J
Explanation:
Given: force = 100,000 kg ms^2 or N; time = 0.5 s
Unknown: impulse = ?
Equation: impulse = (force)(time)
Substitute: impulse = (100,000 kg ms^2 or N)(0.5 s)
Solution: impusle = 50,000 kg ms^2 or J
A beam strikes an irregular mirror and is reflected as shown below. The angle formed by the incoming beam and the reflected beam is 10 degrees. What is the angle of the mirror's surface (measured against vertical) at the location where the beam strikes the mirror?
A. The mirror's surface is angled 5 degrees upward where the laser strikes it.
B. The mirror's surface is angled 5 degrees downward where the laser strikes it.
C. The mirror's surface is angled 10 degrees upward where the laser strikes it.
D. The mirror's surface is angled 10 degrees downward where the laser strikes it.
Answer:
the correct answer is B
Explanation:
The law of reflection states that the angles of incidence and reflection on a surface are the same, the two rays and the normal are in the same part of the surface.
In this exercise indicate that the angle between the incident and reflected ray is 10, therefore the angle with respect to the normal that is a vertical line at the point of contact of the ray must be 5
Consequently the correct answer is B
Quanto tempo deve ficar ligado um ferro eletrico de 1000 w para que tenha o mesmo consumo de energia que um chuveiro de 4400 w que fica ligado 10 minutos
Answer:
Thus, the time for the first lamp is 44 minutes.
Explanation:
Power of first lamp, P' = 1000 W
Power of second lamp, P'' = 4400 W
time for second lamp, t'' = 10 minutes
Let the time for first lamp is t'.
As the energy is same, so,
P' x t' = P'' x t''
1000 x t' = 4400 x 10
t' = 44 minutes
Determine the amount of work done by the applied force when a 87 N force is applied to move a 15 kg object a horizontal
distance of 4.5 meters at a constant speed.
Answer:
391.5 J
Explanation:
The amount of work done can be calculated using the formula:
W = F║d where the force is parallel to the displacementLooking at the formula, we can see that the mass of the object does not affect the work done on it.
Substitute the force applied and the displacement of the object into the equation.
W = (87 N)(4.5 m) W = 391.5 JThe amount of work done on the object is 391.5 J in order to move it 4.5 meters with an applied force of 87 Newtons.
Mass doesn't matter on amount of work done .We can calculate amount of work done through Force and Displacement
Force=87NDisplacement=4.5m[tex]\boxed{\sf W=Fd}[/tex]
W denotes to work doneF denotes to forced denotes to displacement[tex]\\ \sf\longmapsto Work\:Done=87(4.5)[/tex]
[tex]\\ \sf\longmapsto Work\:done=391.5J[/tex]
Q:
The kinetic energy of an object can be determined
using the equation KE = 1/2mv^2, where ke is the
kinetic energy, m is the mass, and v is the speed of
the object. What is the mass of an object moving at
a speed of 12m/s that has 36 J of kinetic energy?
A:
0
2.0 kg
8.0 kg
0.13 kg
0.50 kg
Answer:
Explanation:
We will use the KE equation you wrote here and fill in what we are given:
[tex]36=\frac{1}{2}m(12)^2[/tex] and isolating the m:
[tex]m=\frac{2(36)}{12^2}[/tex] which gives us
m = .50 kg
What kind of waves are present during an earthquake?
a. Transverse
b. Longitudinal
c. Both transverse and longitudinal
Answer:
Explanation:
During an earthquake, there are both transverse and longitudinal wave. The answer is c.
Answer:
Explanation:
ans is c. Both transverse and longitudinal
The equation provided (from the textbook) first defines the elastic potential energy of a spring as ΔUsp = −(WB + WW), where WB is work the spring does on an attached block and WW is work the spring does on the wall to which it is attached. But WW is ignored in the next step. Why?
Answer:
The given potential energy of the spring is expressed as follows;
ΔUsp = -(WB + WW)
Where;
WB = Th work done by the spring on the block to which it is attached
WW = The work done by the spring on the wall
We recall that work done, W = Force applied × Distance moved in the direction of the force
The work done by the spring on the block, WB = The spring force × The distance the block moves
The work done by the spring on the wall, WW = The spring force × The distance the wall moves
However, given that the wall does not move, we have;
The distance the wall moves = 0
∴ The work done by the spring on the wall, WW = The spring force × 0 = 0 J
Therefore, WW = 0 J, and the spring does not do work on the wall, and WW can be ignored in the next subsequent) steps
Explanation:
PLS SHOW WORK ON HOW YOU GOT IT PLS I BEG.
top ( 25 m ) | 75000 j | 0 j | 75000 j
-------------------|---------------|------------|-------------
bottom ( 0 ) | 0 j | 75000j | 75000 j
-------------------|---------------|------------|-------------
h d ( 12.5 m ) | 37500 j | 37500j | 75000 j
-------------------|---------------|------------|-------------
f d ( 5 m ) | 15000 j | 60000j | 75000 j
-------------------|---------------|------------|-------------
Does a feather fall as fast as a rock in a vacuum? If so why?
