A uniform metal rod, with a mass of 3.7 kg and a length of 1.2 m, is attached to a wall by a hinge at its base. A horizontal wire bolted to the wall 0.51 m above the base of the rod holds the rod at an angle of 25 degrees above the horizontal. The wire is attached to the top of the rod.

Find the tension in the wire.

a.Find the horizontal component of the force exerted on the rod by the hinge.
b.Find the vertical component of the force exerted on the rod by the hinge.

Final answer:

To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. Using the given equation, we can solve for the specific heat capacity of the metal, which is found to be 3.75 J/g°C.

Explanation:

To find the final temperature of the system, we can use the principle of conservation of energy. The heat lost by the hot metal is equal to the heat gained by the water. We can use the equation:m1c1(T1 - T) = m2c2(T - T2)

where:

m1 is the mass of the metal (34.5 g)

c1 is the specific heat capacity of the metal

T1 is the initial temperature of the metal (75°C)

T is the final temperature of the system (39°C)

m2 is the mass of the water (64.0 g)

c2 is the specific heat capacity of water (4.18 J/g°C)

T2 is the initial temperature of the water (25°C)

Plugging in the given values, we can solve for c1: 34.5g × c1 × (75°C - 39°C) = 64.0g × 4.18 J/g°C × (39°C - 25°C)

Simplifying the equation gives:34.5g × c1 × 36°C = 64.0g × 4.18 J/g°C × 14°C

Dividing both sides by 34.5g × 36°C gives:c1 = (64.0g × 4.18 J/g°C × 14°C) / (34.5g × 36°C)

Calculating the value gives:c1 = 3.75 J/g°C

The final temperature for both the aluminum and water at thermal equilibrium is 17.3 °C. This is calculated using the principle of conservation of energy. The heat lost by aluminum equals the heat gained by water.

To find the final temperature of a 32.5 g cube of aluminum and 105.3 g of water at thermal equilibrium, we use the principle of conservation of energy. The heat lost by the aluminum will be equal to the heat gained by the water.

Let us simplify: (32.5 g) x (0.900 J/g°C) x (T_final - 45.8 °C) = - (105.3 g) x (4.186 J/g°C) x (T_final - 15.4 °C)

Therefore, the final temperature of both substances at thermal equilibrium is **17.3 °C**.

Answer:

(a.) 4z

(b.) 4w

Explanation:

From the equation y=4zcos(8πwt), where z and w are positive constants.

Comparing this equation to the equation of a wave y = Acos(Wt), where A is the amplitude (largest distance from equilibrium) and W is the angular frequency (W=2πf)

(a.) Comparing our wave equation with the given equation, we see that A = 4z in this case (furthest distance of the mass from equilibrium)

(b.) Comparing similarly we can see from our given equation that angular frequency W =8πw we also know that W = 2πf from our wave equation, therefore 2πf = 8πw

Solving for f we have f = 8πw÷2π

f = 4w (Proves our second answer because the frequency is the number of oscillations completed per second)

Answer:

68 db

Explanation:

Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.

Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,

β(db)=10×㏒(2)

=3 db

So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db

The intensity level of two identical dogs barking together would be approximately 68 dB, as an increase in

sound intensity leads to an increase in the decibel level.

The intensity level of a sound is determined by its amplitude and is measured in decibels (dB). The relationship between intensity and decibel level is logarithmic, which means that an increase of 10 dB corresponds to a tenfold increase in intensity. Hence, if two identical dogs are barking very close to each other their combined sound would be twice as intense as one dog barking alone. However, this does not mean that the decibel level would double. In fact, when the intensity of a sound is doubled, the increase in decibel level is about 3 dB. Thus, if a single dog barking has an intensity level of 65 dB, two dogs barking together would have an estimated intensity level of 68 dB.

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Answer:

Cold front

Explanation:

The occurrence of cold front takes place when colder air at the surface pushes up the warmer into the atmosphere.

The movement of these cold fronts takes place at a faster pace as compared to the other fronts and are these are linked with the most violent types of weather such as severe thunderstorms are produced.

Answer:

The  maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]

Explanation:

We are given that

Frequency of light source=[tex]5.32\times 10^{23}[/tex] Hz

Work function of arsenic=[tex]7.67\times 10[/tex] eV

We have to find the maximum kinetic energy of ejected electron.

We know that the maximum kinetic energy of ejected electron

[tex]K.E=h\nu-w_o[/tex]

Where h=Plank's constant=[tex]6.63\times 10^{-34} J.s[/tex]

[tex]\nu[/tex] =Frequency of light source

[tex]w_o[/tex]=Work function

Substitute the values in the given formula

Then, the maximum kinetic energy of ejected electron

[tex]K.E=6.63\times 10^{-34}\times 5.32\times 10^{23}-7.67\times 10\times 1.6\times 10^{-19}[/tex]

Because 1 e V=[tex]1.6\times 10^{-19} J[/tex]

[tex]K.E=35.2716\times 10^{-11}-12.272\times 10^{-18}[/tex]

[tex]K.E=3.53\times 10^{-10}[/tex] J

Hence, the maximum kinetic energy of the ejected electron=[tex]3.53\times 10^{-10} J[/tex]

The new level of sound β2 = β1 + 10 log10(f).

Suppose that a sound has initial intensity level β1 measured in decibels.

This sound now increases in intensity level by a factor f. What is the new level of sound β2?

Formula to calculate the new sound intensity β2 = β1 + 10 log10(f)

The new level of sound β2 can be calculated using the formula β2 = β1 + 10 log10(f)

Here,β1 is the initial sound intensity levelβ2 is the new sound intensity level if is the factor by which the sound intensity level is increased by.

Sound intensity, also known as acoustic intensity, is defined as the power carried by sound waves per unit area in a direction perpendicular to that area.

The SI unit of intensity, which includes sound intensity, is the watt per square meter (W/m2)

Hence, The new level of sound β2 = β1 + 10 log10(f).

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The new level of the sound, β2, can be found by using the property of logarithms. The formula to calculate the new level is β2 = β1 + 10 log10(f). If one sound is twice as intense as another, it has a sound level about 3 dB higher.

The new level of the sound, β2, can be found by using the property of logarithms. When the intensity of a sound increases by a factor of f, the sound level increases by 10 times the logarithm (base 10) of f. So, in this case, if the initial sound has intensity β1, the new level of sound β2 can be obtained by:

β2 = β1 + 10 log10(f)

For example, if the initial sound has intensity β1 = 60 dB and it increases by a factor of f = 2, then:

β2 = 60 + 10 log10(2) = 60 + 3 = 63 dB

Answer:

0 N

Explanation:

The elevator is under free fall. So, when a body is under free fall, the acceleration is only due to gravity. So, the acceleration of the elevator or the woman inside it, is acceleration due to gravity in the downward direction.

The spring scale gives the value of the normal force acting on the woman and doesn't give the exact weight of the woman. Under normal conditions, when the spring scale is at rest, then the upward normal force equals the weight and hence weight of a body is equal to the normal force acting on the body.

But, here, the body is not at rest. Weight\tex](mg)[/tex] acts in the downward direction and normal force[tex](N)[/tex] acts in the upward direction. The woman is moving down with acceleration equal to acceleration due to gravity[tex](g)[/tex]

So, we apply Newton's second law on the woman.

[tex]\textrm{Net force} = \textrm{mass}\times \textrm{acceleration}\\F_{net}=ma[/tex]

Net force is equal to the difference of the downward force and upward force.

[tex]F_{net}=mg-N[/tex]

Now, [tex]F_{net}=ma[/tex]

[tex]mg-N=mg\\N=mg-mg=0[/tex]

Therefore, the reading on the spring scale is 0 N.

Answer:

(a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

Explanation:

Given that,

The deceleration is

[tex]a=(-2v^3)\ m/s^2[/tex]

We need to calculate the velocity at t = 25 s

The acceleration is the first derivative of velocity of the particle.

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]\dfrac{dv}{dt}=-2v^3[/tex]

[tex]\dfrac{dv}{-v^3}=2dt[/tex]

On integrating

[tex]int{\dfrac{dv}{-v^3}}=\int{2dt}[/tex]

[tex]\dfrac{1}{2v^2}=2t+C[/tex]

[tex]v^2=\dfrac{1}{4t+2C}[/tex]....(I)

At t = 0, v = 10 m/s

[tex]10^2=\dfrac{1}{4\times0+2C}[/tex]

[tex]C=\dfrac{1}{200}[/tex]

Put the value of C in equation (I)

[tex]v^2=\dfrac{1}{4\times25+2\times\dfrac{1}{200}}[/tex]

[tex]v=\sqrt{\dfrac{1}{4\times25+2\times\dfrac{1}{200}}}[/tex]

[tex]v=0.099\ m/s[/tex]

The velocity is 0.099 m/s.

