Answer:
1 +2=3 so thats it
Explanation:
Because i need points
Yellow light shines on a sheet of paper containing a blue pigment. Determine the appearance of the paper.
I dont need google answers if i get google answers i will delete it
Black I think
this is due to the fact that blue lights only reflect blue things so if yellow is shone on it it will reflect black appearance.
how it useful
How long does it take a plane, traveling at a constant speed of 123 m/s, to fly once around a circle whose radius is 4330 m?
Answer:
3.7 minExplanation:
Step one:
given data
speed = 123m/s
radius of circle= 4330m
Step two:
We need to find the circumference of the circle, it represents the distance traveled
C=2πr
C= 2*3.142*4330
C= 27209.72m
Step three:
We know that velocity= distance/time
time= distance/velocity
time= 27209.72/123
time=221.2 seconds
in minute = 221.2/60
time= 3.7 min
what is the potential energy of a 30kg rock that falls 15 meters
Answer:
4500 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 10 m/s²
From the question we have
PE = 30 × 10 × 15
We have the final answer as
4500 JHope this helps you
Help ASAP plz and thx u
Answer:
a). a = F/m
Explanation:
Formula is F=ma
What is the Basic SI unit for distance/length
A. Meters
B. Liters
C. Grams
D. Millimeters
Can someone help with my physics homework? please
A go cart engine applies a force of 888N and moves the cart forward 22m.
a) How much work is done?
b) What is doing the work?
c) If the driver wants to go further will the amount of work increase or decrease? Do you need a bigger engine to go
further?
d) We put on a bigger engine (1111N) but the cart still moves forward 22m. How much work is done now?
e) Why would you put on a bigger engine if you are still moving 22m?
f) Work requires a change in energy, which engine uses more gas to go 22m?
g) Even an empty semi truck uses much more gas than a car. Why?Find the soultions
Answer:
a) 19536 joules of work are done.
b) The work is done by the engine on the structure of the cart.
c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.
d) 24442 joules of work are done.
e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.
f) The bigger engine uses more gas to go 22 meters.
g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.
Explanation:
a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:
[tex]W = F\cdot \Delta s[/tex] (1)
Where:
[tex]F[/tex] - Force applied by the cart, measured in newtons.
[tex]\Delta s[/tex] - Distance travelled by the car, measured in meters.
[tex]W[/tex] - Work done on the cart, measured in joules.
If we know that [tex]F = 888\,N[/tex] and [tex]\Delta s = 22\,m[/tex], then the work done on the cart is:
[tex]W =(888\,N)\cdot (22\,m)[/tex]
[tex]W = 19536\,J[/tex]
19536 joules of work are done.
b) The work is done by the engine on the structure of the cart.
c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.
d) If we know that [tex]F = 1111\,N[/tex] and [tex]\Delta s = 22\,m[/tex] , then the work on the cart is:
[tex]W = (1111\,N)\cdot (22\,m)[/tex]
[tex]W = 24442\,N[/tex]
24442 joules of work are done.
e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.
f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.
g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.
a. The amount of work done by the go cart engine is 19,536 Nm.
b. Work is being done by the go cart engine i.e the engine installed on the go cart.
c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.
d. The amount of work done when a bigger engine is used is 24,442 Nm.
e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.
f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.
g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.
Given the following data:
Force A = 888 NewtonDistance = 22 meterForce B = 1111 Newtona. To determine the amount of work done by the go cart engine:
Mathematically, the work done by an object is given by the formula;
[tex]Work\;done = Force \times distance[/tex]
Substituting the given parameters into the formula, we have;
[tex]Work\;done = 888 \times 22[/tex]
Work done = 19,536 Nm.
b. Work is being done by the go cart engine i.e the engine installed on the go cart.
c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.
d. To determine the amount of work done when a bigger engine is used:
[tex]Work\;done = Force \times distance[/tex]
[tex]Work\;done = 1111 \times 22[/tex]
Work done = 24,442 Nm.
e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.
f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.
g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.
