An object with a mass of 7.8 kg is pulled on a horizontal surface by a horizontal pull of 50 N to the right. The friction force on this object is 30 N to the left. What is the acceleration of the object? 0.16 m/s^2
10 m/s^2
2.6 m/s^2
6.4 m/s^2

Answer:

[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s

[tex]E = 3.68 * 10^{-21}[/tex] J

Explanation:

given data

uncertainty in proton position [tex]\Delta x[/tex] = 0.015 nm

according to Heisenberg's principle of uncertainty

[tex]\Delta x \Delta P_{x}= \frac{\frac{h}{2\pi }}{2}[/tex]

Where h is plank constant = [tex]6.6260 * 10^{-34}[/tex] j-s

[tex]\Delta P_{x}= \frac{\frac{h}{2\pi }}{2\Delta x}[/tex]

[tex]\Delta P_{x}= \frac{\frac{6.6260 * 10^{-34}}{2\pi }}{2*0.015*10^{-9}}[/tex]

[tex]\Delta P_{x}=3.51 *10^{-24}[/tex] kg m/s

b) kinetic energy  of proton whose momentum

[tex]P =\Delta p[/tex]

[tex]E =\frac{\Delta p^{2}}{2m}[/tex]

where m is mass is proton

[tex]E =\frac{(3.51*10^{-24})^{2}}{2*1.67*10^{-27}}[/tex]

[tex]E = 3.68 * 10^{-21}[/tex] J

Answer:

443.3 N

567.4 N

Explanation:

Consider the triangle ABC

AC = hypotenuse = magnitude of force vector = F = 720 N

AB = adjacent = Component of force along east-west line = [tex]F_{x}[/tex]

BC = Opposite = Component of force along north-south line = [tex]F_{y}[/tex]

θ = Angle = 38 deg

In triangle ABC

[tex]Sin38 = \frac{BC}{AC}[/tex]

[tex]Sin38 = \frac{F_{y}}{F}[/tex]

[tex]0.616 = \frac{F_{y}}{720}[/tex]

[tex]F_{y}[/tex] = 443.3 N

Also, In triangle ABC

[tex]Cos38 = \frac{AB}{AC}[/tex]

[tex]Cos38 = \frac{F_{x}}{F}[/tex]

[tex]0.788 = \frac{F_{x}}{720}[/tex]

[tex]F_{x}[/tex] = 567.4 N

Answer:

Linear momentum, p = 50 kg-m/s

Explanation:

It is given that,

Mass of an object, m = 5 kg

Velocity, v = 10 m/s

We need to find the linear momentum of that object. It is given by :

[tex]p=m\times v[/tex]

[tex]p=5\ kg\times 10\ m/s[/tex]

p = 50 kg-m/s

So, the linear momentum of that object is 50 kg-m/s. Hence, this is the required solution.

The linear momentum of an object with a mass of 5 kg and velocity of 10 m/s is calculated using the formula P = mv, which results in 50 kg*m/s.

The student has asked to calculate the linear momentum of an object that has a mass of 5 kg and a velocity of 10 m/s. To find the linear momentum, we use the formula:

P = mv

Where P is the momentum, m is the mass, and v is the velocity. By plugging in the numbers:

P = 5 kg \\* 10 m/s

P = 50 kg\\*m/s

Therefore, the linear momentum of the object is 50 kg\\*m/s, which corresponds to option (C).

Answer:

4557.88 m/s

Explanation:

M = mass of earth = 5.98 x 10²⁴ kg

R = radius of earth = 6.4 x 10⁶ m

r = orbital radius of satellite = 3R = 3 x 6.4 x 10⁶ = 19.2 x 10⁶ m

[tex]v[/tex] = speed of the satellite

speed of the satellite is given as

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

[tex]v=\sqrt{\frac{(6.67\times 10^{-11})(5.98\times 10^{24})}{19.2\times 10^{6}}}[/tex]

[tex]v[/tex] = 4557.88 m/s

Answer:

19.06 Volt

Explanation:

r =0.25 m, l = 11 m, B = 0.17 T, f = 13 Hz

length for one turn = 2 x 3.14 x r = 2 x 3.14 x 0.25 = 1.57 m

Total number of turns in 11 m

N = 11 / 1.57 = 7

Peak emf, e0 = N x B x A x 2 x 3.14 x f

e0 = 7 x 0.17 x 3.14 x 0.25 x 0.25 x 2 x 3.14 x 13 = 19.06 Volt

Answer:

The length of the simple pendulum is 2.4 meters.

Explanation:

Time period of simple pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]

L is the length of pendulum

The time period of the rope is given by :

[tex]T=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]

L' is the length of the rod, L' = 3.6 m

It is given that, the rod have the same period as a simple pendulum and we need to find the length of simple pendulum i.e.

[tex]2\pi\sqrt{\dfrac{L}{g}}=2\pi\sqrt{\dfrac{2L'}{3g}}[/tex]

On solving the above equation as :

[tex]\dfrac{L}{g}=\dfrac{2L'}{3g}[/tex]

L = 2.4 m

So, the length of the thin rod that is hung vertically from one end and set into small amplitude oscillation 2.4 meters. Hence, this is the required solution.