(b). We need to calculate the position at t = 25 sec

The velocity is the first derivative of position of the particle.

[tex]\dfrac{ds}{dt}=v[/tex]

On integrating

[tex]\int{ds}=\int(\sqrt{\dfrac{200}{800t+1}})dt[/tex]

[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800t+1}}{800}+C'[/tex]

At t = 0, s = 15 m

[tex]15=\dfrac{200}{800}+C'[/tex]

[tex]C'=15-\dfrac{200}{800}[/tex]

[tex]C'=14.75[/tex]

Put the value in the equation

[tex]s=\dfrac{\sqrt{200}\times2\sqrt{800\times25+1}}{800}+14.75[/tex]

[tex]s=19.75\ m[/tex]

The position is 19.75 m.

Hence, (a). The velocity is 0.099 m/s.

(b). The position is 19.75 m.

The velocity of an object depends on its position.

Part A: The velocity of the moving particle at t = 25 s is 0.099 m/s.

Part B: The position of the particle at t = 25 s is 15.475 m.

Velocity is defined as the rate at which the position of an object changes.

Given that the deacceleration of the particle is [tex]a = (-2v^3)[/tex]. The velocity of the particle v = 10 m/s and a position s = 15 m when t = 0.

The acceleration of the particle is given below.

[tex]a = \dfrac {dv}{dt}[/tex]

[tex]-2v^3 = \dfrac {dv}{dt}[/tex]

[tex]- 2 dt = \dfrac {1}{v^3}dv[/tex]

Integrating the above equation, we get,

[tex]\int (-2 dt) = \int \dfrac {1}{v^3} dv[/tex]

[tex]2t + C = \dfrac {1}{2v^2}[/tex]

[tex]v^2 = \dfrac {1}{4t + 2C}[/tex]

Putting t=0 s, v = 10 m/s

[tex]100 = \dfrac {1}{4\times 0 + 2C}[/tex]

[tex]100 = \dfrac {1}{2C}[/tex]

[tex]C = 5 \times 10^{-3}[/tex]

Part A: Velocity

The velocity of the particle at t = 25 s is given as below.

[tex]v^2 = \dfrac { 1}{(4\times 25) + (2\times 5\times 10^{-3})}[/tex]

[tex]v^2 = \dfrac{1}{100.01}[/tex]

[tex]v = 0.099 \;\rm m/s[/tex]

Hence the velocity of the moving particle at t = 25 s is 0.099 m/s.

Part B:  Position

The velocity of the particle is given as,

[tex]v = \dfrac {ds}{dt}[/tex]

[tex]v dt = ds[/tex]

Integrating the above equation, we get,

[tex]\int vdt = \int ds[/tex]

[tex]\int \sqrt{\dfrac {1}{4t + (\dfrac {1}{100})}} dt = s[/tex]

[tex]\int\sqrt{ {\dfrac {100}{400 t + 1}}}dt = s[/tex]

[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 t + 1}}{400} + C'[/tex]

Put t = 0 s, and s = 10 m,

[tex]15=\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 0 + 1}}{400} + C'[/tex]

[tex]15 = \dfrac {\sqrt{100}\times 2}{400} + C'[/tex]

[tex]C' = 14.95[/tex]

Now the position at t = 25 s is,

[tex]s =\dfrac {\sqrt{100} \times 2 \sqrt{400 \times 25 + 1}}{400} + 14.95[/tex]

[tex]s =15.475 \;\rm m[/tex]

Hence the position of the particle at t = 25 s is 15.475 m.