Read more: https://brainly.com/question/22599382
. What is the barycenter of the Moon and Earth?
Answer:
About 1000 miles.
Explanation:
A statement of the second law of thermodynamics is that:__________.
a) spontaneous reactions are always exothermic.
b) energy is conserved in a chemical reaction that has a decrease in entropy.
c) spontaneous reactions are always endothermic.
d) in a spontaneous process, the entropy of the universe increases.
Answer:
in a spontaneous process, the entropy of the universe increases.
Explanation:
Entropy is a measure of of the degree of randomness or disorderliness in a system.
The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."
The universe here refers to the system's disorder and the disorder of the surroundings. Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.
For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the edge of the disc is ?m/s2.
Answer:
Centripetal acceleration = 83.77m/s²
Explanation:
Given the following data;
Radius, r = 0.13m
Velocity, v = 3.3m/s
To find centripetal acceleration;
Centripetal acceleration is given by the formula;
[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]
Substituting into the equation, we have;
[tex] Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}[/tex]
[tex] Centripetal \; acceleration, a = \frac {10.89}{0.13}[/tex]
Centripetal acceleration = 83.77m/s²
Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s².
Answer:
the answer is 84
Explanation:
In the past, Africa used to be further away from Europe than it is now
(shown below). What could explain why Africa is closer to Europe now than
it was before? *
Answer: Plates shifting
Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).
Answer:
nvudbwasivnjlscv bwbfvsz
Explanation:
A monatomic ideal gas with an initial pressure of 500 kPa and an initial volume of 1.80 L expands isothermally to a final volume of 5.20 L. How much work is done on the gas in this process?
A) 1700J
B) 875J
C) 1570J
D) 900J
E) 955J
Answer:
955 J
Explanation:
PV = nRT
500 x 10³ x 1.8 x 10⁻³ = nRT
= 900 J
work done by gas in isothermal expansion
= nRT lnV₂ / V₁
= 900 ln 5.2 / 1.8
= 900 x ln 2.89
= 900 x 1.06
= 955 J
pplzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helppppp
Answer:
D. The airplane will turn towards the east
Explanation:
If the airplane is thrown straight towards the north, the window which is moving from left (west) to right (east) the wind will knock the plane towards the right (east) since thats the way it is blowing.
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm.
Complete Question
What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm. The student has 70-cm-long arms
Answer:
The value is [tex]w__{rpm} } = 29.17 \ rpm[/tex]
Explanation:
From the question we are told
The distance from the handle to the bottom of the bucket is [tex]d = 35 \ cm = 0.35 \ m[/tex]
The length of the students arm is L = 70 cm = 0.70 m
Generally the acceleration due to gravity experienced by the bucket of water is mathematically represented as
[tex]g = w^2 * r[/tex]
Here is is the radius of the circle which swinging of the bucket makes and this is mathematically represented as
[tex]r = L + d[/tex]
So
[tex]g = w^2 * ( L + d )[/tex]
= > [tex]w = \sqrt{\frac{g }{ L + d } }[/tex]
= > [tex]w = \sqrt{\frac{ 9.8}{ 0.7 + 0.35} }[/tex]
= > [tex]w = 3.055 \ rad/s[/tex]
Generally the angular speed in revolution per minute is mathematically represented as
[tex]w__{rpm} } = \frac{w * 60 }{2 \pi }[/tex]
=> [tex]w__{rpm} } = \frac{3.055 * 60 }{2 * 3.142 }[/tex]
=> [tex]w__{rpm} } = 29.17 \ rpm[/tex]
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.80 V/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Answer:
The value is [tex]|\vec B| = 1.267 *0^{-8} \ T[/tex]
Explanation:
From the question we are told that
The magnitude of the electric fields is [tex]E = 3.80 V/m[/tex]
Generally speed of light is mathematically represented as
[tex]c = \frac{|\vec E|}{ |\vec B|}[/tex]
Here c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
[tex]|\vec B |[/tex] is the magnitude of the magnetic field so
=> [tex]|\vec B| = \frac{|\vec E|}{c}[/tex]
=> [tex]|\vec B| = \frac{ 3.80 }{3.0*10^{8}}[/tex]
=> [tex]|\vec B| = 1.267 *0^{-8} \ T[/tex]
A bullet with an initial kinetic energy of 400 J strikes a wooden block where a 8000 N resistive force stops the bullet. What is the distance the bullet travels into the block?
