Where do magnetic fields occur?

Answer:

2008

Explanation:

2000+3+5======2008

Answer:

[tex]8\hat i-2\hat j-2\hat k[/tex]

Explanation:

Vectors in 3D

Given a vector

[tex]\vec P = P_x\hat i+P_y\hat j+P_z\hat k[/tex]

A vector [tex]\vec Q[/tex] parallel to [tex]\vec P[/tex] is:

[tex]\vec Q = k.\vec P[/tex]

Where k is any constant different from zero.

We are given the vectors:

[tex]\vec A = \hat i+4\hat j-2\hat k[/tex]

[tex]\vec B = 3\hat i-5\hat j+\hat k[/tex]

It's not specified what the 'resultant' is about, we'll assume it's the result of the sum of both vectors, thus:

[tex]\vec A +\vec B = \hat i+4\hat j-2\hat k + 3\hat i-5\hat j+\hat k[/tex]

Adding each component separately:

[tex]\vec A +\vec B = 4\hat i-\hat j-\hat k[/tex]

To find a vector parallel to the sum, we select k=2:

[tex]2(\vec A +\vec B )= 8\hat i-2\hat j-2\hat k[/tex]

Thus one vector parallel to the resultant of both vectors is:

[tex]\mathbf{8\hat i-2\hat j-2\hat k}[/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

Answer:

A

Explanation:

i think

Answer:

Buoyant Force = 200N

Explanation:

Given

[tex]Object = 200N[/tex]

Required

Determine the buoyant force?

Using Archimedes principle:

When an object is immersed in a fluid, the object is acted upon on by an upward force (Buoyant force) which equals the weight of the object.

In other words;

Buoyant Force = Weight of object

Hence:

Buoyant Force = 200N

The buoyant force on the object which floats with three-fourths of its volume beneath the surface of the water is 200 N.

The buoyant force is the force which is applied by the fluid in the upward direction when a object is placed over it. This buoyant force can be calculated with the following formula.

[tex]F_b=-\rho gV[/tex]

Here, (ρ) is the density of the fluid, (g) is the gravitation force and (V) is the fluid volume.

The buoyant force is equal to the weight of the liquid displace by the object which is placed on it.

It is given that the 200-N object floats with three-fourths of its volume beneath the surface of the water.

This force applied on the object balance the 200-N object and floats it. For this case, the value of buoyant force will be equal to the weight of the object.

[tex]F_b=W\\F_b=200\rm\; N[/tex]

Thus, the buoyant force on the object which floats with three-fourths of its volume beneath the surface of the water is 200 N.

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Answer:

W = 6642 J

Explanation:

Given that,

Mass of a crate, m = 67 kg

Force with which the crate is pulled, F = 738 N

It is moved 9 m across a frictionless floor

We need to find the work done in moving the crate. Let the work done is W. It is given by :

W = F d

W = 738 N × 9 m

= 6642 J

So, the work done is 6642 J.

Answer:

Component

Explanation:

All circuits have some basic parts, called components. One component is the power source, also called a voltage source. The power source is what pushes the electricity through the circuit.

Answer:

v=59[m/s]

Explanation:

To solve this problem we must use the principle of conservation of energy, which tells us that energy is transformed from Kinetic to potential or vice versa. At the moment when the car is at the top before falling down the cliff, we have the car moving at speed 50 [m/s] (kinetic energy) also it is 50 [m] above ground level (potential energy).

[tex]E_{k1}+E_{p1}=E_{k2}\\[/tex]

where:

Ek1 = kinetic energy before falling [J]

Ep1 = potential energy before falling [J]

Ek2 = kinetic energy in the ground [J]

The potential energy can be calculated by means of the following equation.

[tex]E_{p}=m*g*h[/tex]

where:

m = mass = 500 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 50 [m]

Whereas the kinetic energy can be calculated by means of the following equation.