Answer:
No.
Explanation:
A feather is less dense and thus less force exerted while a rock is very dense thus exerting more force .
How do you find a wavelength?
We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of the spring, at what speed will a 2 g projectile be ejected from the gun
Answer:
[tex]31.6\:\mathrm{m/s}[/tex]
Explanation:
The elastic potential energy of a spring is given by [tex]Us=\frac{1}{2}kx^2[/tex], where [tex]k[/tex] is the spring constant of the spring and [tex]x[/tex] is displacement from point of equilibrium.
When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the object's velocity.
Thus, we have:
[tex]Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2[/tex]
Substituting given values, we get:
[tex]\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}[/tex]
The relationship between the decay constant (λ) and the half-life (t1⁄2) is given by the equation t(1/2) = 0.693/λ. Use the equation for N to derive this relationship.
Hint: when t = t(1/2), N/No = 0.5 (1 point)
A sample of carbon-14 initially consists of 5 × 1024 particles. Carbon-14 has a half-life of 5730 years.
a. What is the decay constant for carbon-14? (Answer in units of yr–1.) (1 point)
b. How many radioactive particles of the sample remain after 100 years? (1 point)
c. What percentage of the radioactive particles remains after 500 years? (1 point)
d. How many radioactive particles of the sample remain after 1000 years? (1 point)
e. How much time will it take for 50% of the particles to decay? (1 point)
f. How much time will it take for 99% of the particles to decay? (1 point)
g. How many half-lives will it take for 99% of the sample to decay? (1 point)
h. What is the initial decay rate of the sample? (Answer in decays/yr.) (1 point)
i. After 200 years, what is the decay rate of the sample? (1 point)
j. How long will it take for the decay rate to decrease to 1015 decays/year? (1 point)
k. How many half-lives have passed after the time you found in part (j)? (1 point)
Answer: A: 70/2=35
B: 35/2=17
C: 17.5/2=8.75
D: 8.75 of C-14 will be left
E: 5,730 years
F: 5,631
Explanation:
that’s all I got, hope I helped kinda
Nitrogen gas at standard atmospheric pressure 101.3 kPa, has a volume of 0.080 m3. If there are 3.6 mol of the gas, what is the temperature?
Answer:
271 K
Explanation:
Answer:
The temperature of Nitrogen is 270.85 K.
Explanation:
What is perfect gas ?A gas that follows Boyle's law , Charle's law and law of pressure , is called perfect gas.
Given data;
[tex]P =101.3 KPa , V = 0.08 m^{3} , \\ n = 3.6 mol , R = 8.31\\T = ?[/tex]
By the formula,
P.V = n .R.T
On putting values
[tex]1.013[/tex] x [tex]10^{5}[/tex] x [tex]0.08[/tex] = [tex]3.6[/tex] x [tex]8.31[/tex] x [tex]T[/tex]
So [tex]T =\frac{1.013*10^{5} * 0.08}{3.6 * 8.31}[/tex]
[tex]T =\frac{8104}{29.92}[/tex]
[tex]T = 270.85 K[/tex]
Hence the temperature of Nitrogen is 270.85 K
For more details also read -
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_____ Rocks are porous.
A) Sedimentary
B) Metamorphic
C) igneous
Answer:
Sedimentary rocks are porous.
Explanation:
So option A) Sedimentary is your answer. Have a great summer
An aluminum-alloy rod has a length of 10.0 cm at 20°C and a length of 10.015 cm at the boiling point of water (1000C). (a) What is the length of the rod at the freezing point of water (0 0C)? (b) What is the temperature if the length of the rod is 10.009 cm? /4mks
Answer:
a. 9.99625 cm b. 68 °C
Explanation:
(a) What is the length of the rod at the freezing point of water (0 0C)?
