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]
what is the initial position of the object?
a. 2m b. 4m c. 6m d. 8m e. 10 m
what is the velocity of the object?
a. -10 m/s b. -5 m/s c. 0 m/s d. 5 m/s e. 10 m/s
which of the following is true?
a. the object increase its velocity.
b. the object decrease its velocity.
c. the objects velocity stays unchanged.
d. the object stays at rest.
e. more information is required.
Answer:
e:10 m b: -5 m/s b.The object decreases its velocity.
Explanation:
Ask For
6.
A horizontal 100 N force is applied to a 50 kg classmate resting on a level tile
floor. The coefficient of kinetic friction is 0.15.
a. Draw a force diagram to represent this situation.
b. What is the acceleration of the classmate?
c. Suppose the classmate was resting on a carpet where the coefficient of static
friction is 0.25. Is the horizontal 100 N force sufficient to cause the classmate to
accelerate? Draw a force diagram, and then explain why or why not.
A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the block. What is the new speed of the block?
Please include an explanation.
Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]
[tex]E_{1}=E_{pot}+E_{kin}+E_{elas}\\[/tex]
In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]
[tex]E_{2}=E_{kin}+E_{pot}[/tex]
Now we can use the first statement to get the first equation:
[tex]E_{1}+W_{1-2}=E_{2}[/tex]
where:
W₁₋₂ = work from the state 1 to 2.
[tex]E_{k}=\frac{1}{2} *m*v^{2} \\[/tex]
[tex]E_{pot}=m*g*h[/tex]
where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]
[tex]70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5[/tex]
[tex]58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s][/tex]
Which scenario is an example of the transfer of thermal energy by radiation?
A. Water boils in a pan.
B. Hot air circulates in an oven.
C. An ice cube melts in a person's hand.
D. A frozen lake melts under the Sun.
Correct answer is D
Answer:
its D: A frozen lake melts under the sun.
Explanation:
Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the earth from the sun is invisible. Only a small portion comes as visible light. Light is made of waves of different frequencies.
Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt
Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.
The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.
Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have
• net parallel force
∑ Force (//) = W (//) - F = m a
(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)
• net perpendicular force
∑ Force (⟂) = W (⟂) + N = 0
Notice that
W (//) = W sin(θ) … … … which is positive since it points down the plane
W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N
So the equations become
W sin(θ) - F = m a
-W cos(θ) + N = 0
Solving for a gives
a = (W sin(θ) - F ) / m
which is good enough if you know the magnitude of the friction force.
If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as
F = µ N
so that
a = (W sin(θ) - µ N ) / m
and the normal force itself has a magnitude of
N = W cos(θ)
so that
a = (W sin(θ) - µ W cos(θ) ) / m
The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so
a = (m g sin(θ) - µ m g cos(θ) ) / m
a = g (sin(θ) - µ cos(θ))
A 100 kg. football player and a 20 kg. child sit on rolling carts and push off
from each other. Which person applies more force?*20 kg. child applies more force
100 kg. football player applies more force
Same force is applied
Answer:
the forces are the same
Explanation:
one newton equals 0.225
Among the elements potassium, lithium, and iron, the metallic bonds are likely to be strongest in
彼が好きな男のような話は妻と関係があったのでそう
Explanation:これが難しい場合はコメントで教えてくださいが何でも
lol okay
from: a random person :)
Answer: Iron
Explanation: Gradpoint
Please help
1. Who was Ptolemy and Copernicus?
2. Describe their ideas about the planets and the sun
Answer:
1 +2=3 so thats it
Explanation:
Because i need points
A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?
Answer:
Fx = 467.65N
Fy = 270N
Explanation:
Given
Force = 540N
angle of inclination = 40 degree
Horizontal component Fx = Fcos 30
Fx = 540cos30
Fx = 540(0.8660)
Fx = 467.65N
Hence the horizontal component is 467.65N
Vertical component Fy = Fsin 30
Fy = 540sin30
Fy = 540(0.5)
Fy = 270N
Hence the vertical component is 270N
. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it by 10.0 cm?
Ans 4 more to be exact
Explanation:
Given values are:
Spring constant,
k = 40 N/mand,
x = 10 mThe force will be:
→ [tex]F = kx[/tex]
By substituting the values, we get
[tex]= 40\times 10[/tex]
[tex]= 400 \ N[/tex]
Thus the above answer is right.
Learn more about spring constant here:
https://brainly.com/question/15277652
Calculate the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
Answer:
RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]
Explanation:
We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.