Answer:

work done by applied force = 1120J

work done by frictional force = 943.25J

work done by the gravitational force = work done by the normal force = 0J

Explanation:

Given:

Mass of the box = 55kg

Displacement of the box = 7.0m across the floor

Applied force = 160N

Coefficient of kinetic friction ([tex]\mu_k[/tex]) = 0.25

Now the work done (W) is given as:

W = F×d

Where,

F = Applied force on the body

d = Displacement of the body in the direction of the applied force

Forces acting on the box are:

1) Applied force

2) Frictional Force

3) Gravitational force

4) Normal force due to the weight

Now the work done by the respective forces:

1) By the applied force

W = 160 kg × 7 m = 1120 J

2) By the frictional force

[tex]W_f =-\mu mg\times displacement[/tex]

Where

g = acceleration due to the gravity

and the negative sign here depicts that the frictional force is acting the against the applied force and the direction of displacement.

thus,

[tex]W_f =-0.25\times 55\times 9.8\times 7[/tex]

or

[tex]W_f =-943.25J[/tex]

The work done by the normal force and the gravitational force will be zero as there is no displacement in the direction of the application of both the forces

The work done by both the gravitational force and normal force is equal to zero (0).

Given the following data:

In this scenario, there are four forces acting on the box and these include:

For the applied force:

Mathematically, the work done due to an applied force is given by this formula:

[tex]Work\;done = force \times distance\\\\Work\;done = 160 \times 7[/tex]

Work done = 1,120 Nm.

For the frictional force:

Mathematically, the work done due to a frictional force is given by this formula:

[tex]Work\;done = -\mu mg \times distance\\\\Work\;done = -0.25 \times 55 \times 9.8 \times 7[/tex]

Work done = -943.25.

However, the work done by both the gravitational force and normal force is equal to zero (0) because there isn't any displacement or distance covered in the direction of the application of these forces.

Read more on work done here: brainly.com/question/22599382

Answer:

Correct option is 'd' Surroundings

Explanation:

The space surrounding a thermodynamic system is known as Surroundings

Explanation:

It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them

[tex]F=k\dfrac{q_1q_2}{1^2}[/tex]...............(1)

We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.

[tex]F'=k\dfrac{q_1q_2}{0.5^2}[/tex]...............(2)

On dividing equation (1) and (2) we get :

[tex]\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}[/tex]

[tex]\dfrac{F}{F'}=0.25[/tex]

F' = 4 F

So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.

Final answer:

The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. The force is inversely proportional to the square of the distance, so if the distance is increased, the force decreases. In this case, if the electron moves 0.5 m away from the proton, the new force can be calculated using Coulomb's Law.

Explanation:

The electrostatic force between two charges of 1 C each, separated by a distance of 0.5 m can be calculated using Coulomb's Law. Coulomb's Law states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

So, if the distance between the charges is increased to 1 m, the force between them will decrease. The force is inversely proportional to the square of the distance, which means that if the distance is doubled, the force will decrease by a factor of four.

In this case, if the electron moves 0.5 m away from the proton, the new distance becomes 1.5 m. Using Coulomb's Law, we can calculate the new force:

F = k * (q1 * q2) / r^2

Where F is the force, k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), q1 and q2 are the charges (in this case, both 1 C), and r is the distance (1.5 m). Plugging in these values, we get:

F = (9 x 10^9 Nm^2/C^2) * (1 C * 1 C) / (1.5 m)^2

F ≈ 4 x 10^9 N

Answer:

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

Explanation:

Joule -Thompson effect

 Throttling phenomenon is called Joule -Thompson effect.We know that throttling is a process in which pressure energy will convert in to thermal energy.

Generally in throttling exit pressure is low as compare to inlet pressure but exit temperature maybe more or less or maybe remains constant depending upon flow or fluid flow through passes.

Now lets take Steady flow process  

Let

 [tex]P_1,T_1 [/tex] Pressure and temperature at inlet and

 [tex]P_2,T_2 [/tex] Pressure and temperature at exit

We know that Joule -Thompson coefficient given as

[tex]\mu _j=\left(\frac{\partial T}{\partial p}\right)_h[/tex]

Now from T-ds equation

dh=Tds=vdp

So

[tex]Tds=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p\right]dp[/tex]

⇒[tex]dh=C_pdt-\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

So Joule -Thompson coefficient

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

This is Joule -Thompson coefficient for all gas (real or ideal gas)

We know that for Ideal gas Pv=mRT

[tex]\dfrac{\partial v}{\partial T}=\dfrac{v}{T}[/tex]

So by putting the values in

[tex]\mu _j=\dfrac{1}{C_p}\left [T\left(\frac{\partial v}{\partial T}\right)_p-v\right]dp[/tex]

[tex]\mu _j=0[/tex] For ideal gas.