Before we find the length of the rod, we need to find the coefficient of linear expansion, α = (L - L₀)/[L₀(T - T₀)] where L₀ = length of rod at temperature T₀ = 10.0 cm, T₀ = 20 °C, L = length of rod at temperature T = 10.015 cm and T = 100 °C
Substituting the values of the variables into the equation, we have
α = (L - L₀)/[L₀(T - T₀)]
α = (10.015 cm - 10.0 cm)/[10.0 cm(100 °C - 20 °C)]
α = 0.015 cm/[10.0 cm × 80 °C]
α = 0.015 cm/[800.0 cm °C]
α = 0.00001875 /°C
We now find the length L₁ at T₁ = 0 °C from
L₁ = L₀(1 + α(T₁ - T₀))
So, substituting the values of the variables into the equation, we have
L₁ = L₀(1 + α(T₁ - T₀))
L₁ = 10.0 cm[1 + 0.00001875 /°C(0° C - 20 °C)]
L₁ = 10.0 cm[1 + 0.00001875 /°C × -20° C]
L₁ = 10.0 cm[1 - 0.000375]
L₁ = 10.0 cm[0.999625]
L₁ = 9.99625 cm
(b) What is the temperature if the length of the rod is 10.009 cm?
With length L₃ = 10.009 cm at temperature T₃, using
L₃ = L₀(1 + α(T₃ - T₀))
making T₃ subject of the formula, we have
L₃/L₀ = 1 + α(T₃ - T₀)
L₃/L₀ - 1 = α(T₃ - T₀)
T₃ - T₀ = (L₃/L₀ - 1)/α
T₃ = T₀ + (L₃/L₀ - 1)/α
substituting the values of the variables into the equation, we have
T₃ = 20 °C + (10.009 cm/10.0 cm - 1)/0.00001875 /°C
T₃ = 20 °C + (1.0009 - 1)/0.00001875 /°C
T₃ = 20 °C + 0.0009/0.00001875 /°C
T₃ = 20 °C + 48 °C
T₃ = 68 °C
A tennis ball hits with an initial velocity of 25 m/s [10 o above the
horizontal] starting from a height of 2.1 m, lands on the ground. There
is a net 15 m away and 0.9 m high. Will the ball clear the net at its
peak, and by how much?
First, calculate the components of its initial velocity v0:
v0x = v0cos10 = 24.6 m/s
v0y = v0sin10 = 4.34 m/s
The ball reaches its peak when vy = 0. Let's calculate the time it takes for vy to become zero:
vy = v0y - gt ---> t = v0y/g = 0.44 s
The horizontal distance it travels in this time is
x = v0xt = (24.6 m/s)(0.44 s)
= 10.8 m
Note that the net is 15 m away. After traveling a horizontal distance of 10.8 m, the height of the ball is
y = -(1/2)gt^2 + v0yt + 2.1
= -(4.9 m/s^2)(0.44 s)^2 + (4.34 m/s)(0.44 s) + 2.1 m
= -0.95 m + 1.9 m + 2.1 m
= 3.05 m
Note that this is the height of the ball at its peak. While the ball is well above the net at its peak, it is well short of its required horizontal distance to clear it. Instead, let's find the time it takes for the tennis ball to travel a horizontal distance of 15 m first:
x = v0xt ----> t = x/v0x = (15 m)/(24.6 m/s) = 0.61 s
Then calculate the height y when t = 0.61 s. If y > 0.9 m (height of the net), then the ball will clear the net.
y = -4.9t^2 + v0yt + 2.1
= -4.9(0.61 s)^2 + (4.34 m/s)(0.61 s) + 2.1 m
= -1.82 m + 2.65 m + 2.1 m
= 2.93 m
Yes, the ball will clear the net by 2.03 m.
Which two changes would decrease the gravitational force between two
objects?
A. Increase the distance between the objects.
B. Increase the mass of one of the objects.
C. Decrease the distance between the objects.
D. Decrease the mass of one of the objects.
E. Increase the mass of both objects.
Answer:
A. Increase the distance between the objects.
D. Decrease the mass of one of the objects.
Explanation:
The smaller the mass of an object the less gravity it has and the farther away two objects are the less gravitational force
How to keep my family safe during a hurricane?
If you are not advised to evacuate or are unable to do so safely, stay indoors, away from windows, skylights and doors. Continue to monitor weather reports and do not go outside until the storm has passed. And, to protect your children from seeing too many sights and images of the hurricane, limit media exposure.
please help, thank you
4 N to right
( the third option )
...................................
Look at the graph below.
