if A=3i +2j+3k ,find the magnitude of A+B and A-B​

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

Answer:

Plants the chloroplasts

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.

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

Answer:

e:10 m   b: -5 m/s b.The object decreases its velocity.

Explanation:

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

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

Answer:

the forces are the same

Explanation:

Answer

I Think Its 150

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

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))

Answer:98

Explanation:hope this helps!

Answer:

a) 4.0 rad/s2

Explanation:

       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.

       [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.

       [tex]\alpha = \frac{F*r}{I} = \frac{5.0N*1.6m}{2.0 kg*m2} = 4.0 rad/s2 (3)[/tex]        

Answer:

Explanation:

[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]

Answer:

x=-3/4

Explanation:

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

Answer:

1 +2=3 so thats it

Explanation:

Because i need points

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.

Answer:

1.898 × 10^27 kg

Explanation:

thats how much it ways

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

Ans    4 more to be exact

Explanation:

Given values are:

Spring constant,

and,

The 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

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;

The 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

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.

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]

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