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Answer: The answer is 225 joules

Answer:

a)

2.5 rads⁻²

b)

1.5 rads⁻²

Explanation:

a)

[tex]R[/tex] = distance of the center of mass from the wire = 5 cm = 0.05 m

[tex]F_{g}[/tex] = Force of gravity on walker

[tex]M[/tex] = mass of  walker = 71 kg

Force of gravity on walker is given as

[tex]F_{g} = Mg\\F_{g} = (71)(9.8)\\F_{g} = 695.8 N[/tex]

[tex]I[/tex] = Rotational inertia of walker = 14 kgm²

[tex]\alpha[/tex] = Angular acceleration about the wire

Torque equation is given as

[tex]F_{g} R = I \alpha \\(695.8) (0.05) = (14) \alpha \\\alpha = \frac{(695.8) (0.05)}{14} \\\alpha= 2.5 rads^{-2}[/tex]

b)

[tex]F_{p}[/tex] = Force of gravity on pole

[tex]m[/tex] = mass of  walker = 14 kg

Force of gravity on pole is given as

[tex]F_{p} = mg\\F_{g} = (14)(9.8)\\F_{g} = 137.2 N[/tex]

[tex]r[/tex] = distance of the center of mass of pole from the wire = 10 cm = 0.10 m

Torque equation is given as

[tex]F_{g} R - F_{p} r = I \alpha \\(695.8) (0.05) - (137.2)(0.10) = (14) \alpha \\\alpha= \frac{21.07}{14} \\\alpha=1.5 rads^{-2}[/tex]

Answer:

Vacuum lamp experiment

Explanation:

Edison began his work on electricity the year 1878, where he made use of platinum wire to make filaments for his lamp as the melting point of Platinum is.

However, later he concluded that Platinum pores absorbed air while heating thus reducing its melting point.

So he came up with the idea of vacuum lamp and developed quite a good vacuum lamp with the help of carbon.

Being highly resistive in nature, carbon would end up burning at quite a rapid rate.

Along with his staff, in the year 1879, Edison conducted his first fruitful experiment by using vacuum lamp with carbon filament.

Answer:

All these is caused by the repulsion force.

Explanation:

The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.

As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.

Answer:

Oceanic crust and continental crust

Explanation:

A subduction zone is normally between oceanic crust which is made of basalt and continental crust which is made of granite. Oceanic crust is denser than continental crust. So when oceanic crust collides with continental crusts, it subsducts underneath the continental crust since it is denser.

Final answer:

In subduction zones, the denser oceanic crust is pushed beneath another plate, which may be either oceanic or continental. This process results in the formation of ocean trenches and volcanic activity. Subduction zones play a crucial role in the cycle of plate tectonics.

Explanation:

The two kinds of crust involved in a subduction zone are the oceanic crust and the continental crust. In a subduction zone, one tectonic plate, usually the denser oceanic plate, is forced beneath another plate, which can be either oceanic or continental. Often it is the older and colder oceanic plate that subducts beneath the younger and warmer one. The subduction process is marked by features such as ocean trenches, volcanic island arcs, and is associated with powerful earthquakes and volcanic eruptions.

Subduction zones are integral to the Earth's plate tectonics cycle, balancing the creation of new crust at mid-ocean ridges by destroying the old crust as it descends into the mantle. Such areas of crust destruction are most commonly found around the edges of the Pacific Ocean. The descending plate undergoes changes due to high pressures and temperatures, leading to the release of water and the melting of the crust, forming magma that can result in volcanic activity.

Answer:

m = 4.65 kg

Explanation:

As we know that the mass of the water that evaporated out is given as

[tex]m = 0.0250 kg[/tex]

so the energy released in form of vapor is given as

[tex]Q = mL[/tex]

[tex]Q = (0.0250)(2.25 \times 10^6)[/tex]

[tex]Q = 56511 J[/tex]

now the heat required by remaining water to bring it from 15 degree to 100 degree

[tex]Q_2 = ms\Delta T[/tex]

[tex]Q_2 = (4 - 0.025)(4186)(100 - 15)[/tex]

[tex]Q_2 = 1.41\times 10^6J[/tex]

total heat required for above conversion

[tex]Q = 56511 + 1.41 \times 10^6 = 1.47 \times 10^6 J[/tex]

now by heat energy balance

heat given by granite = heat absorbed by water

[tex]m(790)(500 - 100) = 1.47 \times 10^6[/tex]

[tex]m = 4.65 kg[/tex]

Answer:

Minimum required energy

The correct orientation

Explanation:

A collision is described as successful or effective if the particles:

The success of a collision between reactant molecules in causing a reaction depends on the molecules colliding with enough energy to overcome the activation energy, proper orientation during the collision, and a higher collision frequency influenced by reactant concentration.