Answer:

[tex]i_2 = 7.6 A[/tex]

Explanation:

As we know that the force per unit length of two parallel current carrying wires is given as

[tex]F = \frac{\mu_o i_1 i_2}{2\pi d}[/tex]

here we know that

[tex]F = 3.6 \times 10^{-5} N/m[/tex]

[tex]i_1 = 0.52 A[/tex]

d = 2.2 cm

now from above equation we have

[tex]3.6 \times 10^{-5} = \frac{4\pi \times 10^{-7} (0.52)(i_2)}{2\pi (0.022)}[/tex]

[tex]3.6 \times 10^{-5} = 4.73 \times 10^{-6} i_2[/tex]

[tex]i_2 = 7.6 A[/tex]

Answer:

A. Both spheres land at the same time.

Explanation:

The horizontal motion doesn't affect the vertical motion.  Since the two spheres have the same initial vertical velocity and same initial height, they land at the same time.

Answer:

A. Both spheres land at the same time.

Explanation:

If air resistance and friction are ignored, then both spheres land at the same time. Falling objects fall toward the center of the Earth with the same constant acceleration, independent of their mass, when they are defined to be in free-fall. For an object to be in free-fall, it has to be in a situation in which both air resistance and friction are considered negligible. This is true regardless of the direction of the fall.

To answer this question, let us break down the initial velocity into its horizontal and vertical components.

v_{ix} = vcos(θ)

v_{iy} = vsin(θ)

v_{x} is the horizontal component of the initial velocity, and this value will stay constant over time because we assume only gravity acts on the stone, therefore only the vertical component of the stone's velocity will change over time

v_{iy} is the vertical component of the initial velocity, and this will change over time due to gravity.

v is the magnitude of the initial velocity.

θ is the angle the velocity vector is oriented at with respect to the horizontal.

Given values:

v = 19.6m/s

θ = 40.8°

Plug in these values and solve for v_{ix} and v_{iy}:

v_{ix} = 19.6cos(40.8°) = 14.8m/s

v_{iy} = 19.6sin(40.8°) = 12.8m/s

To find both the horizontal and vertical displacement at any time, we will use this kinematics equation:

D = X + Vt + 0.5At²

When solving for the horizontal displacement, the following values are:

t = elapsed time

D = horizontal displacement

X = initial horizontal displacement

V = initial horizontal velocity

A = horizontal acceleration

There is no initial horizontal displacement, so X = 0m

The initial horizontal velocity V = v_{ix} = 14.8m/s

Assuming we don't care about air resistance, no force has a component acting horizontally on the stone, so A = 0m/s²

Therefore the equation for the stone's horizontal displacement is given by:

D = 14.8t

When solving for the vertical displacement, the following values are:

t = elapsed time

D = vertical displacement

X = initial vertical displacement

V = initial vertical velocity

A = vertical acceleration

There is no initial vertical displacement, so X = 0m

The initial vertical velocity V = v_{iy} = 12.8m/s

Gravity acts downward on the stone, therefore A = -9.81m/s²

Therefore the equation for the stone's vertical displacement is given by:

D = 12.8t - 4.905t²

Now we just plug in various values of t...

a) At t = 1.03s, the horizontal displacement is D = 14.8(1.03) = 15.2m

b) At t = 1.03s, the vertical displacement is D = 12.8(1.03)-4.905(1.03)² = 7.98m

c) At t = 1.73s, the horizontal displacement is D = 14.8(1.73) = 25.6m

d) At t = 1.73s, the vertical displacement is D = 12.8(1.73)-4.905(1.73)² = 7.46m

Before you write down the following results, read the following explanation.

e) At t = 5.05s, the horizontal displacement is D = 14.8(5.05) = 74.7m

f) At t = 5.05s, the vertical displacement is D = 12.8(5.05)-4.905(5.05)² = -60.4m

Notice that we have a negative value for the vertical displacement. This isn't possible within the context of the problem, so the vertical displacement at t = 5.05s is actually 0m

The problem hasn't stated whether the ground is frictionless or not. I'm going to assume it is frictionless and therefore the stone will keep moving across the ground after landing, so the horizontal displacement at t = 5.05s is 74.7m

Answer:

Centripetal force, F = 9.81 N

Explanation:

It is given that,

Mass of the object (yo-yo), m = 0.12 kg

It is whirled around by its string in a circle

Radius of circle, r = 0.44 m

Speed of object, v = 6 m/s

We need to find the required centripetal force in Newtons to keep this yo -yo whirling in a circle. The centripetal force is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]F=\dfrac{0.12\ kg\times (6\ m/s)^2}{0.44\ m}[/tex]

F = 9.81 Newton

So, the centripetal force required to keep this yo -yo whirling in a circle is 9.81 N. Hence, this is the required solution.