A graph is shown with Absolute magnitude shown on y axis and Surface temperature in degree Celsius shown on x axis. The Dwarf stars are shown along a slanting line from coordinates 30,000 and minus 3 to 10,000 and minus 4. The Main Sequence stars are shown along a slanting line from coordinates 20,000 and minus 2 to 2,000 and minus 6. The giants are shown along a line parallel to the x axis from coordinates 5,000 and 2 to 2,000 and 3. The supergiants are shown along a line parallel to the x axis from coordinates 7,500 and 4 to 2,500 and 4. Point A has coordinates 20,000 and minus 4. Point B has coordinates 2,500 and minus 4. Point C has coordinates 5,000 and 2. Point D has coordinates 7,000 and 4.
Which statement is true about Star A?
a
It is red in color.
b
It is yellow in color.
c
It is four times brighter than the sun.
d
It is four times dimmer than the sun.
Answer:
I'm pretty sure its C
Explanation:
The Keck Observatory is home to the largest Earth-based twin telescopes. The telescopes are located away from city lights, above the clouds where the atmosphere is clear, calm, and dry. In 1999, additional steps were taken to help the telescopes overcome any distortions caused by changes in the atmosphere by adding a laser sighting system. Now, the Keck telescopes can probe distant galaxies and capture images with more detail than even the Hubble Space Telescope. Scientists can now use the Keck telescopes to investigate many questions. What is one question the Keck telescopes would NOT help scientists answer
Answer:
Option D
Explanation:
The options for the question are
a) How do galaxies rotate?
b) What is the weather on Neptune?
c) What is the core of Saturn made of?
d) What other solar systems have planets?
Solution
The Hubble space telescope was designed and integrated into the extraterrestrial system in order to capture information about the surrounding universe. If the Keck Observatory has a better observation capacity than the Hubble space telescope then the scientist would be interested to know the surrounding planets in the solar system.
Hence, option D is correct
What if We Nuke the Moon?
Answer:
chonks of rock will hit the earth and if one astoroid is big enough it can distroy all liofe on earth
Explanation:
Answer:
If we nuke the moon, it won't have any sound and will quickly be stopped from travelling any further than the place we nuked it. If some parts of the moon crack and break, it will be forced down on us (like an asteroid).
Many possibilities can happen but I'm just listing one of them.
An extremely fast-moving charged particle traveling in a magnetic field can radiate X-rays, a phenomenon known as synchrotron radiation. Why is the magnetic field essential to this emission
Answer:
The magnetic field will help to create an electromagnetic radiation which will prevent the charged particle from moving in a constant direction in a straight path.
Explanation: A magnetic field is an area around a particle where magnetic energy is felt or experienced, this ensures that all the charged particle within the magnetic field will be influenced by magnet.
In the highlighted situation, the magnetic field will help to ensure that an electromagnetic radiation is created to synchrotron radiation occurs effectively.
what is the size of the electron charge?
Answer:
It’s like medium uk
Explanation:
The size of electron charge is estimated to be 1.6 × 10-¹⁹
A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed
Answer:
Explanation:
Magnitude of frictional force = μ mg
μ is either static or kinetic friction.
To start the crate moving , static friction is calculated .
a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .
force required =.517 x 56.6 x 9.8 = 286.76 N .
b ) to slide the crate across the dock at a constant speed , force required
= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction
= .26 x 56.6 x 9.8 = 144.21 N .
name few biodegradable materials that get decomposed in a week
Answer:
Cardboard, Paper towels, Food waste, and wooden based items
Answer:
How fast do things biodegrade?
Vegetables 5 days –1 month
Aluminium cans 80–100 years
Glass bottles 1 million years
Styrofoam cup 500 years to forever
Plastic bags 500 years to forever
which of the two bodies a and b in the following graph is moving with higher speed and why
.........................................
pls say it fst its very urgent
In which stage of a star's life cycle, does gravity and fusion become balanced, and an adult star forms?
A. Main-sequence
B. Red supergiant
C. Protostar
D. White dwarf
Answer:
im guessing red giant-
Suppose that scientists observe violent gas eruptions on a planet with an acceleration due to gravity of 3.9 m/s2. The jets throw sand and dust about 75 m above the surface. What is the speed i of the material just as it leaves the surface?
Answer:
The velocity with which the sand throw is 24.2 m/s.
Explanation:
acceleration due to gravity, a = 3.9 m/s2
height, h = 75 m
final velocity, v = 0
Let the initial velocity at the time of throw is u.
Use third equation of motion
[tex]v^{2} = u^{2} - 2 g h \\\\0 = u^{2} -2\times 3.9\times 75\\\\u=24.2m/s[/tex]
The velocity with which the sand throw is 24.2 m/s.