A source emits sound at a fixed constant frequency f. If you run towards the source, the frequency you hear is

Answer:

Explanation:

There are all kinds of things floating around in B. The large dark blue and the smaller dark blue are one kind of compound.

The yellow by itself could be from column 18.

So could the smaller dark blue circle (be from column 18).

There are Big blue ones that have only 1 small blue one associated with it and others with one Big blue and two smaller light blues.

B is the answer to your question.

Answer:

For example, the pressure acting on a dam at the bottom of a reservoir is ... pressure = height of column × density of the liquid × gravitational field ... The density of water is 1,000 kg/m 3.

Explanation:

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Answer:

Greatest potential: moment before being dropped

Zero Kinetic: when it comes to rest

Greatest Kinetic: moment before first bounce

Explanation:

Answer:

The incoming white light is composed of light of different colors,

Since these different colors have different refractive indices they are refracted at different angles from one another.

The output light is then separated by color creating a color spectrum.

Since n is greater for shorter wavelengths  (violet colors) these wavelengths are refracted thru the larger angles.

Answer:

Time

Explanation:

The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.

Explanation:

Given

Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]

time taken to reach ground is given by

[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]

(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]

From the figure, it can be written

[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]

Considering horizontal motion

[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]

(c) The vertical velocity with which it strikes the ground is given by

[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]

Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]

Explanation:

Given

Ball is projected horizontally from a building of height  

time taken to reach ground is given by

(b) Line joining the point of projection and the point where it hits the ground makes an angle of  

From the figure, it can be written

Considering horizontal motion

(c) The vertical velocity with which it strikes the ground is given by

Thus, the ball strikes with a vertical velocity of

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

Answer:

ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

Explanation:

Given the data in the question;

T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol

since helium is monoatomic;

Cv = (3/2)R, Cp = (5/2)R

W₁ = 0 J [ at constant volume or ΔV = 0]

Now for the first cylinder; from the first law of thermodynamics;

Q₁ = ΔE[tex]_{int[/tex],₁ +  W₁

Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT

we substitute  

Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )

Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112

Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J

Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J

Answer:

I = 0.25 A

Explanation:

Given that,

Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.

In series combination, the equivalent resistance is given by :

[tex]R=R_1+R_2+R_3+....[/tex]

So,

[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]

The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.

V = IR

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]

So, the current of 0.25 A passes through each bulb.

Answer:

a) [tex]R=126Km[/tex]

b) [tex]\theta=74.6\textdegree[/tex]

Explanation:

From the question we are told that:

1st segment

243km at Angle=30

2nd segment

178km West

Resolving to the X axis

[tex]F_x=243cos30+178[/tex]

[tex]F_x=33.44Km[/tex]

Resolving to the Y axis

[tex]F_y=243sin30+178sin0[/tex]

[tex]R=\sqrt{F_y^2+F_x^2}[/tex]

[tex]F_y=121.5Km[/tex]

Therefore

Generally the equation for Directional Angle is mathematically given by

[tex]\theta=tan^{-1}\frac{F_y}{F_x}[/tex]

[tex]\theta=tan^{-1}\frac{121.5}{33.44}[/tex]

[tex]\theta=74.6\textdegree[/tex]

Generally the equation for Magnitude is mathematically given by

[tex]R=\sqrt{F_y^2+F_x^2}[/tex]

[tex]R=\sqrt{33.44^2+121.5^2}[/tex]

[tex]R=126Km[/tex]

Answer:

a)     a = 27.44 m / s²,  b) a = 5.39 m / s², c)  a = 156.8 m / s², cabinet maximum acceleration does not change

Explanation:

a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.

Let's apply Newton's second law

we set a regency hiss where the x axis is in the direction of movement of the truck

Y axis y

        N- W = 0

        N = W = m g

X axis

       2fr = m a

the expression for the friction force is

      fr = μ N

      fr = μ m g

we substitute

      2 μ m g = m /2   a

     a = 4 μ g

      a = 4 0.7 9.8

      a = 27.44 m / s²

b) let's look for the maximum acceleration that can be applied to the cabinet

       fr = m a

       μ N = ma

       μ m g = m a

       a = μ g

       a = 0.55  9.8

       a = 5.39 m / s²

as the acceleration of the platform is greater than this acceleration the cabinet must slip

c) the friction force is in the four wheels as well

With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all

applying Newton's second law

         4 fr = (m/4) a

         16 mg = (m) a

         a = 16 g

         a = 16 9.8

         a = 156.8 m / s²

cabinet maximum acceleration does not change

Answer:

how to solve this problem ???????

The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.

The resultant of the two forces is determined by resolving the force into x and y component as shown below;

[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]

where;

The value of Angle θ is determined from equation (1)

-500sinθ + 600sin(30) = 0

500sinθ = 600sin(30)

500sinθ = 300

sinθ = 3/5

θ = 36.87⁰

The resultant of the forces is determined using the second equation;

500cosθ + 600cos(30) = R

500 x cos(36.87) + 600 x cos(30) = R

919.6 N = R

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Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

(a) As it gets compressed by a distance x, the spring does

W = - 1/2 (52.1 N/m) x ²

of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so

W = ∆K

- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.118 m

(b) Taking friction into account, the only difference is that more work is done on the object.

By Newton's second law, the net vertical force on the object is

∑ F = n - mg = 0

where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives

n = mg = 2.45 N

and from this we get the magnitude of kinetic friction,

f = µn = 0.120 (2.45 N) = 0.294 N

Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:

W (friction) = - (0.294 N) x

W (spring) = - 1/2 (52.1 N/m) x ²

==>   W (total) = W (friction) + W (spring)

Solve for x :

- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.112 m

For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:

a) The spring's maximum compression when the track is frictionless is 0.118 m.

b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.

a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex]  (1)

Where:

[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg

[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s

k: is the spring constant = 52.1 N/m

x: is the distance of compression

After solving equation (1) for x, we have:

[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]

Hence, the spring's maximum compression is 0.118 m.

b) When the track is not frictionless, we can calculate the spring's compression by work definition:

[tex] W = \Delta E = E_{f} - E_{i} [/tex]

[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex]   (2)

Work is also equal to:

[tex] W = F*d = F*x [/tex]     (3)

Where:  

F: is the force

d: is the displacement = x (distance of spring's compression)  

The force acting on the object is given by the friction force:

[tex] F = -\mu N = -\mu m_{o}g [/tex]   (4)

Where:

N: is the normal force = m₀g

μ: is the coefficient of kinetic friction = 0.120

g: is the acceleration due to gravity = 9.81 m/s²

The minus sign is because the friction force is in the opposite direction of motion.

After entering equations (3) and (4) into (2), we have:

[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]

[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]

[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]        

Solving the above quadratic equation for x

[tex] x = 0.112 m [/tex]  

Therefore, the spring's compression is 0.112 m when the track is not frictionless.

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Answer:

n × t_1/2

Exmplanation:

The age of meteorite is calculated by multiplying it's quantity n with the half life . This means that the formula is for age of this meteorite is;

Age of meteorite= n × t_1/2

where;

n = quantity of the meteorite

t_/2 = half life of the meteorite

Thus:

The correct formula is; n × t_1/2

The image is missing and so i have attached it.

Answer:

A) E = 8740 N/C

B) E = -6536 N/C

Explanation:

The formula for electric field is;

E = kq/r²

Where;

q is charge

k is a constant with value 8.99 x 10^(9) N•m²/C²

A) Now, to find the net electric field at point A, the formula would now be;

E = (kq1/(r1)²) - (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point A

r2 is distance from charge q2 to point A.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m

r2 = 10 cm = 0.1 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))

E = 8740 N/C

B) similarly, electric field at point B;

E = (kq1/(r1)²) + (kq2/(r2)²)

Where;

r1 is distance from charge q1 to point B

r2 is distance from charge q2 to point B.

q1 = -6.25 nC = -6.25 × 10^(-9) C

q2 = -12.5 nC = -12 5 × 10^(-9) C

From the attached image, r1 = 10 cm = 0.1 m

r2 = 25cm + 10 cm = 35 cm = 0.35 m

Thus;

E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))

E = -6536 N/C

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]

Velocity acquired during this time

[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]

Consider vertical motion

[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]

Net velocity is

[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]

Angle made is

[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]

Answer:

Well a meter stick has increments of a centimeter, and since 1 cm=0.01m he should record it as 4.02m(b)

Explanation:

Answer:

13.33 K

Explanation:

Given that,

Heat absorbed, Q = 30 J

Mass of nail, m = 5 g = 0.005 kg

The specific heat capacity of the nail is 450 J/kg-°C.