Answer:

37.91594 keV

Explanation:

[tex]E_i[/tex] = Incident energy = 400 keV

θ = 30°

h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s

Incident photon wavelength

[tex]\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m[/tex]

Difference in wavelength

[tex]\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m[/tex]

[tex]\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}[/tex]

Final photon wavelength

[tex]\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV[/tex]

Energy of the recoiling electron

[tex]\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV[/tex]

Energy of the recoiling electron is 37.91594 keV

Answer:

The energy of recoiling electron=192.44 keV

Explanation:

Energy of x-ray[tex]E_o=400 keV[/tex]

Web know that compton shift is

[tex]\Delta \lambda =\dfrac{h}{m_eC(1-cos\theta)}[/tex]

[tex]m_e[/tex] is the mass of electron and C is the velocity of sound.

Given that θ=30°

Now by putting the values

[tex]\Delta \lambda =\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8(1-cos30)}[/tex]

[tex]\Delta \lambda =2.8\times 10^{-3}nm[/tex]

[tex]\Delta \lambda_o=\dfrac{hc}{E_o}[/tex]

By putting the values

[tex]\Delta \lambda_o=\dfrac{6.63\times 10^{-34}3\times 10^8}{400\times 1.602\times 10^{-16}}[/tex]

[tex]\Delta \lambda_o=3.31times 10^{-3}nm[/tex]

[tex]\lambda =\Delta \lambda +\lambda _o[/tex]

[tex]\lambda=2.8\times 10^{-3}+3.31\times 10^{-3}nm[/tex]

[tex]\lambda=6.11`\times 10^{-3}nm[/tex]

Energy [tex]E=\dfrac{hC}{\lambda }[/tex]

So [tex]E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.11\times 10^{-12}}[/tex]

E=207.55 keV

The energy of recoiling electron=400-207.55 keV

The energy of recoiling electron=192.44 keV

The power drawn by the air conditioner to cool the house from 35°C to 20°C in 38 minutes, given a COP of 2.8, is calculated to be 1.35 kW.

To calculate the power drawn by the air conditioner, we first determine the energy needed to cool the house from 35°C to 20°C. The energy required (Q) can be calculated using the formula Q = m⋅cv⋅ΔT, where m is the mass of the air, cv is the specific heat capacity at constant volume, and ΔT is the change in temperature.

Plugging the given values into the formula: Q = 800 kg ⋅ 0.72 kJ/kg°C ⋅ (35°C - 20°C) = 8,640 kJ.

Next, we use the Coefficient of Performance (COP) of the air conditioner to find the work input (W) required for this cooling process. The formula relating these quantities is COP = Q/W, rearranging for W gives W = Q/COP.

Therefore, W = 8,640 kJ / 2.8 = 3,085.71 kJ. To find the power drawn, we need the work in kilowatts (kW), knowing that 1 kW = 1 kJ/s, and that the total duration is 38 min or 2,280 seconds. Thus, the power drawn is Power = W / time = 3,085.71 kJ / 2,280 s = 1.35 kW.

Power drawn by the air conditioner is approximately 6.76 kW, calculated from COP, heat transfer, and time taken for cooling.

To solve this problem, we can use the energy balance equation for the air within the house:

[tex]\[ Q = mc\Delta T \][/tex]

where:

- [tex]\( Q \)[/tex] is the heat transfer,

- [tex]\( m \)[/tex] is the mass of air,

- [tex]\( c \)[/tex] is the specific heat capacity of air,

- [tex]\( \Delta T \)[/tex] is the change in temperature.

We also know that the Coefficient of Performance (COP) is defined as:

COP = Q_cooling/W_input

where:

- [tex]\( Q_{\text{cooling}} \)[/tex] is the heat removed from the house (in this case),

- W_input is the work input (power consumed) by the air conditioner.

We are given that the COP [tex](\( \text{COP} = 2.8 \))[/tex] and the initial and final temperatures [tex](\( T_{\text{initial}} = 35^\circ C \)[/tex], [tex]\( T_{\text{final}} = 20^\circ C \))[/tex].

First, let's calculate the heat transfer using the energy balance equation:

[tex]\[ Q = mc\Delta T \][/tex]

[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(35 - 20) \][/tex]

[tex]\[ Q = (800 \, \text{kg})(1.0 \, \text{kJ/kg}^\circ C)(15) \][/tex]

Q = 12000kJ

Now, we can find the work input using the COP formula:

COP = Q_cooling/{W_input

W_input = [tex]\frac{Q_{\text{cooling}}}{\text{COP}}[/tex]

W_input = [tex]\frac{12000 \, \text{kJ}}{2.8}[/tex]

W_input ≈ 4285.71 kJ

To find the power drawn by the air conditioner, we need to convert the energy to power. Since the time taken for cooling is 38 minutes, or [tex]\( \frac{38}{60} \)[/tex] hours:

Power = W_input /Time

[tex]\[ \text{Power} = \frac{4285.71 \, \text{kJ}}{\frac{38}{60} \, \text{hours}} \][/tex]

Power ≈ 6762.45 W

So, the power drawn by the air conditioner is approximately [tex]\( 6762.45 \, \text{W} \)[/tex] or [tex]\( 6.76 \, \text{kW} \)[/tex].