How do you answer this question?
Answer:
d = 0.05 [m] = 50 [mm]
Explanation:
We must remember the principle of conservation of energy which tells us that energy is transformed from one way to another. For this case, the initial kinetic energy is transformed into useful work that is equal to the product of force by distance.
[tex]E_{k}=F*d\\400 = 8000*d\\d = 0.05 [m] = 50 [mm][/tex]
What is the error in this representation of the steps involved in gene therapy?
Answer:
a
Explanation:
If an ocean wave passes a stationary pointevery 4 s and has a velocity of 7 m/s, what isthe wavelength of the wave?Answer in units of m.
Answer:
28mExplanation:
Step one:
given data
period T= 4seconds
velocity v= 7m/s
wave lenght λ=?
Step two:
we know that f=1/T
the expression relating period and wave lenght is
v=λ/T
λ=v*T
λ=7*4
λ=28m
The wavelength of the wave is 28m
A diffraction grating with 68 slits per cm is used to measure the wavelengths emitted by hydrogen gas.
A. At what angles in the fourth-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm?
B. What are the angles if the grating has 12,800 slits per cm?
Answer:
a
[tex]\theta _1 =0.687 ^o[/tex]
[tex]\theta _2 =0.630 ^o[/tex]
b
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.1 will not be valid
Explanation:
From the question we are told that
The slit grating is [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]
The order of spectrum is [tex]n = 4[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 6800}[/tex]
=> [tex]a = 0.000147 \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _2 =0.630 ^o[/tex]
Gnerally if grating is [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 1280000}[/tex]
=> [tex]a = 7.813 *10^{-7} \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
[tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Objects accelerate because
PLEASE HELP!
I don't even know what Science is I'm so dumb lol XD
Answer:
C
Explanation:
Sled A has more potential energy because it's mass is 100 kg, and it is higher up than Sled B. The more high up the sled is and the lighter it is, the faster it gets, it creates more and more potential energy.
Hope this helps!
One of the harmonics of a column of air in a tube that is open at both ends has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column in this tube?
Answer:
The fundamental frequency is [tex]f_1 =128 \ Hz[/tex]
Explanation:
From the question we are told that
The frequency of one harmonics is [tex]f_x= 448 \ Hz[/tex]
The next higher harmonic is [tex]f_z = 576 \ Hz[/tex]
Generally the frequency of an air column open at both ends is mathematically represented as
[tex]f_n = \frac{nv }{ 2 L }[/tex]
Here n is the order of the harmonics (frequency)
v is the velocity of the sound
L is the length of the column
So for one harmonics we have that
[tex]f_k = \frac{n v }{2L}[/tex]
Then for the next higher harmonics
[tex]f_x = \frac{n+1 ) v}{2 L }[/tex]
Generally the difference between these frequencies is mathematically represented as
[tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]
=> [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]
=> [tex]\frac{ v }{2L} = 128[/tex]
Generally for fundamental frequency n = 1
So
[tex]f_1 = n * \frac{v}{2L}[/tex]
So
[tex]f_1 =1 * 128[/tex]
=> [tex]f_1 =128 \ Hz[/tex]
Forces of 70 N at 130 degrees, and 20 N at an angle of 280 degrees, measured counter-clockwise from the positive x-axis, act on an object.