We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:

[tex]Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K[/tex]

So, the increase in temperature is 13.33 K.

Answer:

Due to the applied filed the electrons move in a particular direction.

Explanation:

Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity.  So, tat the net flow is almost zero.

When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.

Answer:

North

Explanation:

Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.

Answer:

South To North

Explanation:

Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north

Answer:

5N

Explanation:

We have a simple problem of momentum here.

ΔMomentum= mΔv= FΔt

Solve for F

mΔv/Δt=F

Plug in givens

1*(2-1.5)/0.1=F

F=5N

The amount of force that the wall imparts on the ball is 5.0N

According to Newton's second law, the formula for calculating the force applied is expressed as:

[tex]F=ma[/tex]

m is the mass of the object

a is the acceleration of the object

Since acceleration is the change in velocity of an object, hence [tex]a=\frac{\triangle v}{t}[/tex]

The applied force formula becomes [tex]F=\frac{m\triangle v}{t}[/tex]

Given the following parameters

m = 1.0kg

[tex]\triangle v=2.0-1.5\\\triangle v=0.5m/s[/tex]

t = 0.1sec

Substitute the given parameter into the formula

[tex]F=\frac{1.0\times 0.5}{0.1}\\F=\frac{0.5}{0.1}\\F=5N[/tex]

Hence the amount of force that the wall imparts on the ball is 5.0N

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Answer:

The angular acceleration is 4.5 rad/s^2.

Explanation:

Acceleration, a = 26.5 m/s2

length, L = 5.89 m

The angular acceleration is

[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]

Answer:

Option (a), (d) are correct.

Explanation:

Frequency, f = 220 Hz

Resonant frequency = 660 Hz

The next frequency of stopped organ pipe is

2f, 3 f, 4 f ....

= 2 x 220 , 3 x 220 , 4 x 220 ....

= 440 Hz, 660 Hz, 880 Hz

So, the option (a) is correct.

The next resonant frequency of open organ pipe is

3 f, 5 f,...

= 3 x 220, 5 x 220 , ..

= 660 Hz, 1100 Hz,...

So, option (d) is correct.

Answer:

frequency

Explanation:

The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.

So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is  

[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.

It does not change the chemical composition of water.

Answer:

a. The average speed on her way to Grandmother's house is 48.08 mph

b. The average speed in the return trip is 50 mph.

Explanation:

The average speed (S) can be calculated as follows:

[tex] S = \frac{D}{T} [/tex]

Where:

D: is the total distance

T: is the total time

a. To find the total distance in her way to Grandmother's house, we need to find the total time:

[tex]T_{i} = t_{1_{i}} + t_{2_{i}} = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}}[/tex]

Where v is for velocity

[tex] T = \frac{d_{1_{i}}}{v_{1_{i}}} + \frac{d_{2_{i}}}{v_{2_{i}}} = \frac{(100/2) mi}{40.0 mph} + \frac{(100/2) mi}{60.0 mph} = 1.25 h + 0.83 h = 2.08 [/tex]    

Hence, the average speed on her way to Grandmother's house is:

[tex]S_{i} = \frac{D}{T_{i}} = \frac{100 mi}{2.08 h} = 48.08 mph[/tex]

b. Now, to calculate the average speed of the return trip we need to calculate the total time:                        

[tex]D = v_{1_{f}}\frac{T_{f}}{2} + v_{2_{f}}\frac{T_{f}}{2} = \frac{T_{f}}{2}(v_{1_{f}} + v_{2_{f}})[/tex]

[tex]100 mi = \frac{T_{f}}{2}(40 mph + 60 mph)[/tex]

[tex] T_{f} = \frac{200 mi}{40 mph + 60 mph} = 2 h [/tex]

Therefore, the average speed of the return trip is:

[tex]S_{f} = \frac{D}{T_{f}} = \frac{100 mi}{2 h} = 50 mph[/tex]

I hope it helps you!