Answer:

220.8 m

30.5 m/s

Explanation:

a = acceleration of the car = 2.1 m/s²

t = time of travel = 14.5 s

x = displacement of the car

v₀ = initial velocity of the car = 0 m/s

Displacement of the car is given as

x = v₀ t + (0.5) a t²

Inserting the values

x = (0) (14.5) + (0.5) (2.1) (14.5)²

x = 220.8 m

v = final velocity of the car

Final velocity of the car is given as

v = v₀ + at

v = 0 + (2.1) (14.5)

v = 30.5 m/s

The approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]

To calculate the magnitude of the pressure on the sail due to radiation pressure, we can use the formula for pressure:

[tex]\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \][/tex]

The force due to radiation pressure is given by:

[tex]\[ \text{Force} = 2 \times \text{radiative momentum} \][/tex]

The radiative momentum can be calculated using the energy intensity of sunlight:

[tex]\[ \text{Radiative momentum} = \text{Energy intensity} \times \text{Time} \][/tex]

Given:

- Energy intensity of sunlight [tex](\( I \))[/tex] is approximately [tex]\( 1300 \, \text{W/m}^2 \),[/tex]

- Time [tex](\( t \)) is \( 1 \, \text{s} \)[/tex] (per second),

- Atmospheric pressure is about [tex]\( 105 \, \text{N/m}^2 \).[/tex]

Let's calculate the radiative momentum:

[tex]\[ \text{Radiative momentum} = 1300 \times 1 \, \text{N/s} = 1300 \, \text{N/s} \][/tex]

Now, let's calculate the force:

[tex]\[ \text{Force} = 2 \times 1300 \, \text{N/s} = 2600 \, \text{N/s} \][/tex]

Finally, let's calculate the pressure on the sail:

[tex]\[ \text{Pressure} = \frac{2600 \, \text{N/s}}{1 \, \text{m}^2} = 2600 \, \text{N/m}^2 \][/tex]

So, the approximate magnitude of the pressure on the sail due to radiation pressure is [tex]\( 2600 \, \text{N/m}^2 \).[/tex]

Answer:

a) [tex]V=0.314 m/s[/tex]

b) [tex]\omega=2.09rad/s[/tex]

c) [tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]

Explanation:

Given:

Radius of the circle, r = 15cm = 0.15m

Time period, T  =3.0s

a) The velocity (V) of a particle moving in the circular motion is given as:

[tex]V=\frac{2 \pi r}{T}[/tex]

substituting the given values in the above equation we get

[tex]V=\frac{2\times \pi \times 0.15m}{3.0s}[/tex]

or

[tex]V=0.314 m/s[/tex]

b) Angular speed ([tex]\omega[/tex]) is given as:

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

or

[tex]\omega=\frac{2\times \pi}{3.0s}[/tex]

or

[tex]\omega=2.09rad/s[/tex]

c) The position of the particle on the x-position is given as:

[tex]x(\theta)=rcos(\theta)[/tex]         (reffer the attached figure)

now the relation between the Θ and the time T is given as:

[tex]\omega = \frac{2\pi}{T}=\frac{\theta}{t}[/tex]

or

[tex]\theta= \frac{2\pi}{T}\times t[/tex]

or

[tex]\theta= \frac{2\pi}{3}\times t[/tex]

substituting the values of r and Θ, we get

[tex]x(\theta)=0.15\times cos(\frac{2\pi}{3.0} t)[/tex]

Final answer:

The particle's speed is 0.1π m/s, its angular speed is 2/3π rad/s, and the x-component of the position function is x(t) = 0.15 * cos((2/3π) * t + π).

Explanation:

(a) Speed of the particle: The speed is the distance traveled by a particle along a circular path per unit time. Given a particle completes 1 revolution (which is 2π times the radius r) every 3 seconds, the speed v is given by: v = (2π * 15 cm) / 3 s. Convert 15 cm to meters by dividing by 100 to get 0.15 meters. So, v = (2π * 0.15 m) / 3 s = 0.1π m/s.

(b) Angular speed of the particle: The angular speed ω is the rate of change of the angular displacement and is measured in radians per second. Since the particle completes one revolution every 3 seconds and one revolution is 2π radians, ω = 2π radians / 3 s = 2/3π rad/s.

(c) Position function for the x-component: Assume the particle starts on the negative x-axis, so the angle θ at t=0 is π. The particle moves in a circle with radius r = 0.15 m. The x-component of the position as a function of time t, using the cosine function is x(t) = r * cos(ω*t + π), where ω = 2/3π rad/s. Plugging values, we get x(t) = 0.15 * cos((2/3π) * t + π).