A. What are the components (F1x, F1y) of the first force force (in Newtons)?
B. What are the components (F2x, F2y) of the second force force (in Newtons)?
C. What are the components (Fx, Fy) of the resultant force (in Newtons)?
D. What is the magnitude of the resultant force (in Newtons)?
E. What is the angle of the resultant force with respect to x-axis?
Answer:
A. ) F₁ₓ = -45.0 N F₁y = 53.6 N
B.) F₂ₓ = 3.48 N F₂y = -19.7 N
C.) Fₓ = -41.5 N Fy = 33.9 N
D) F = 53.6 N
E) θ = -39. 2º (320.8º)
Explanation:
A)
Applying simple trig, like definitions of cos and sin of an angle, we can get the x- and y- components of F₁, as follows:[tex]F_{x1} = F_{1} * cos (130) = 70 N * cos (130) = -45 N (1)\\F_{y1} = F_{1} * sin (130) = 70 N * sin (130) = 53.6 N (2)[/tex]
B)
Repeating for F₂:[tex]F_{x2} = F_{2} * cos (280) = 20 N * cos (280) = 3.48 N (3)\\F_{y2} = F_{2} * sin (280) = 20 N * sin (280) = -19.7 N (4)[/tex]
C)
The x- and y- components of the resultant force, are just the algebraicsum of the x- and - y components of F₁ and F₂:
Fₓ = Fₓ₁ + Fₓ₂ = -45 N + 3.48 N = -41.5 N (5)By the same token, Fy can be written as follows:Fy = Fy₁ + Fy₂ = 53.6 N + (-19.7 N) = 33.9 N (6)D)
The magnitude of the resultant force can be obtained applying the Pythagorean Theorem to Fx and Fy, as follows:[tex]F_{t} =\sqrt{F_{x} ^{2} + F_{y} ^{2} } = \sqrt{(-41.5N)^{2} +(33.9N)^{2}} = 53.6 N (7)[/tex]
E)
Finally the angle regarding the x- axis of the resultant force vector, can be obtained using the definition of the tangent of an angle, as follows:[tex]\theta = arc tg \frac{33.9N}{(-41.5N)} = arc tg (-0.817) = -39. 2 \deg[/tex]
Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?
Answer:
The value is [tex]p = 0.7556 c[/tex]
Explanation:
From the question we are told that
The speed at which galaxy B moves away from galaxy A is [tex]v = 0.577c[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
The speed at which galaxy C moves away from galaxy B is [tex]u = 0.731 c[/tex]
Generally from the equation of relative speed we have that
[tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]
Here p is the velocity at which galaxy C recede from galaxy A so
[tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]
=> [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]
=> [tex]0.731c - 0.4218 p = p - 0.577c[/tex]
=> [tex]0.731c + 0.577c = p + 0.4218 p[/tex]
=> [tex]1.308 c = 1.731 p[/tex]
=> [tex]p = 0.7556 c[/tex]
A 500 kg car is moving at 30 m/s. The driver sees a barrier ahead. If the car takes 100 m to come to rest, what is the magnitude of the force necessary to stop the car?
How do you solve this question?
Answer:
F = 2250 [N]
Explanation:
In order to solve this problem, we must first use the following equation of kinematics.
[tex]v_{f}^{2} =v_{o}^{2}-2*a*x[/tex]
where:
Vf = final velocity = 0 (come to rest)
Vo = initial velocity = 30 [m/s]
a = acceleration or desaceleration [m/s²]
x = distance = 100 [m]
[tex](0)=30^{2} -2*a*100\\900 = 200*a\\a = 4.5 [m/s^{2}][/tex]
Now we must use the following equation of kinetics, which is based on Newton's second law that explains that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
F = force [N]
m = mass = 500 [kg]
a = acceleration = 4.5 [m/s²]
[tex]F = 500*4.5\\F = 2250 [N][/tex]
if A=3i +2j+3k ,find the magnitude of A+B and A-B
Since vector B was not specified, I'll assume one at random. You can later answer your own question.