To answer your question about the factors that determine whether a collision between two reactant molecules results in a reaction, we need to consider several key factors. First, reactant molecules must collide with sufficient energy; this energy must be equal to or higher than the activation energy of the reaction. Second, the molecules must have proper orientation during the collision to allow the necessary rearrangement of electrons and atoms. Lastly, the collision frequency is also essential, which is influenced by the concentration of the reactants; higher concentration generally leads to more collisions and an increased likelihood of reaction.

The deformation of molecules during a collision can also be a contributing factor, as more violent collisions tend to lead to greater deformation and enhance the probability of reaction. Additionally, external factors such as temperature and the presence of catalysts can increase the kinetic energy of the molecules or reduce the activation energy, both of which can lead to a higher rate of successful collisions and reactions.

Answer:

Boyle's Law

[tex]\therefore P_1.V_1=P_2.V_2[/tex]

Explanation:

Given that:

initially:

pressure of gas, [tex]= P_1[/tex]

volume of gas, [tex]= V_1[/tex]

finally:

pressure of gas, [tex]= P_2[/tex]

volume of gas, [tex]= V_2[/tex]

To solve for final volume [tex]V_2[/tex]

According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.

According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.

But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.

Here nothing is said about the temperature, so we consider the Boyle's Law which states that at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.

Mathematically:

[tex]P_1\propto \frac{1}{V_1}[/tex]

[tex]\Rightarrow P_1.V_1=k\ \rm(constant)[/tex]

[tex]\therefore P_1.V_1=P_2.V_2[/tex]

Final answer:

To calculate the new volume V2 in Boyle's Law, use the formula V2 = P1V1/P2, ensuring that P1 and P2 are in the same pressure units and that V1 and V2 are in the same volume units.

Explanation:

To determine the new volume V2 after a change in pressure of an ideal gas, one can utilize Boyle's Law, which states that at a constant temperature, the pressure of a gas is inversely proportional to its volume. The formula representing Boyle's Law is P1V1 = P2V2. In this equation, P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. To solve for V2, you simply rearrange the equation to V2 = P1V1/P2.

Remember that in order for Boyle's Law to be applied correctly, both volumes should be in the same unit. If you solve for V2 and find it to be greater than V1, it indicates that the pressure has decreased, consistent with the inverse proportionality.

Answer:

[tex]v_f=8.17\frac{m}{s}[/tex]

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

[tex]W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m[/tex]

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

[tex]W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}[/tex]

The box is initially at rest, so [tex]v_i=0[/tex]. Solving for [tex]v_f[/tex]:

[tex]v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}[/tex]

Answer:

574.25 °

Explanation:

To know this, we need to get the expressions to calculate °F and °K.

In the case of Kelvin:

°K = °C + 273

in fahrenheit:

°F = 9/5(°C + 32)

Now, in order to know the temperature which both Kelvin and Fahrenheit are the same, we need to use both above equations, and solve for °C.

°C = °K - 273 (1)

°C = 5/9(°F - 32) (2)

Using 1 and 2 into a same expression:

°K - 273 = 5/9(°F - 32)

With this, we need to know now the moment which K = F, so, all we need to do is replace the F for the K in the above expression. Doing this, we have:

°K - 273 = 5/9(°K - 32)

°K = 0.555(°K - 32) + 273

°K = 5/9K - 17.78 + 273

°K - 5/9K = 255.22

4/9°K = 255.22

°K = 255.22 * 9 / 4

°K = 574.25 °

And this is the temperature in which both Kelvin and Fahrenheit are the same.

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure,  41.0  PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be  14.6959  P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂    ⇒    V₂ = P₁*V₁ / P₂

where  

1  and  2  are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

P₁ = 14.6959  P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒  V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A  = (13.2985 in³) / (3.00 in²)

⇒  h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

Answer:

The magnitude of the resultant is 30.4 N.

The resultant angle direction is 25.3°.

Explanation:

To find the resultant of the magnitude and direction for given forces “P” and “Q” are 20 N and 15 N respectively, the angle (θ) between them is 60°.

We know that from triangle law of forces,  

[tex]R=\sqrt{P^{2}+2 P Q \cos \theta+Q^{2}}[/tex]

Substitute the given values in the above formula,

[tex]R=\sqrt{20^{2}+2 (20)(15) Q \cos 60+15^{2}}[/tex]

[tex]R=\sqrt{400+600(0.5) + 225}[/tex]

[tex]R=\sqrt{400+300 + 225}[/tex]

[tex]R=\sqrt{925}[/tex]

R = 30.4 N

The magnitude of the resultant is 30.4 N.

To find the direction of the resultant we know that [tex]\text {Resultant angle}=\tan ^{-1} \frac{Q \sin \theta}{P+Q \cos \theta}[/tex]

Substitute the given values in the above formula,

[tex]\text {Resultant angle}=\tan ^{-1} \frac{15 \sin 60}{20+15 \cos 60}[/tex]

[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{20+7.5}[/tex]

[tex]\text {Resultant angle}=\tan ^{-1} \frac{12.99}{27.5}[/tex]

[tex]\text { Resultant angle }=\tan ^{-1} 0.472[/tex]

Resultant angle=25.3°

The resultant angle direction is 25.3°.

Answer:

v₂ = 3.44 m/s

Explanation:

given,                                                

Dan is gliding on skateboard at = 4 m/s

speed of skateboard = 8 m/s              

mass of Dan = 50 Kg                                

mass of skateboard = 7 Kg                            

Speed of Dan when his feet hit the ground = ?

Using conservation of momentum            

       PE(initial) = PE(final)                              

       (M+m)V = m v₁ + M v₂                  

       (50+7)x 4 = 7 x 8 + 50 x v₂        

       50 x v₂ =172                              

            v₂ = 3.44 m/s                  

speed of Dan when his feet hit the ground = v₂ = 3.44 m/s

The block of ice would have lost half of its gravitational potential energy after sliding halfway down the frictionless inclined plane. Therefore, its gravitational potential energy will be 2.5 Joules.

The question is about the concept of energy conservation in physics, specifically the gravitational potential energy. The gravitational potential energy (U) of an object is given by the formula U = mgh, where 'm' is the mass of the object, 'g' is the acceleration due to gravity, and 'h' is the height from the reference point.

In this case, we know that the block of ice has a gravitational potential energy of 5.0 Joules when it's at the top of the incline. When the block of ice has slid halfway down the incline, it will have lost half of its height. Therefore, it would have lost half of its gravitational potential energy. Hence, after sliding halfway down the incline, the block's gravitational potential energy will be 2.5 Joules.

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Answer:

Friction force is 0.1375 N

Solution:

As per the question:

Radius of the metal disc, R = 4.0 cm = 0.04 m

Magnetic field, B = 1.25 T

Current, I = 5.5 A

Force on a current carrying conductor in a magnetic field is given by considering it on a differential element:

[tex]dF = IB\times dR[/tex]

Integrating the above eqn:

[tex]\int dF = IB\int_{0}^{R}dr[/tex]

[tex]\tau = IB\times \frac{R^{2}}{2} = \frac{1}{2}IBR^{2}[/tex]          (1)

Now the torque is given by:

[tex]\tau = F\times R[/tex]                        (2)

From eqn (1) and (2):

[tex]F\times R = \frac{1}{2}IBR^{2}[/tex]

Thus the Frictional force is given by:

[tex]F = \frac{1}{2}\times 5.5\times 1.25\times 0.04 = 0.1375\ N[/tex]

Answer:

Bulb will remain at constant brightness for 6 hour

Explanation:

We have given resistance of the bulb R = 4 ohm

Potential difference of the battery V = 2 volt

So current [tex]i=\frac{V}{R}=\frac{2}{4}=0.5A[/tex]

Capacity of the battery is given as 3 amp hour

We have to find the time by which bulb will remain at constant brightness

So time will be [tex]=\frac{3}{0.5}=6hour[/tex]

So bulb will remain at constant brightness for 6 hour

Answer:

Part a)

[tex]W = 2716.5 J[/tex]

Part B)

[tex]W = -2716.5 J[/tex]

Explanation:

Part A)

While escalator is moving up work done to move the person upwards is given as

[tex]W = mgh[/tex]

here we know that

m = 63 kg

h = 4.4 m

now we have

[tex]W = 63 \times 9.8 \times 4.4[/tex]

[tex]W = 2716.5 J[/tex]

Part B)

Work done while we move from up to down

So we have

[tex]W = - mgh[/tex]

so we have

[tex]W = -2716.5 J[/tex]