Answer:

The initial temperature of copper is 96.3 °C

Explanation:

Specific heat capacity is the energy needed to raise the temperature of one gram of material by one degree celsius. The energy absorbed or released by a material during temperature change can be calculated by the below formula:

[tex]Q=mc(T_{2}-T_{1})[/tex] , where:

Q = energy (J)

m = mass (g)

c = specific heat capacity (J/g·°C)

[tex]T_{1}[/tex] = initial temperature (°C)

[tex]T_{2}[/tex] = final temperature (°C)

In an ideal situation, it can be assumed that all the energy lost by the piece of copper is gained by the water, resulting in its temperature rise and that there is no change of mass/state for either material. Thus, the equation can be written as below:

[tex]+Q_{c}=-Q_{w}[/tex]

[tex]+m_{c}c_{c}(T_{c2}-T_{c1})=-m_{w}c_{w}(T_{w2}-T_{w1})[/tex]

It can also be assumed that the final temperature of both the copper and water are the same. Thus substituting below values in above equation will give:

[tex]m_{w}[/tex] = 400 g

[tex]c_{w}[/tex] = 4.18 J/g·°C

[tex]T_{w1}[/tex] = 20.7 °C

[tex]T_{w2}[/tex] = 22.2 °C

[tex]m_{c}[/tex] = 89.1 g

[tex]c_{c}[/tex] = 0.38 J/g·°C

[tex]T_{c1}[/tex] = ? °C

[tex]T_{c2}[/tex] = 22.2 °C

[tex]+89.1*0.38*(22.2-T_{c1})=-400*4.18(22.2-20.7)[/tex]

Solving for [tex]T_{c1}[/tex] gives:

[tex]T_{c1}[/tex] = 96.3 °C

Here, we are required to determine the initial temperature of the copper.

The initial temperature of the copper is;

T(c1) = 96.27 °C

The energy required to raise the temperature of one gram of a material by one degree celsius is termed the Specific Heat Capacity of that material.

Mathematically, we have;

Q=mc{T(2) -T(1)}

Q=mc{T(2) -T(1)} where:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

Q=mc{T(2) -T(1)} where:Q = energy (J)m = mass (g)c = specific heat capacity (J/g·°C)T1 = initial temperature (°C)T2 = final temperature (°C)By the law of energy conservation, Energy can neither be created nor destroyed. Thus, the equation can be written as below:+Q(c) =−Q(w)

m(c) × c(c) × (T2(c) - T1(c)) = -{m(w) × c(w) × (T2(w) - T1(w))}

However, the final temperature of the copper piece and water can be assumed to be equal

i.e. T2(c) = T1(w)

+89.1*0.38*(22.2-T(c1)) = -400*4.18(22.2-20.7)

(22.2−T(c1) ) = −2508/33.858

(22.2−T(c1) ) = -74.07

T(c1) = 74.07 + 22.2

T(c1) = 96.27 °C.

Therefore, the initial temperature of the copper is; T(c1) = 96.27 °C

Read more:

https://brainly.com/question/19863538&

To find the speed at the bottom of the ramp, we can use the principles of physics. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp. Using the given values and trigonometry, we can calculate the height of the ramp and then determine the velocity.

To find the speed of the skateboarder at the bottom of the ramp, we can use the principles of physics. Since we are neglecting friction, we can assume that the mechanical energy of the skateboarder is conserved. The mechanical energy consists of both kinetic energy and potential energy. At the top of the ramp, the skateboarder has only potential energy, which is given by the formula: potential energy = mass * gravity * height. At the bottom of the ramp, the skateboarder has only kinetic energy, which is given by the formula: kinetic energy = 1/2 * mass * velocity^2. By equating the potential energy at the top to the kinetic energy at the bottom, we can solve for the velocity at the bottom of the ramp.

In this case, the mass of the skateboarder is not provided, but it cancels out when equating the potential and kinetic energies. So we can assume any mass value as long as it is consistent throughout the calculation. Let's assume the mass of the skateboarder is 1 kg for simplicity. The height of the ramp is not provided, but we can calculate it using trigonometry. The height is the vertical component of the ramp's length, which is given by: height = length * sin(angle). Plug in the values and calculate the height.

Next, we calculate the potential energy at the top of the ramp using the calculated height. The potential energy is then equated to the kinetic energy at the bottom of the ramp and solve for the velocity. Plug in the values and calculate the velocity at the bottom of the ramp.

Answer:

156.3 J

Explanation:

E = Amplitude of the electric field = 25000 V/m

t = time interval = 1 minute = 60 sec

d = diameter of the spot = 2 mm = 2 x 10⁻³ m

Area of the spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Energy delivered to the spot is given as

U = (0.5)ε₀ E² c A t

Inserting the values

U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)

U = 156.3 J

To find the energy delivered by the laser beam in 1 minute, we need to calculate the power of the laser beam and then multiply it by the duration.

E = Amplitude of the electric field = 25000 V/m

t = time interval = 1 minute = 60 sec

d = diameter of the spot = 2 mm = 2 x 10⁻³ m

Area of the spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Energy delivered to the spot is given as

U = (0.5)ε₀ E² c A t

Inserting the values

U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)

U = 156.3 J

Answer:

one ninth

Explanation:

d = 1 cm , v = v

D = 3d, V = ?