The formula for RMS speed of a gas is given by :
[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]
Where
R is ideal gas constant, R = 8.314 J /mol K
T = 5300 K
m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]
Substituting all the values in above formula :
[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]
So, the RMS speed Helium atoms 5748.75 m/s.
Where do magnetic fields occur?
Answer:
The Magnetosphere and MagnetsExplanation:
[tex]--------------------------------------------[/tex]
According to the National Geographic, "Earth’s magnetic field dominates a region called the magnetosphere, which wraps around the planet and its atmosphere. Solar wind, charged particles from the sun, presses the magnetosphere against the Earth on the side facing the sun and stretches it into a teardrop shape on the shadow side. The magnetosphere protects the Earth from most of the particles, but some leak through it and become trapped. When particles from the solar wind hit atoms of gas in the upper atmosphere around the geomagnetic poles, they produce light displays called auroras. These auroras appear over places like Alaska, Canada and Scandinavia, where they are sometimes called “Northern Lights.” The “Southern Lights” can be seen in Antarctica and New Zealand. The magnetic field is the area around a magnet that has magnetic force. All magnets have north and south poles."
[tex]--------------------------------------------[/tex]
Hope this helps! <3
[tex]--------------------------------------------[/tex]
If you lie on the ground at night with no cover, you get cold rather quickly. Much of this is due to energy loss by radiation. At night in a dry climate, the temperature of the sky can drop to −40∘C. If you are lying on the ground with thin clothing that provides little insulation, the surface temperature of your skin and clothes will be about 30∘C. Estimate the net rate at which you body loses energy by radiation to the night sky under these conditions. Hint: What area should we use?
Answer:
301.48 J/s
Explanation:
We are given;
Temperature of the sky dropping to −40∘C: T_o = -40°C = -40 + 273 = 233 K
Temperature of your skin and clothes: T = 30°C = 30 + 273 = 303 K
Body surface area of human body is around 2 m². But here only half of the body is facing the sky, Thus Area is: A = 2/2 = 1 m²
To solve this, we will use the equation for thermal heat transfer known as the Stefan bolt Mann equation.
ΔQ/Δt = εσA(T⁴ - (T_o)⁴)
Where;
ΔQ/Δt is the rate at which you body loses energy by radiation
ε is the emissivity of the human body with a value of 0.97
σ is Stefan boltzmann constant with a value of 5.67 X 10^(-8) W/m².K⁴
Thus;
ΔQ/Δt = 0.97 × 5.67 X 10^(-8) × 1(303⁴ - 233⁴)
ΔQ/Δt = 301.48 J/s
How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?
A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km
Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
The Slingshot is a ride for two people. It consists of a single passenger cage, two towers, and two elastic bands. Potential energy is stored in the elastic bands and the passenger cage is released. On the way up, this potential energy in the elastic bands is converted into the kinetic energy of the cage. At the maximum height of the ride, the energy has been converted into gravitational potential energy of the cage. The slingshot has two towers of height h = 76 m. The towers are a distance d = 31 m apart. Each elastic band has an unstretched length of L0 = 41 m and a spring constant of k = 310 N/m. The total mass of the passengers and cage is m = 410 kg. The car is pulled down to the ground in the middle of the two towers.
Calculate the maximum height, in meters, of the ride.
my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,
Explanation:
first its going to say 10 points for my question but after that I well make answer a small question and give you 30points. like whats your favorite color . stay tuned .
Answer:
The maximum height reached by the ride after it was pulled to the ground is 51.6 m.
The given parameters;
Distance between the two towers, d = 31 mHeight of the tower, h = 76 mUnstretched length of the band, L₀ = 41 mElastic constant of the band, k = 310 N/mThe distance half-way between the bands;
[tex]\frac{d}{2} = \frac{31}{2} = 15.5 \ m[/tex]
The maximum length of the band when stretched is calculated as;
[tex]c^2 = 15.5^2 + 76^2\\\\c^2 = 6016.25\\\\c = \sqrt{6016.25} \\\\c = 77.57 \ m[/tex]
The extension of the elastic band;
x = 77.57 m - 41 m
x = 36.37 m
The elastic potential energy stored in the band;
[tex]E = \frac{1}{2} kx^2\\\\E = \frac{1}{2} \times 310 \times (36.57)^2\\\\E = 207,291.56 \ J[/tex]
The elastic potential energy of the elastic band will be converted into kinetic energy of the ride and the speed of the ride is calculated as;
[tex]E = \frac{1}{2} mv^2\\\\207,291.56 = \frac{1}{2} \times 410 \times v^2\\\\v^2 = \frac{207,291.56}{(0.5\times 410)} \\\\v^2 = 1011.178\\\\v = \sqrt{1011.178} \\\\v = 31.8 \ m/s[/tex]
The maximum height reached by the ride is calculated as;
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{(31.8)^2}{(2\times 9.8)} \\\\h = 51.6 \ m[/tex]
Thus, the maximum height reached by the ride is 51.6 m.
