suppose the tank is open to the atmosphere instead of being closed. how does the pressure vary along

Answer:

a. True

Explanation:

A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;

1. Extrinsic semiconductor.

2. Intrinsic semiconductor.

An intrinsic semiconductor is a crystalline solid substance that is in its purest form and having no impurities added to it. Examples of intrinsic semiconductor are Germanium and Silicon.

Basically, the number of free electrons in an intrinsic semiconductor is equal to the number of holes. Also, the number of holes and free electrons in an intrinsic semiconductor is directly proportional to the temperature; as the temperature increases, the number of holes and free electrons increases and vice-versa.

In an intrinsic semiconductor, each free electrons (valence electrons) produces a covalent bond.

Generally, a process referred to as doping can be used to increase the conductivity of an intrinsic semiconductor such as silicon or germanium, by adding small amounts of impurities found in group 3A or group 5A elements.

Answer:

7.9 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]

Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:

[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]

Plug everything in:

7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]

(1590kg the initial weight plus the weight of the added passenger)

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

Answer:

135J

Explanation:

So we know ΔKinetic Energy= ΔWork

Kinetic energy=1/2mv²

So Kf-Ki=ΔK

ΔK=1/2*0.45(25²-5²)=135J

135J=ΔWork

Answer:

31250 meters

Explanation:

Given data

Intitially at rest, the velocity will be

u= 0m/s

acceleration a= 25m/s^2

Time= 50s

We know that the expression for the displacement is given as

S=U+ 1/2at^2

S= 0+ 1/2*25*50^2

S= 12.5*2500

S=31250 meters

Hence the displacement is 31250 meters

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

[tex]V_A = \frac{1}{2}V_B[/tex]

Now, from Charle's Law:

[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]

Hence, the correct option is:

A. TA = TB/2.

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

Answer:

Explanation:

6cos28

=5.3 N

Explanation:

someone to check if the answer is correct

Answer:

Electromechanical transducer and Electrical receiver.

Explanation:

Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.

Answer:

sup qwertyasdfghjk

Explanation:

Answer:

Measurements allow people to find their way to new places. Measurements such as miles or kilometers are used by GPS systems to give directions. Time measurements help to create schedules so tasks get done on time. Measurements are used in food as well. Ingredients in recipes have to be measured to make the dish correctly. Serving sizes are a measurement that keep people healthy by showing how much of each food you should eat.

Answer:

a

Explanation:

you take 32,000kg ÷2.0m

South

Explanation:

The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by

[tex]\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}[/tex]

and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.

Answer:

Power = Energy/time

Energy = Power xtime.

Time= 20hrs

Power = 100Watt =0.1Kw

Energy = 0.1 x 20 = 2Kwhr.

This Answer is in Kilowatt-hour ...

If the one given to you is in Joules

You'd have to Change your time to seconds

Then Multiply it by the power of 100Watts.

Answer:

A(3.56m)

Explanation:

We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.

We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.

Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2

Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2=  v^2).

Ef= -mgh+3/4mv^2

Since Ef=Ei=0

Mgh=3/4mv^2

gh=3/4v^2

h=0.75v^2/g

plug in givens to get h= 3.57m

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

Answer:

w = 1,066 rad / s

Explanation:

For this exercise we use Newton's second law

         F = m a

the centripetal acceleration is

         a = w² r

indicate that the force is the mass of the body times the acceleration

        F = m 0.58g = m 0.58 9.8

        F = 5.684 m

we substitute

       5.684 m = m w² r

       w = [tex]\sqrt{5.684/r}[/tex]

To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m

       w = [tex]\sqrt{ 5.684/5}[/tex]

       w = 1,066 rad / s

Answer:

a) [tex]a_{avg}=1.25ft/s^2[/tex]

b) [tex]v_b=13.35ft/s[/tex]

Explanation:

From the question we are told that:

Speed at point A [tex]v_A=10.6ft/s[/tex]

Speed at point C [tex]v_C=15.6ft/s[/tex]

Time from Point A to C [tex]T_{ac}=4.00s[/tex]