Answer:

it will need 0.06 hectare of land

So option (c) is correct

Explanation:

We have given that 12000000 kilo calories of chicken can be produced per hectare

It is given that someone with 2000 kilo calories / day for a year

Total number of days in a year = 365

So total kilo calories of chicken = 365×2000 = 730000 kilo calories

As 12000000 kilo calories of chicken can be produced per hectare

So 730000 kilo calories need a land of [tex]=\frac{730000}{12000000}=0.06hectare[/tex]

So it will need 0.06 hectare of land

Final answer:

To provide someone with 2,000 kilocalories/day for a year, 0.06 hectares of land used for chicken production is needed, as the closest correct answer.

Explanation:

The question asks how much land is needed to provide someone with 2,000 kilocalories/day for a year if 12,000,000 kilocalories of chicken can be produced per hectare. First, calculate the total kilocalories needed for a year by multiplying 2,000 kcal/day by the number of days in a year (365), which equals 730,000 kcal/year. Then, find how much land is needed by dividing the total annual kilocalorie requirement (730,000 kcal) by the amount of kilocalories produced per hectare (12,000,000 kcal).

This calculation gives us:

730,000 kcal / 12,000,000 kcal/ha = 0.0608333 ha

Therefore, the closest answer is c. 0.06 ha.

Answer:

c. the hole in the plate and the opening in the ring get larger.

Explanation:

The hole in the plate and the ring both of them get enlarged.

The thermal expansion is very much similar to the optical expansion of the shadow of an object.

The change in length of an object is given by the relation:

[tex]\Delta l=l.\alpha.\Delta T[/tex]

where:

l = original length of the object

[tex]\alpha=[/tex] coefficient of linear expansion

ΔT = change in temperature

When heated from 20 degrees C to 100 degrees C, the diameter of the hole in a steel plate and the inner diameter of a steel ring both increase due to thermal expansion. So the correct option is C.

The correct answer to the question about what happens to the diameter of a hole in a steel plate and the inner diameter of a steel ring when both are heated is that both will get larger. This is due to the thermal expansion of materials, which states that objects expand in all dimensions as their temperature increases. When the steel plate and ring are heated from 20 degrees C to 100 degrees C, the particles within the steel vibrate more and push each other farther apart, causing both the solid material and the empty space within, like the hole and the opening of the ring, to increase in size.

Complete Question:
Two uniform spheres,each of mass 0.260kg are fixed at points A and B

A.)Find the magnitude of the initial acceleration of a uniform sphere with mass 0.010 kg if released from rest at point P and acted on only by forces of gravitational attraction of the spheres at A and B.

B.) Find the direction of the initial acceleration of a uniform sphere with mass 0.010 kg.

Answer:

343m

Explanation:

From the question Sally hits one blow every second

However, the sound is delayed by that same amount of time (1 sec)

If the speed of sound is 343m/s

Speed = Distance/Time

Speed = 343m/s

Time = 1s

Distance = Speed x Time

= 343 x 1 = 343m

The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation.

The magnitude of the acceleration (ae) of the Earth due to the gravitational pull of the Moon can be calculated using Newton's law of universal gravitation. The formula is F = G × (m1 × m2) / r^2, where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers. In this case, the Earth is the object experiencing the acceleration and the Moon is the object exerting the gravitational force.

Given the distance between the center of the Earth and the center of the Moon as r = 4700 km (or 4.7 x 10^6 m) and the gravitational constant F = G × (m1 × m2) / r^2[tex]F = G × (m1 × m2) / r^2[/tex], the formula becomes:

[tex]F = G × (m1 × m2) / r^2[/tex]

Since we know the mass of the Earth, we can calculate the acceleration by rearranging the formula to solve for ae:

ae = F / m1

Substituting the values, we get:

[tex]ae = (G × (m1 × m2) / r^2) / m1[/tex]

Simplifying the equation, we are left with:

[tex]ae = G × m2 / r^2[/tex]

Now, we can plug in the known values. The mass of the Moon (m2) is approximately 7.35 x 10^22 kg:

[tex]ae = (6.67 x 10^-11 N · m^2/kg^2) × (7.35 x 10^22 kg) / (4.7 x 10^6 m)^2[/tex]

Calculating this expression will give us the value for the magnitude of the acceleration of the Earth due to the gravitational pull of the Moon.