A 66 g66 g ball is thrown from a point 1.05 m1.05 m above the ground with a speed of 15.1 ms/15.1 ms . When it has reached a height of 1.59 m1.59 m , its speed is 10.9 ms/10.9 ms . What was the change in the mechanical energy of the ball-Earth system because of air drag

Answer:

Velocity is 34.42 m/s at an angle of 56.91° below horizontal

Speed is 34.42 m/s

Explanation:

Velocity = 20.0 m/s at an angle of 30° below horizontal

Vertical velocity = 20 sin 20 = 6.84 m/s downward.

Horizontal velocity = 20 cos 20 = 18.79 m/s towards right.

Let us consider the vertical motion of ball we have equation of motion

               v² = u² + 2as

We need to find v,  u = 6.84 m/s, a = 9.81 m/s² and s = 40 m

Substituting

              v² = 6.84² + 2 x 9.81 x 40 = 831.59

               v = 28.84 m/s

So on reaching ground velocity of ball is

        Vertical velocity =  28.84 m/s downward.

        Horizontal velocity = 18.79 m/s towards right.  

Velocity

        [tex]v=\sqrt{28.84^2+18.79^2}=34.42m/s[/tex]

        [tex]tan\theta =\frac{28.84}{18.79}\\\\\theta =56.91^0[/tex]

So velocity is 34.42 m/s at an angle of 56.91° below horizontal

Speed is 34.42 m/s

Answer:

The correct option is : (a) Create a strong magnetic field.

Explanation:

An electromagnet is a substance which produces magnetic field when electric current is passed through it. The magnetic field produced by them disappears when the electric current is turned off.

A Ferromagnetic substance is a substance that gets magnetized when kept in an external magnetic field. Such substances remain magnetized even after the external magnetic field is removed.

When an electromagnet and a ferromagnet is combined, it results in the production of a strong magnetic field.

Answer:3.98cm

Explanation:

given data

mass of object[tex]\left ( m\right )[/tex]=0.5kg

intial extension=5 cm

elevator acceleration=2 m/[tex]s^2[/tex]

From FBD of intial position

Kx=mg

K=[tex]\frac{0.5\times 9.81}{0.05}[/tex]

k=98.1 N/m

From FBD of second situation

mg-k[tex]x_0[/tex]=ma

k[tex]x_0[/tex]=m(g-a)

[tex]x_0[/tex]=[tex]\frac{0.5(9.81-2)}{98.1}[/tex]

[tex]x_0[/tex]=3.98cm

Answer:

QC = 122 KJ

QH = 2.64 x 122 = 322 KJ

Explanation:

TH = 500 Degree C = 500 + 273 = 773 K

TC = 20 degree C = 20 + 273 = 293 K

W cycle = 200 KJ

Use the formula for the work done in a cycle

Wcycle = QH - QC

200 = QH - QC    ..... (1)

Usse

TH / TC = QH / QC

773 / 293 = QH / QC

QH / QC = 2.64

QH = 2.64 QC     Put it in equation (1)

200 = 2.64 QC - QC

QC = 122 KJ

So, QH = 2.64 x 122 = 322 KJ

Answer:

1454.54 kJ/h or 0.404 kW

Explanation:

Given:

Removal of heat by the refrigerator = 800 kJ/h

Coefficient of performance, COP of the refrigerator = 2.2

given that the  refrigerator runs one-fourth of the time only

therefore, refrigerator removes the heat, Q = [tex]4\times 800 kJ/h[/tex] = 3200 kJ/h

Now, the Power drawn by the refrigerator while running = [tex]\frac{Q}{COP}[/tex]

=  [tex]\frac{3200 kJ/h}{2.2}[/tex]

= 1454.54 kJ/h

Hence, the power drawn by the refrigerator while running is = 1454.54 kJ/h

or 0.404 kW      (as 1 kW = 3600 kJ/h)

To determine the power the refrigerator draws when running, use the formula for the coefficient of performance (COP) and substitute the given values

To determine the power the household refrigerator draws when running, we need to use the formula for the coefficient of performance (COP) of a refrigerator. The COP is given by the ratio of the heat removed from the cold reservoir to the work done on the engine's working substance. In this case, the COP is given as 2.2.

The power drawn by the refrigerator can be calculated using the formula:

Power = COP x Heat removed from the cold reservoir / Time

Given that the refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of 800 kJ/h, we can substitute these values into the formula to find the power drawn by the refrigerator.

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Answer:

Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.