Time from Point B to C [tex]T_{bc}=1.80s[/tex]

Generally the equation for acceleration From A to B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{\triangle t}[/tex]

 [tex]a_{avg}=\frac{15.6-10.6}{4.0 }[/tex]

 [tex]a_{avg}=1.25ft/s^2[/tex]

Generally the equation for cart speed at B is mathematically given by

 [tex]a_{avg}=\frac{v_c-v_a}{T_{bc}}[/tex]

 [tex]v_b=v_c-a_{avg}*T_{bc}[/tex]

 [tex]v_b=15.6ft/s-(1.25ft/s^2)(1.80)[/tex]

  [tex]v_b=13.35ft/s[/tex]

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

Answer:

Hence the amount of heat transfer is 918.75 Btu.

Explanation:

Now,

Answer:

5.76 μC

Explanation:

The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T

So, ε = -ΔΦ/Δt

ε = -NAΔB/Δt

ε = -NAΔB/Δt

Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)

So, iR = -NAΔB/Δt

iΔt = -NAΔB/R

Δq = -NAΔB/R where Δq = charge = iΔt

substituting the values of the variables into the equation, we have

Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω

Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω

Δq = 0.011π × 10⁻¹ m²T/600 Ω

Δq = 0.03456 × 10⁻¹ m²T/600 Ω

Δq = 5.76 × 10⁻⁶ C

Δq = 5.76 μC

Answer:

They established the laws of planetary motion. They explained how the Sun rises and sets. They made astronomy accessible to people who spoke Italian.

Explanation:

(b) Use Newton's second law. The net forces on block M are

• ∑ F (horizontal) = T - f = Ma … … … [1]

• ∑ F (vertical) = n - Mg = 0 … … … [2]

where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.

Right away, we see n = Mg, and so f = µn = 0.2Mg.

The net force on block m is

• ∑ F = mg - T = ma … … … [3]

You can eliminate T and solve for a by adding [1] to [3] :

(T - 0.2Mg) + (mg - T ) = Ma + ma

(m - 0.2M) g = (M + m) a

a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)

a = 1.96 m/s²

We can get the tension from [3] :

T = m (g - a)

T = (10 kg) (9.8 m/s² - 1.96 m/s²)

T = 78.4 N

(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of

1/2 (1.96 m/s²) t

(e) Assuming block M starts from rest, its velocity at time t is

(1.96 m/s²) t

(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have

∑ F = -f = Ma

The effect of friction is constant, so that f = 0.2Mg as before, and

-0.2Mg = Ma

a = -0.2g

a = -1.96 m/s²

Then block M slides a distance x such that

0² - (1.96 m/s²) = 2 (-1.96 m/s²) x

x = (1.96 m/s²) /  (2 (1.96 m/s²))

x = 0.5 m

(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)

Meanwhile, block m would be in free fall, so after 1 s it would fall a distance

x = 1/2 (-9.8 m/s²) (1 s)

x = 4.9 m

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

v = 45m/s

t = 5s

Therefore,

d = v*t

= 45*5

= 225m

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, [tex]$\omega = \frac{1}{8}$[/tex]  rev/sec

                             [tex]$=\frac{2 \pi \times 7.5}{8}$[/tex]  rad/sec

                              = 5.89 rad/sec

Therefore, force required,

[tex]$F=m.\omega^2.r$[/tex]

   [tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]  

   = 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

   = 427126.9 x 7.5

   = 3,203,451.75 J

Answer:

changing the N to S. that's how the error will be corrected

Answer:

C is the correct answer

Explanation:

i took the test

Answer:

The direction of magnetic field is along + Z axis.

Explanation:

The direction of motion of electron is along y axis.

The magnetic  force is along - X axis.

The force on the charged particle moving in the magnetic field is

[tex]\overrightarrow{F} = q (\overrightarrow{v}\times \overrightarrow{B})\\\\- F \widehat{i} = - q (v \widehat{j}\times \overrightarrow{B})\\[/tex]

So, the direction of the magnetic field is along + Z axis.

Answer:

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces