if p=2i+4j+3k and q=I+5j-2k,find P×q.
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Answer:

[tex]V=8.08m/s[/tex]

Explanation:

From the question we are told that:

Height[tex]h=5.00m[/tex]

Mass [tex]m=0.750kg[/tex]

Radius [tex]r=4.00cm=>0.04m[/tex]

Generally the equation for Total energy is mathematically given by

  [tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]

Therefore

 [tex]V=\sqrt{\frac{4gh}{3}}[/tex]

 [tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]

 [tex]V=8.08m/s[/tex]

Answer:

T = 2398 K

Explanation:

To calculate the emission of the light bulb we use the law is Stefan

           P = σ A e T⁴

as they indicate that the filament is a black body, the emissivity is equal to 1 (e = 1)

           T = [tex]\sqrt[4]{\frac{P}{ \sigma A} }[/tex]

    let's calculate

           T =[tex]\sqrt[4]{\frac{120}{5.67 \ 10^{-8} \ 6.4 \ 10^{-5}} }[/tex]

           T = [tex]\sqrt[4]{33.06878 \ 10^{12} }[/tex]

           T = 2,398 10³ K

           T = 2398 K

Answer:

a) F₁ = F₂,  F₃ = F₄,  b)  the correct answer is 3

Explanation:

a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies

Forces 1 and 2 are action and reaction forces F₁ = F₂

Forces 3 and 4 are action and reaction forces F₃ = F₄

as it indicates that the

b) how the car increases if speed implies that force 1> force3

      F₁ > F₃

therefore the correct answer is 3

Answer:

For liquids: A measuring cylinder is used.

For solid: Over flow can is used

Answer:

i think a measuring cylinder

Answer:

The kinetic energy and potential energy lost to friction is 2,420 J.

Explanation:

Given;

total mass, m = 40 kg

initial velocity of the girl, Vi = 5 m/s

hight of the hill, h = 10 m

length of the hill, L = 100 m

initial kinetic energy of the girl at the top hill:

[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]

initial potential energy of the girl at the top hill:

[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]

Total energy at the top of the hill:

E = 500 J + 3920 J

E = 4,420 J

At the bottom of the hill:

final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s

hight of the hill = 0

final kinetic energy of the girl at the bottom of the hill:

[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]

final potential energy of the girl at the bottom of the hill:

[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]

Based on the principle of conservation of energy;  

the sum of the energy at the top hill = sum of the energy at the bottom hill

The energy at the bottom hill is less due to energy lost to friction.

[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]

Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.

Answer:

The maximum height reached by the rocket is 486.53 m

Explanation:

Given;

initial velocity of the rocket, u = 0

acceleration of the rocket, a= 20 m/s²

duration of the rocket first motion, t = 4 s

The distance traveled by the rocket before its thrust failed

h₁ = ut + ¹/₂at²

h₁ = 0 + ¹/₂ x 20 x 4²

h₁ = 160 m

The second distance moved by the rocket is calculated as follows;

The velocity of the rocket before its thrust failed;

v = u + at

v = 0 +  20 x 4

v = 80 m/s

This becomes the initial velocity for the second stage

At maximum height, the final velocity = 0

[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]

The maximum height reached by the rocket = h₁ + h₂

                                                                          = 160 + 326.53

                                                                          = 486.53 m

Answer:

A) 20 N, B) 20 N, & C) 8 N

Explanation:

For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.

1. Determination of A and B.

Forward forces = Backward forces

A + 10 + B = 25 + 25

A + 10 + B = 50

Collect like terms

A + B = 50 – 10

A + B = 40

Assume A and B to be equal. Thus, A is 20 N and B is 20 N.

2. Determination of C

Upward forces = Downward forces

C + 112 = 20 + 100

C + 112 = 120

Collect like terms

C = 120 – 112

C = 8 N

Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.

It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.

Answer:

you drove 50km

Explanation:

10×5 hope this helps

Answer:

50 Km

Explanation:

This is how far you have got on your journey if traveling like this.

Please Mark as Brainliest

Hope this Helps

Answer:

Chelsea F.C

Explanation:

Chelsea F.C

Soccer

Answer:

THE ANSWER IS SOMETHING LIKE 55

Answer:

72  is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72  is the premimum of the insurance.

Answer:

francium

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

Answer:

The same as its translational KE.

The easy way to do this is to make up numbers and use them.

So, I'll say m=2 and r=3. I will also say v=3 .

Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)

note: (1/2*I*w^2)

Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)

Now, lets plug our made up values in:

Rot Ke : 1/2 (9*2)(3/3) *note w = v/r

Tran Ke: 1/2(2)(9)

Rot Ke: 9

Tran Ke: 9

9=9, same.

Answer:

9.42 m/s

Explanation:

Applying,

V' = Aω.............. Equation 1

Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.

But,

ω = 2πf................. Equation 2

Where f = frequency, π = pie

And,

f = v/λ................ Equation 3

Where, λ = wave length, v = velocity

Also,

v = √(T/μ)................. Equation 4

Where T = Tension, μ = linear density.

From the question,

Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m

Substitute into equation 4

v = √(50/0.005)

v = √(10000)

v = 100 m/s

Also Given: λ = 2.0 m

Substitute into equation 3

f = 100/2

f = 50 Hz.

Substitute the value of f into equation 2

Where π = constant = 3.14

ω = 2(3.14)(50)

ω = 314 rad/s

Finally,

Given: A = 3.0 cm = 0.03 m

Substitute into equation 1

V' = 0.03(314)

V' = 9.42 m/s

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

Answer:

Frequency is the number of complete oscillations or cycles or revolutions made in one second.

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

The distance of the final image from the first lens will be is 6 cm.

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

https://brainly.com/question/3229491

Answer:

The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.

Explanation:

Answer:

English please

Explanation:

I don't understand the question

Answer:

[tex]V_{rms}=5.6*10^3m/s[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=5100K[/tex]

Generally the equation for RMS Speed is mathematically given by

 [tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]

Where

 [tex]K=Boltzman's constant[/tex]

 [tex]K=1.38*10^{-23}[/tex]

And

 [tex]M=molecular mass[/tex]

 [tex]M=4*1.67*10^{-27}[/tex]

 [tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]

 [tex]V_{rms}=5.6*10^3m/s[/tex]

Answer:

potential energy is a type of energy an object has because of it's position

Answer:

hindi ko maintindihan teh

Answer:

Explanation:

The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation

[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...

0 = 21.5 + (-9.8)t and

-21.5 = -9.8t so

t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)

Now we will use that time to find out the max height of the object in the equation

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:

Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to

Δx = 47.1 - 23.5 so

Δx = 23.6 meters.

Now we can plug that in to the PE equation to find the PE of the object:

PE = (.19)(9.8)(23.6) so

PE = 43.9 J

Answer:

[tex]v=3.49\times 10^4\ m/s[/tex]

Explanation:

Given that,

Mass of Venus, [tex]M_V=4.87\times 10^{24}\ kg[/tex]

We know that,

Mass of Sun, [tex]M_s=1.98\times 10^{30}\ kg[/tex]

The distance between the center of Sun and the center of Venus is [tex]1.08\times 10^{11}\ m[/tex]

We need to find the peed of the planet Venus in its essentially circular orbit around the sun. using the formula,

[tex]v=\sqrt{\dfrac{GM_s}{r}}[/tex]

Put all the values,

[tex]v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.98\times 10^{30}}{1.08\times 10^{11}}}\\\\v=3.49\times 10^4\ m/s[/tex]

So, the speed of the planet venus is [tex]3.49\times 10^4\ m/s[/tex].

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg