A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the mass of the beaker? The specific heat of glass is 0.18 cal/g ∙ °C, and that of water is 1.0 cal/g ∙ C°.

Answer:

Explanation:

350 cm / 25 cm = 14 radians

14 rad / 2π rad/rev = 2.23 revolutions

Answer:

The answer for the question should be True

Answer:

Transmission

Explanation:

It’s the passage of electromagnetic waves through a medium.

The passage of a wave through a medium is known as transmission.

The transmission of waves may be characterized as a wave traveling through a medium or into a new medium. Not all waves transmit through the same materials. It significantly depends on the type of wave, the type of material, and even the frequency of the wave.

Transmission is generally the passage of electromagnetic radiation through a medium. While Reflection is the process by which electromagnetic radiation is returned either at the boundary between two media (surface reflection) or at the interior of a medium (volume reflection).

Waves can go through things. Waves go through water, light can pass through the glass, and sound can pass through walls. The material the wave passes through is called a medium. A single disturbance passing through a medium is called a pulse.

Therefore, the passage of a wave through a medium is known as transmission.

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Answer:

v=u+at

v= 10+100×10

v = 10+1000

v = 1010 m/s

Answer:

By definition the acceleration due to gravity at the surface is g:

The altitude above the surface is zero for an acceleration of g.

The increase in tension on the steel wire is 8,484.75 N.

The given parameters;

The area of the steel wire is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]

The extension of the steel wire is calculated as follows;

[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]

The increase in tension on the steel wire is calculated as follows;

[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]

Thus, the increase in tension on the steel wire is 8,484.75 N.

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Answer:

Explanation:

v² = u² + 2as

s = (v² - u²)/2a

s = (0² - 75²) / (2(-14))

s = 200.8928

s = 200 m

Answer:

Identification

Explanation:

:-;..........

Following are the calculation to the friction force:

The substring tension force is provided as follows:  

[tex]\to T-m_{1}g=0 \\\\\to T = m_{1}g \\\\[/tex]

       [tex]= (68\ kg) (9.81 \ \frac{m}{s^2} ) (\frac{ 1\ N}{ 1\ kg \cdot \frac{m}{s^2}}) \\\\ = (68 ) (9.81 ) ( 1\ N) \\\\= 667.08\ N \\\\[/tex]

Therefore, the friction force on the box is "667 N".

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Answer:

Explanation:

I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was

Conservation of momentum

14000(8.1) + m(0) = (14000 + m)(3.6)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

The mass of the second car is 17500 kg.

The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.

Newton's third law of motion has a direct impact on the idea of momentum conservation.

mass of the first car = 14000 kg.

Initial speed of first car = 8.1 m/s.

Final speed of the two cars = 3.6 m/s.

From law of conservation of momentum; we can write:

Total initial momentum = total final momentum

14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

Hence, the mass of the second car is 17500 kg.

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At the equilibrium position of the spring, the mass has kinetic energy

1/2 (1.00 kg) (0.0400 m/s)² = 0.000800 J = 0.800 mJ = 800 µJ

and when the spring is maximally compressed, this kinetic energy is converted into potential energy.

The maximum spring potential energy of the simple harmonic motion is approximately 7.48 millijoules.

To find the maximum spring potential energy (Us) in simple harmonic motion, we need to determine the maximum displacement (A) of the mass from its equilibrium position.

In simple harmonic motion, the maximum displacement is equal to the amplitude of the oscillation.

The formula for maximum displacement (A) is given by:

A = v0 / ω

where v0 is the initial velocity (4.00 cm/s) and ω is the angular frequency of the oscillation.

The angular frequency (ω) is given by:

ω =[tex]\sqrt{(k / m)[/tex]

where k is the spring constant (9.00 N/m) and m is the mass (1.00 kg).

Substitute the values:

ω = [tex]\sqrt{(9.00 N/m / 1.00 kg)} = \sqrt{9.00 rad/s[/tex] ≈ 3.00 rad/s

Now, calculate the maximum displacement:

A = 4.00 cm/s / 3.00 rad/s ≈ 1.33 cm

Next, calculate the maximum spring potential energy:

Us =[tex](1/2) * k * A^2[/tex]

Us = (1/2) * 9.00 N/m * [tex](1.33 cm)^2[/tex]≈ 7.48 mJ

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--The compleete Question is,  A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m. If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm, then what is the maximum spring potential energy of the simple harmonic motion--

Hi there!

In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.

The net force is equivalent to the centripetal force, so:

∑F = T

mv²/r = T

Solve for v:

v = √rT/m

v = 13.96 m/s

Answer:

180 LUV:)

Explanation:

Given our understanding of the situation, for this vehicle with four wheels is traveling in a linear path through soft dirt, we would expect for only two wheels to leave tire track impressions.

The vehicle has four wheels, and one might think initially that it makes sense for all four wheels to create tire track impressions on the dirt. This assessment is correct, all four tires will create impressions, However, not all four tires will leave a said impression.

This is due to the fact that as the vehicle advances, the tires in the back will erase the impressions left by the front tires. Therefore, only two tire track impressions will be left.

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Answer:

Con el cielo del agua

Explanation:

espero que te ayude

Recall the definition of the cross product with respect to the unit vectors:

i × i = j × j = k × k = 0

i × j = k

j × k = i

k × i = j

and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)

Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have

a × b = (8i + j - 2k) × (5i - 3j + k)

a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)

… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)

… … … … + 8 (i × k) + (j × k) - 2 (k × k)

a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)

a × b = - 5k - 10j - 24k - 6i - 8j + i

a × b = -5i - 18j - 29k

Answer:

Explanation:

If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k​

Answer:

i think its his law of inertia

Explanation:

this law is about motion

The distance the car has skidded if it had been moving at 40.0 m/s is 27.2m

The linear force acting on the car is opposed by the frictional force. Hence;

[tex]F=F_f[/tex]

[tex]ma = -\mu R\\ma =-\mu mg[/tex]

m is the mass of the car

a is the acceleration

[tex]\mu[/tex] is the coefficient of friction

R is the normal force

Given the following parameters

[tex]a=-\mu g[/tex]

[tex]a =-0.75(9.8)\\a=-7.35m/s^2[/tex]

Get the distance the car has skidded if it had been moving at 40.0 m/s

[tex]v^2=u^2+2as[/tex]

[tex]0^2=20^2+2(-7.35)s\\0=400-14.7s\\s=\frac{400}{14.7}\\s= 27.2m[/tex]

Hence the distance the car has skidded if it had been moving at 40.0 m/s is 27.2m

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The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.

The relationship between the shutter speed and the focal ratio is an inverse relationship.

Ratio of Areas = 5.6²/4²

Since; T1A1 = T2A2

In essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.

Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.

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The actual weight of the gas = apparent weight + weight.

The actual weight = [tex]W_{A}[/tex] + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/[tex]m^{2}[/tex]

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight =  [tex]W_{A}[/tex] + W

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Answer:

1

+

1

is

2

Explanation:

because if you add one to a one it would end up being 2 <3

Answer: 2

Explanation: 1+1=2

This was THE hardest question I've ever decided to answer

The value of the constant K is [tex]\mathbf{K = \dfrac{1}{LC}}[/tex]

According to Kirchhoff's loop rule, the total algebraic sum of potential differences in any loop, combining voltage provided by voltage sources as well as resistive components, must equal zero.

Thus, the relation for Kirchhoff's loop rule can be expressed as:

[tex]\mathbf{\dfrac{q}{c}- L\dfrac{dI}{dt} = 0}[/tex]

We all know that the current in the nonconstant charge flow can be written as:

[tex]\mathbf{I = \dfrac{dq}{dt}}[/tex]

∴

Replacing the current (I) into Kirchhoff's loop rule, we have:

[tex]\mathbf{ L\dfrac{d}{dt} ( \dfrac{dq}{dt})= -\dfrac{q}{c}}[/tex]

[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -\dfrac{q}{Lc} \ \ ---(1)}[/tex]

From the given question, when the switch is closed

[tex]\mathbf{ \dfrac{d^2q}{dt^2}= -kq\ \ ---(2)}[/tex]

Then, the charges on the capacitor start to b, resulting in the rise of the current in the circuit.

∴

By equating both equations (1) and (2);

[tex]\mathbf{K = \dfrac{1}{LC}}[/tex]

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Answer:

[tex]E=mgh[/tex]

[tex]m=3kg[/tex]

[tex]h=4m[/tex]

[tex]g=9.8m/s^{2}[/tex]

[tex]E= 3*4*9.8=117.6J[/tex]

Explanation:

Only substitute amounts to formula.

Hope this helps ;)

Cheers :D

A. psychodynamic is the personality best explains Sam's behavior

The psychology of mental or emotional forces or processes developing especially in early childhood and their effects on behavior and mental states is called psychodynamic .

Since , any behavior  is not genetical or it never get inherited hence it is not biological . This has been  grown with time and this behavior have been grown in Sam by observing his parents doing the same hence correct option should be A. psychodynamic

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The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:

       a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm

     b) For time zero, the displacement at position x = 0.40 m is: y = 0

      c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:

          v = 9.11 cm / s

      d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:

          v = 9.65 cm / s

The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:

         y = A sin (kx - wt)

Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.

Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship

           [tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]  

Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.

Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore

               [tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]  

              λ = 4L

              λ = 4 0.20

              λ = 0.80 m

The wave number.

              k = [tex]\frac{2\pi }{\lambda }[/tex]  

              k = [tex]\frac{2 \pi }{0.80 }[/tex]  

              k = 2.5π i m⁻¹

In the associated traveling wave, from the graph we can see that the period of the wave is:

             T = 2.8 s

the angular velocity is related to the period.

             [tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]  

             w = 0.714π  rad/s

indicate the maximum displacement that is the amplitude of the wave.

              A = [tex]y_s[/tex]  

             A = 4.3 cm

Let's write the equation of the traveling wave.

              y = 4.3 sin [π (2.5 x - 0.714 t)]

with this expression we can answer the questions.

a) the displacement of the particle for x = 0.30 m

            y = 4.3 sin (π (2.5 0.30 - 0.714 t))

            y = 4.3 sin π( 0.75 - 0.714 t(

Remember that the angles must be in radians.  For time t = 0 the displacement is

              y = 4.3  0.707

              y = 3.04 cm

b) The displacement for x = 0.4m

              y = 4.3 sin (π 2.5 0.4)

              y = 0 cm

c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s

the speed of the wave is

              [tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]  

              v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)

              v = 9.65 cos π(0.75 - 0.714 t)

For time t = 0.90 s the velocity is:

            v = 9.65 cos π(0.75 - 0.714 0.9)

            v = 9.65 0.9436

            v = 9.11 cm / s

d) The velocity for time t = 1.3 s

           v = 9.65 cos π(0.75 - 0.714 1.3)

           v = 9.65 0.9999

           v = 9.65 cm / s

In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:

      a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm

     b) For time zero, the displacement at position x = 0.40 m is: y = 0

      c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:

          v = 9.11 cm / s

      d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:

          v = 9.65 cm / s

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Answer:

Explanation:

F = ma

3(9.81) - 3.8(9.81sin36) = (3 + 3.8)a

a = 1.10566...

a = 1.1 m/s²

the 3.8 kg mass will move up slope and the 3 kg mass will fall at that acceleration.