By the equation of continuity,

A V = a v

3.14 x D^2 / 4 x V = 3.14 x d^2 / 4 x v

9d^2 x V = d^2 x v

V = v / 9

Thus, the velocity becomes one ninth the initial velocity

The energy  stored in the spring when it is compressed 20.0 cm is 15.54J, the maximum height of the ball is 1.056 meters. while the muzzle velocity of the ball is  4.55 m/s.

a) Potential Energy in the Spring

Potential energy = 1/2 * spring constant * (compression distance)^2

Potential energy = 1/2 * 777 N/m * (0.2 m)^2 = 15.54 J

b)  Maximum Height of the Ball

Maximum height = potential energy / weight of the ball

Maximum height = 15.54 J / (1.5 kg * 9.81 m/s^2) = 1.056 m

Therefore, the maximum height reached by the ball is 1.056 meters.

(c) Muzzle Velocity of the Ball

muzzle velocity = sqrt(2 * potential energy / mass)

muzzle velocity = sqrt(2 * 15.54 J / 1.5 kg) = 4.55 m/s

Hence, the muzzle velocity of the ball is 4.55 m/s.

A) The energy stored in the spring when compressed is 15.54 J. B) The maximum height reached by the ball is 1.05 m. C) The muzzle velocity of the ball is 4.56 m/s.

A) The energy stored in a spring, also known as elastic potential energy, can be calculated using the formula:

Espring = ½ * k * x2

where k is the spring constant, and x is the compression distance. By plugging in the given values (k = 777 N/m and x = 0.20 m), we get:

Espring = ½ * 777 N/m * (0.20 m)2 ≈ 15.54 J

B) At the maximum height, all the elastic potential energy is converted into gravitational potential energy (Egravitational), given by:

Egravitational = m * g * h

Where m is the mass of the ball, g is the acceleration due to gravity (9.81 m/s2), and h is the height. Since Espring = Egravitational, we can solve for h:

h = Espring / (m * g) = 15.54 J / (1.5 kg * 9.81 m/s2) ≈ 1.05 m

C) The muzzle velocity of the ball can be found using the conservation of energy principle. The elastic potential energy stored in the spring is converted entirely into the kinetic energy of the ball at the spring's equilibrium position:

Kinetic energy, Ekinetic = ½ * m * v2

Since Espring = Ekinetic, solving for v gives:

v = √(2 * Espring / m) = √(2 * 15.54 J / 1.5 kg) ≈ 4.56 m/s

We can determine the angular velocity using the formula ω = v/r, where v is the tangential speed (2.2 m/s) and r is the radius of the circle (510 mm). After converting the radius to meters, we can then plug these values into the formula and solve for ω, yielding an angular velocity of  3.9636 rad/s.

To determine the resulting angular velocity ωOA of rod OA when point B has a displacement s = 220 mm, we can use the following formula:

ωOA = v / r

Where:

ωOA = Angular velocity of rod OA

v = Linear velocity of point B

r = Radius or distance from point O to point B

Given:

v = 2.2 m/s

s = 220 mm = 0.22 m

d = 510 mm = 0.51 m

First, we calculate the radius (r) using the Pythagorean theorem because the motion is along a right-angled triangle:

r² = d² + s²

r² = (0.51 m)² + (0.22 m)²

r² = 0.2601 m² + 0.0484 m²

r² = 0.3085 m²

Now, we calculate the square root of r² to get r:

r = √0.3085 m

r ≈ 0.555 m

Now, we can calculate the angular velocity ωOA:

ωOA = v / r

ωOA = 2.2 m/s / 0.555 m

ωOA ≈ 3.9636 rad/s

The angular velocity ωOA is approximately 3.9636 rad/s. Since the motion is counterclockwise (as indicated by the positive linear velocity of point B), the angular velocity is also positive.

Learn more about Angular Velocity here:

https://brainly.com/question/30733452

#SPJ3

Final answer:

The magnetic field at a point on the +x-axis near a wire carrying current in the +y-direction will be directed into the page, as determined by the right-hand rule. To cancel the magnetic field at a specific point on the +x-axis, a second wire carrying equal but opposite current must be arranged parallel to the first.

Explanation:

The question concerns the determination of the magnetic field direction around a long, straight wire carrying a current. According to the right-hand rule, if you point your right thumb in the direction of the current, then the direction in which your fingers curl will show the direction of the magnetic field. In this case, the current is running in the +y direction, so if one were to stand at a point on the +x-axis near the wire and apply the right-hand rule, the fingers would curl into the page, indicating that the magnetic field at that point is directed into the page.

An experimenter wanting to make the total magnetic field at the coordinate x = 1.0 m zero must arrange a second wire parallel to the first with current in the opposite direction. This second wire would also need to be carrying the same magnitude of current but in the -y direction to cancel out the magnetic field from the first wire at the point of interest.

The index of refraction can be calculated using the speed of light in vacuum divided by the speed of light in a specific medium. Given the speed of light in a medium is 2.26 x 10^8 m/s, the index of refraction would be calculated as roughly 1.33.