Learn more here: https://brainly.com/question/156316
A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?
A 300N
B 202N
C 398N
D 98N
Answer:98
Explanation:hope this helps!
What is the frequency of this wave?
can someone pls explain and answer
Answer:
f = 1 Hz
Explanation:
From the attached figure, we find that the time period of the wave is 1 second.
It is a longitudinal wave. It travel in the form of compression and rarefaction. The relation between time period T and frequency f is given by :
[tex]f=\dfrac{1}{T}\\\\\text{As T = 1 s}\\\\f=1\ s^{-1}\\\\=1\ Hz[/tex]
Hence, the frequency of this wave is 1 Hz.
A skier is moving 8.33 m/s when
he starts to slide UP a 8.44°
frictionless slope. How much time
does it take him to come to a stop?
(Unit = s)
Answer:
5.79
Explanation:
Vf=Vi+at
0=8.33+(-9.8)sin8.44t
t=8.33/(9.8sin8.44)=0.83/sin 8.44
=5.79
8x = -6. What does x equal?
Answer:
x=-3/4
Explanation:
Find the state of Georgia in the southeastern U.S. What can you say about the weather in this state?
What would its weight be on Jupiter?
24.9N
Answer:
1.898 × 10^27 kg
Explanation:
thats how much it ways
HELPP physics final will give brainliest
What is the wavelength of a wave whose velocity is 12m/s and has a frequency of .75 Hz?
Answer:
16 meters
Explanation:
Use the formula that relates frequency velocity and wavelength:
velocity = wave-length x frequency
in our case:
12 m/s = wave-length * 0.75 Hz
wave-length= 12/0.75 m = 16 meters
A disk with a rotational inertia of 2.0 kg m2 and a radius of 1.6 m is free to rotate about a frictionless axis perpendicular to the disk's face and passing through its center. A force of 5.0 N is applied tangentially to the rim of the disk. What is the angular acceleration of the disk as a result of this applied force?
a) 4.0 rad/s2
b) 1.0 rad/s2
c) 2.0 rad/s2
d) 0.40 rad/s2
e) 0.80 rad/s2
Answer:
a) 4.0 rad/s2
Explanation:
For rigid bodies, Newton's 2nd law becomes :τ = I * α (1)
where τ is the net external torque applied, I is the rotational inertia
of the body with respect to the axis of rotation, and α is the angular
acceleration caused by the torque.
At the same time, we can apply the definition of torque to the left side of (1), as follows:[tex]\tau = F*r*sin \theta (2)[/tex]
where τ = external net torque applied by Fnet, r is the distance
between the axis of rotation and the line of Fnet, and θ is the
angle between both vectors.
In this particular case, as Fnet is applied tangentially to the disk, Fnet
and r are perpendicular each other.
Since left sides of (1) and (2) are equal each other, right sides are equal too, so we can solve for the angular acceleration as follows:[tex]\alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2 (3)[/tex]
Two objects of equal mass are a distance of 5.0 m apart and attract each other with a gravitational force of 3.0 x 10^-7 N find their mass.
A) 150 kg
B) 9.8 kg
C) 11.000 kg
D) 340 kg
Answer
I Think Its 150
When water freezes, its volume increases by 9.05% (that is, equation). What force per unit area is water capable of exerting on a container when it freezes?
Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²
The 20 oz orange soda you drank at lunch contained 1 oz. Of real orange juice. What percent of the orange soda is real orange juice?
What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.0001 N?
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required:
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance between the compact car and pickup truck is 0.96048 m
Explanation:
The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them. This is shown in equation 1;
[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1
Where F is the gravitational force = 0.0001 N
G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]
[tex]m_{1}[/tex] is the mass of the compact car = 900kg
[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg
d is the distance and its unknown ?
Let us make d the subject formula in equation 1
[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2
Substituting into equation 2 we have
[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]
d = 0.96048m
Therefore the distance between the compact car and pickup truck is 0.96048 m
What is the original source of the energy stored in fossil fuels?
plants
the Sun
air
O water
The sun
Answer:
Plants the chloroplasts