Explanation:

Amount of energy reaching 1 square meter of area = 2278 kJ/hour

Total area of the United States = [tex]9,158,960 km^2=9,158,960\times 10^6 m^2[/tex]

[tex](1 km^2=10^6 m^2)[/tex]

Amount of energy reaching [tex]9,158,960\times 10^6 m^2[/tex] of area:

[tex]2278 kJ/hour\times 9,158,960\times 10^6 kJ/hour[/tex]

[tex]=2.0864\times 10^{16} kJ/Hour[/tex]

Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.

Final answer:

The total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.

Explanation:

To calculate the total solar energy reaching the entire United States per hour, given that 2278 kJ of energy reaches a square meter in one hour, we would first convert the entire area of the United States into square meters. Since the area is given as 9,158,960 km², we convert this to square meters by multiplying by (1000 m/km)²:

9,158,960 km² * (1000 m/km)² = 9,158,960,000,000 m²

Next, we multiply the area in square meters by the energy received per square meter:

9,158,960,000,000 m² * 2278 kJ/m² = 20,866,319,680,000 kJ

To generate an accurate answer, we can express this in megajoules (MJ) by dividing by 1,000 (since 1 MJ = 1,000 kJ):

20,866,319,680,000 kJ / 1,000 = 20,866,319,680 MJ

Therefore, the total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.

Answer:

D. half as much

Explanation:

let m and M be the mass of the planets and r be the distance between them.

then: the force of attraction between them is given by,

F = G×m×M/(r^2)

if we keep one mass constant and double the other and also double the distance between them.

the force of attraction becomes:

F1 = 2G×m×M/[(2×r)^2]

   = 2G×m×M/[4×(r)^2]

   = (1/2)G×m×M/(r^2)

   = 1/2×F

therefore, when you double one mass and keep the other mass constant and double the distance between the masses you decrease the force by a factor of 1/2.

The player's foot must be in contact with the ball for a distance of 1.48 meters to increase the ball's speed from 2.00 m/s to 6.00 m/s.

To increase the ball's speed from 2.00 m/s to 6.00 m/s, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity (6.00 m/s), u is the initial velocity (2.00 m/s), a is the acceleration, and s is the distance.

Rearranging the equation, we have:

s = (v^2 - u^2) / (2a)

Plugging in the values, we get:

s = (6.00^2 - 2.00^2) / (2 * 40.0)

s = 1.48 m

Therefore, the player's foot must be in contact with the ball for a distance of 1.48 meters.

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The physics concepts of force, mass, acceleration, and work were used to determine the distance over which the soccer player's foot needs to be in contact with the soccer ball in order to increase its speed from 2.00 m/s to 6.00 m/s. The resulting distance was approximately 0.09 m or 9 cm.

This question involves the relationship between force, mass, and acceleration, as well as the concept of work. Great! For the student to understand this, we need to look at a couple of formulas from physics. The force exerted on an object equals its mass times its acceleration (F = ma). Also, work done on an object is the force applied to it times the distance over which the force is applied (W = Fd).

Firstly, we need to determine the acceleration of the soccer ball when it is kicked. Using the first formula, we rearrange to find a = F/m. Here, F is the force applied (40.0 N) and m is the mass of the soccer ball (0.42 kg). Substituting these values, we get a = 40.0N / 0.42kg = approximately 95.2 m/s².

The initial speed of the ball is 2.00 m/s and the final speed we want is 6.00 m/s. The change in speed (which is also the change in velocity, as the direction doesn't change) is therefore 6.00 m/s - 2.00 m/s = 4.00 m/s. Using the formula for acceleration, which is a change in velocity divided by time (a = Δv / t), we can plug in our known values to solve for the time of contact: t = Δv / a = 4.00 m/s / 95.2 m/s² = approximately 0.042 s.

With the time of contact known, we can now determine the distance over which the soccer player's foot must be in contact with the ball using the formula for distance traveled during uniformly accelerated motion: d = vt + 0.5at². The initial velocity v is 2.00 m/s, the time of contact t is 0.042 s, and the acceleration a is 95.2 m/s². Substituting these values, we find that the distance is approximately 0.09 m or 9 cm.