To calculate the index of refraction for a certain medium, you would use the speed of light in vacuum (c) divided by the speed of light in that medium (v). This equation is denoted as n = c/v.

The speed of light in vacuum is approximately 3.00×10^8 m/s. Given that the speed of light in a certain substance is 2.26 x 10^8 m/s, we can substitute these values into the equation to find the material's index of refraction.

So, n = (3.00×10^8 m/s) / (2.26 x 10^8 m/s). Therefore, the index of refraction of that substance is approximately 1.33. This property of a substance tells us how much the path of light is bent, or refracted, when entering the material.

https://brainly.com/question/30761100

#SPJ3

Answer:

Magnetic field at given point is along + x direction

Explanation:

As we know that position of the wire is along Y axis

current is flowing along +y direction

so here we will have

[tex]\vec B = \frac{\mu_0 i (\vec {dl} \times \hat r)}{4\pi r^2}[/tex]

now the direction of length vector is along +Y direction

also the position vector is given as

[tex]\hat r = \hat k[/tex]

now for the direction of magnetic field we can say

[tex]\vec B = (\hat j \times \hat k)[/tex]

[tex]\vec B = \hat i[/tex]

so magnetic field is along +X direction

Using the right-hand rule for a wire carrying current in the +y direction along the y-axis, the magnetic field at the point (0,0,1.045 m) on the x-axis would be directed in the +x-direction.

When a current runs through a wire in the +y direction along the y-axis, we can use the right-hand rule to determine the direction of the magnetic field at a point near the wire. For a point on the x-axis such as (0,0,1.045 m), you would point your thumb in the direction of the current (upward along the y-axis) and curl your fingers.

Your fingers will curl in the direction of the magnetic field lines, which at the given point on the x-axis would circle around the wire. This means the magnetic field at that point would be directed in the +x-direction.

Answer:

The initial speed is 7.174 m/s.

Explanation:

Given that,

Mass = 92.0 kg

Coefficient of friction = 0.61

Time = 1.2 s

We need to calculate the acceleration

Using of friction force

[tex]F = \mu mg[/tex]...(I)

Where, [tex]\mu[/tex] =Coefficient of friction

g = acceleration due to gravity

Using newton's law

[tex]F = ma[/tex]....(II)

m = mass of the baseball

a = acceleration of the baseball

From equation (I) and (II)

[tex]ma=\mu mg[/tex]...(I)

Put the value in the equation

[tex]a=0.61\times9.8[/tex]

[tex]a=5.978\ m/s^2[/tex]

We need to calculate the initial velocity

Using equation of motion

[tex]v = u-at[/tex]

Where, v = final velocity

u = initial velocity

a = acceleration

Here, a is negative because the player comes to rest

t = time

Put the value in the equation

[tex]0=u-5.978\times1.2[/tex]

[tex]u=7.174\ m/s[/tex]

Hence, The initial speed is 7.174 m/s.

The initial speed of the baseball player was approximately 7.176 m/s, calculated using the principles of Physics (Kinematics) and understanding of friction.

The subject of the problem involves a concept in Physics known as Kinematics, specifically dealing with friction and motion. The initial speed of the baseball player can be calculated using the equation of motion: final velocity (v_f) = initial velocity (v_i) + acceleration (a) x time (t), where the final velocity in this case is 0 (as the player comes to rest), time is 1.2 seconds, and the acceleration is the frictional force divided by the mass of the player. Frictional force is obtained by multiplying the mass of the player, the gravitational acceleration, and the coefficient of friction. In this case, substituting acceleration = -0.61 * 9.8 m/s² (because it's a decelerating force), the equation becomes 0 = v_i - 0.61 * 9.8 * 1.2 from which initial velocity (v_i) comes out to be approximately 7.176 m/s.

https://brainly.com/question/32983297

#SPJ6

Answer : The total pressure in the container at equilibrium is, 0.4667 atm

Solution :  Given,

Initial pressure of [tex]H_2S[/tex] = 0.259 atm

Equilibrium constant, [tex]K_p[/tex] = 0.841

The given equilibrium reaction is,

                             [tex]H_2S(g)\rightleftharpoons H_2(g)+S(g)[/tex]

Initially                 0.259          0           0

At equilibrium   (0.259 - x)      x           x

Let the partial pressure of [tex]H_2[/tex] and [tex]S[/tex] will be, 'x'

The expression of [tex]K_p[/tex] will be,

[tex]K_p=\frac{(p_{H_2})(p_{S})}{p_{H_2S}}[/tex]

Now put all the values of partial pressure, we get

[tex]0.841=\frac{(x)\times (x)}{(0.259-x)}[/tex]

By solving the term x, we get

[tex]x=0.2077atm[/tex]

The partial pressure of [tex]H_2[/tex] and [tex]S[/tex] = x = 0.2077 atm

Total pressure in the container at equilibrium = [tex]0.259-x+x+x=0.259+x=0.259+0.2077=0.4667atm[/tex]

Therefore, the total pressure in the container at equilibrium is, 0.4667 atm