[tex]E_{k}=\frac{1}{2}*m*v^{2}[/tex]

where:

v = velocity = 50 [m/s]

Now replacing in the general equation:

[tex]\frac{1}{2} *500*(50)^{2} +500*9.81*50=\frac{1}{2} *500*v^{2}\\625000+245250=250*v^{2} \\250*v^{2} =870250\\v=\sqrt{870250/250} \\v=59[m/s][/tex]

Answer:

The half-life is [tex] t_{1/2} = 1.005 h[/tex]

Explanation:

Using the decay equation we have:

[tex]A=A_{0}e^{-\lambda t}[/tex]

Where:

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]

[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]

[tex]0.5=e^{-\lambda*1 h}[/tex]

Taking the natural logarithm on each side we have:

[tex]ln(0.5)=-\lambda[/tex]

[tex]\lambda=0.69 h^{-1}[/tex]

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]

[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]

[tex] t_{1/2} = 1.005 h[/tex]

I hope it helps you!

Answer:

Explanation:

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

Answer:

the air is being stopped in the pocket of the parachutes theys why the parachute has a certain shape so that the air that gets caught inside of it as the skydiver goes down slows down the landing

Answer:

The answer is "[tex]\bold{7.56 \ Me\ V}[/tex]".

Explanation:

calculating the binding energy on per nucleon:

calculating number of proton and neutrons:

proton [tex]P_u=94[/tex]

neutron[tex]= 239-94=145[/tex]

calculating mass:

proton mass [tex]\ m_P=1.007825 \ amu\\\\[/tex]

neutron mass [tex]\ m_n=1.008665 \ amu\\\\[/tex]

neutral atom mass [tex]m = 239.05216 \ amu\\\\[/tex]

mass of prtons[tex]= 94 \times 1.007825 = 94.73555 \ amu\\\\[/tex]  

mass of neutrons[tex]= 145 \times 1.008665= 146.256425 \ amu\\\\[/tex]

Total nucleons mass formula:

[tex]\to m_n = (P+n)[/tex]

          [tex]= 94.73555+ 146.256425\\\\= 240.991975 \ amu[/tex]

calculating the mass of defect:

[tex]\to \Delta m= m_n-m\\\\[/tex]

           [tex]= 240.991975 - 239.05216\\\\= 1.939815 \ amu\\\\[/tex]

calculating the total of the binding energy:

[tex]\to BE=\Delta m\times 931.5 \ mev[/tex]

          [tex]= 1.939815 \times 931.5\\\\=1806.938 \ Me \ V\\\\[/tex]

BE in per nucleon [tex]=\frac{BE}{239}= 7.56 \ Me\ V[/tex]

Answer:

Decreasing

Hope this helps! :)

Answer:

222/88 Ra

Explanation:

We have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.

To find the tike taken for the activity, we need to know about radioactivity and half-life.

                R=R₀e^(-λ×t)

                 t= 1/λ ln(R₀/R)

    t=(half-life/ln2)×ln(R₀/R)

Here R=0.021R₀, so t= (half-life/ln2)×ln(R₀/0.021R₀)=5.57 half-lives

Thus, we can conclude that we have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.

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Answer:

a) h(g) = 358,53 m

b) t = 8,16 s

c) t(t) = 16,71 s

Explanation:

Equations for vertical shooting are:

Vf = V₀ -  g * t   ;     h  =  V₀*t -  (1/2)*g*t²  ;    Vf² = V₀² - 2*g*h

And at maximum heigt Vf = 0 then

0 = V₀ - g * t

t = V₀/g                     V₀  = 80 m/s      and   g = 9,8 m/s²

t  =  80 / 9,8   (s)

t = 8,16 s

Then 8,16 s is the time to get maximum height

If we plug t = 8,16 (s) in equation  h  =  V₀*t -  (1/2)*g*t²

we get:     h (max)  =  (80)*8,16 - 0,5*9,8*(8,16)² (m)

h (max) = 652,8  -  326,27 m

h (max) = 326,53 m

Then relative to ground that height becomes

h(g) = 326,53 + 32

h(g) = 358,53 m

In order to get the time the arrow is in the air we proceed as follows:

a) for the arrow to be at the launched point will take the same time that from the launched point to the maximum height, and after that we have to find out the time the arrow takes from 32 m down to the ground level