Answer:
[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}[/tex]
[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}[/tex]
Explanation:
Given:
[tex]\vec A=3\hat i +2\hat j+3\hat k[/tex]
And (assumed):
[tex]\vec B=-5\hat i +8\hat j-4\hat k[/tex]
Find the magnitude of
[tex]\vec A+\vec B[/tex]
[tex]\vec A-\vec B[/tex]
Given a vector
[tex]\vec P=x\hat i +y\hat j+z\hat k[/tex]
The magnitude of the vector is:
[tex]\mid\mid \vec P\mid \mid=\sqrt{x^2+y^2+z^2}[/tex]
First part:[tex]\vec A+\vec B =3\hat i +2\hat j+3\hat k-5\hat i +8\hat j-4\hat k[/tex]
[tex]\vec A+\vec B =-2\hat i +10\hat j-\hat k[/tex]
The magnitude of the sum is:
[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{(-2)^2+10^2+(-1)^2}=\sqrt{4+100+1}[/tex]
[tex]\mathbf{\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}}[/tex]
Second part:[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k-(-5\hat i +8\hat j-4\hat k)[/tex]
[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k+5\hat i -8\hat j+4\hat k[/tex]
[tex]\vec A-\vec B =8\hat i -6\hat j+7\hat k[/tex]
The magnitude of the difference is:
[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{8^2+(-6)^2+7^2}=\sqrt{64+36+49}[/tex]
[tex]\mathbf{\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}}[/tex]
friction reduces air resistance?
Answer:no
Explanation:
No.....................................
An object with a mass of 32 kg has an initial energy of 500). At the end of the experimentthe velocity of the object is recorded as 5.1 m/s . the object travelled 50 m to get to this point, what was the average force of friction on object during the tripAssume no potential energy Show all work
Answer:
F = 1.68 N
Explanation:
Let's solve this exercise in parts.
Let's use the concept of conservation of the mechanical nerve
initial
Em₀ = 500 J
The energy is totally kinetic
Em₀ = K = ½ m v₀²
v₀ = [tex]\sqrt{\frac{2 Em_{o} }{m} }[/tex]
v₀ = √ (2 500/32)
v₀ = 5.59 m / s
now with kinematics we can find a space
v² = v₀² - 2 a x
the negative sign is because the body is stopping
a =[tex]( \frac{v_{o}^{2} - v^{2} }{2x} )[/tex]
let's calculate
a = (5.59² - 5.1²) / 2 50
a = 0.0524 m / s²
Finally let's use Newton's second law
F = ma
F = 32 0.0524
F = 1.68 N
Pushes and pulls that result from objects that are physically touching
each other
Answer:
That is false. Take a look at this way. You can push a ball with your own breath, you just need to blow it. And you can pull something from afar with a magnet. It is possible to do both.
Explanation:
Not all physical things can be done only physically. Like I just said, it is possible to use other forces (no, not the dark side one), such as a magnetic force, displayed by a magnet or anything with a force like so.
When a neutral atom gains or loses
electrons, it becomes charged
and is called a(n)
Answer:
It is called an ion.
Explanation:
Atoms of elements can lose or gain electrons making them no longer neutral, they become charged. A charged atom is called an ion. When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons.
A 1.00-m3 object floats in water with 30.0% of its volume above the waterline. What does the object weigh out of the water?
Answer:
Object's weight = 6,839.42 N
Explanation:
Given
Above waterline = 30%
Volume of object = 1m^3
Required
Determine the weight of the object
First, we need to calculate its Mass
Mass = Density of Water * Volume of object in water
Density of water = 997kg/m³
If 30% is above waterline, then 70% is in water.
So:
Mass = Density of Water * Volume of object in water
Mass = 997kg/m³ * 70%m³
Mass = 997kg * 70%
Mass = 697.9 kg
The object weight sis then calculated as thus:
Weight = Mass * Acceleration of gravity
Weight = 697.9 kg * 9.8m/s²
Weight = 6 839.42 N