Then  

t(t) = 8,16 + 8,16 + tₓ     (2)

Where tₓ  is the time from 32 m height to ground

h  =  V₀*tₓ -  (1/2)*g*tₓ²  but since the arrow now is going down then we change the sign of the second term on the right side of the equation

32 = (80)*tₓ  +  0,5 * 9,8 * tₓ²     Note that when the arrow is at 32 m height the speed is again V₀ = 80 m/s

32 = 80*tₓ  + 4,9*tₓ²

A second-degree equation for tₓ, solving it

4,9*tₓ² + 80*tₓ - 32 = 0

t₁,₂ = -80 ± √ 6400 + 627,2 / 9,8

t₁,₂ =( - 80 ± 83,8 ) / 9,8

there is not a negative time therefore we dismiss such solution and

t₁ = 3,8 / 9,8

t₁ = 0,39 s

And

t(t) = 8,16 + 8,16 + 0,39  s

t(t) = 16,71 s

Answer:

The end temperature is 174 °C

Explanation:

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P*V = n*R*T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

So, being:

and replacing:

2 atm*V= 2 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] *298 K

you get:

[tex]V=\frac{2 moles* 0.082\frac{atm*L}{mol*K} *298 K}{2 atm}[/tex]

V= 24.436 L

Now, two moles of neon gas is expanded to 3 times the original volume while the pressure is reduced to 1.0 atm. Then you know:

Replacing:

1 atm*73.308 L= 2 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] *T

Solving:

[tex]T=\frac{1 atm*73.308 L}{2 moles* 0.082\frac{atm*L}{mol*K}}[/tex]

T= 447 °K= 174 °C (being 0°C=273 °K)

The end temperature is 174 °C

Answer: 3.

Explanation:

The correct answer is a higher amplitude and lower frequency. Since an opera singer is lowering his pitch it means that he is creating higher amplitude and because he is raising his voice for a song with that higher amplitude he is creating lower frequency.

Answer:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

Explanation:

Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

A collision is any situation in which two or more bodies quickly exert forces on one another. Despite the fact that the most common usage of the word "collision" refers to situations in which two or more objects clash violently, the scientific usage of the word makes no such assumptions.

The following are a few instances of physical encounters that scientists might classify as collisions. Legs of an insect are said to collide with a leaf when it falls on one.

Every contact of a cat's paws with the ground while it strides across a lawn is seen as a collision, as is every brush of its fur with a blade of grass.

Therefore, Ball 1 has a mass of 0.5 kilogram and an initial velocity of 1.00 meter/second. Ball 2 has a mass of 1.5 kg and an initial velocity of 0.00 meters/second.

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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:

It's C. Moving Waterrrr

Answer:

x=-3/4

Explanation:

Answer:

uniformes

Explanation:

Why are u asking this

Answer:

STOP CHETING

Answer:

C is the answer to your question

Explanation:

Have a great day!!! :)

Answer:

3.1

Explanation:

use formula f = v/lambda

Answer:

F = 288 [N]

Explanation:

To solve this problem we must use the following equation of kinematics and find the value of acceleration.

[tex]v_{f}^{2} =v_{o}^{2} +2*a*x[/tex]

where:

Vf =  final velocity = 6 [m/s]

Vo =  initial velocity = 0 (starting from rest)

a = acceleration [m/s²]

x = distance = 5 [m]

Now replacing, we have:

[tex](6)^{2}=0+(2*a*5)\\36=10*a\\a = 3.6 [m/s^{2}][/tex]

Since we already have the value of acceleration, we can use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

[tex]F =80*3.6\\F = 288 [N][/tex]