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Answer:

[tex]B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)[/tex]

Explanation:

As the current density is given as

[tex]J = \frac{J_0}{r_0}r[/tex]

now we have current inside wire given as

[tex]i = \int J(2\pi r)dr[/tex]

[tex]i = \int \frac{J_0}{r_0} r(2\pi r)dr[/tex]

[tex]i = 2\pi \frac{J_0}{r_0} \int r^2 dr[/tex]

[tex]i = \frac{2}{3} \pi \frac{J_0}{r_0} r^3[/tex]

Now by Ampere's law we will have

[tex]\int B. dl = \mu_0 i[/tex]

[tex]B. (2\pi r) = \mu_0(\frac{2}{3} \pi \frac{J_0}{r_0} r^3)[/tex]

[tex]B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)[/tex]

Answer:

The ratio of the pressure at the top to the pressure at the bottom is [tex]\dfrac{701}{1000}[/tex]

Explanation:

Given that,

Number of molecules [tex]n= 10^24[/tex]

Mass [tex]m= 3\times10^{-26}\ kg[/tex]

Temperature = 300 K

Height [tex]h = 5\times10^{3}[/tex]

We need to calculate the  ratio of the pressure at the top to the pressure at the bottom

Using barometric formula

[tex]P_{h}=P_{0}e^{\dfrac{-mgh}{kT}}[/tex]

[tex]\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-mgh}{kT}}[/tex]

Where, m = mass

g = acceleration due to gravity

h = height

k = Boltzmann constant

T = temperature

Put the value in to the formula

[tex]\dfrac{P_{h}}{P_{0}}=e^{\dfrac{-3\times10^{-26}\times9.8\times5\times10^{3}}{1.3807\times10^{-23}\times300}}[/tex]

[tex]\dfrac{P_{h}}{P_{0}}=\dfrac{701}{1000}[/tex]

Hence, The ratio of the pressure at the top to the pressure at the bottom is [tex]\dfrac{701}{1000}[/tex]

Answer:

Top pressure : Bottom pressure = 701 : 1000

Explanation:

Number of molecules = n = 10^24

Height = h = 5 × 10^3 m

Mass = m = 3 × 10^-26 kg  

Boltzman’s Constant = K = 1.38 × 10^-23 J/K

Temperature = T = 300K  

The formula for barometer pressure is given Below:

Ph = P0 e^-(mgh/KT)

Ph/P0 = e^-(3 × 10^-26 × 9.81 × 5 × 10^3)/(1.38 × 10^-23)(300)

Ph/P0 = e^-0.355

Ph/P0 = 1/e^0.355

Ph/p0 =0.7008 = 700.8/1000 = 701/1000

Hence,

Top pressure : Bottom pressure = 701 : 1000

Answer:

[tex]I_{2}=0.27 W/m^2[/tex]

Explanation:

Intensity is given by the expresion:

[tex]I_{2}=Io (\frac{r1}{r2} )^{2}[/tex]

where:

Io = inicial intensity

r1= initial distance

r= final distance

[tex]I_{2}=10 W/m^2 (\frac{30m}{180m} )^{2}[/tex]

[tex]I_{2}=0.27 W/m^2[/tex]

Answer:

25.6 V

Explanation:

The kinetic energy of electron associated with its potential difference is given by eV which is equal to the 1/2 mv^2.

m = 9.1 x 10^-31 kg, v = 3 x 10^6 m/s, e = 1.6 x 10^-19 C

eV = 1/2 m v^2

V = mv^2 / 2 e

V = (9.1 x 10^-31) x (3 x 10^6)^2 / (2 x 1.6 x 10^-19)

V = 25.6 V

Answer:

[tex]8.89\cdot 10^4 V/m[/tex]

Explanation:

The electric field strength between the plates of a parallel plate capacitor is given by

[tex]E=\frac{\sigma}{\epsilon_0}[/tex]

where

[tex]\sigma[/tex] is the surface charge density

[tex]\epsilon_0[/tex] is the vacuum permittivity

Here we have

[tex]A=3.00 \cdot 3.00 = 9.00 cm^2 = 9.0\cdot 10^{-4} m^2[/tex] is the area of the plates

[tex]Q=0.708 nC = 0.708 \cdot 10^{-9} C[/tex] is the charge on each plate

So the surface charge density is

[tex]\sigma=\frac{Q}{A}=\frac{0.708\cdot 10^{-9}}{9.0\cdot 10^{-4} m^2}=7.87\cdot 10^{-7} C/m^2[/tex]

And now we can find the electric field strength

[tex]E=\frac{\sigma}{\epsilon_0}=\frac{7.87\cdot 10^{-7}}{8.85\cdot 10^{-12}}=8.89\cdot 10^4 V/m[/tex]

The electric field strength inside a parallel-plate capacitor with given dimensions and charge is approximately 88.89 kV/m. This is calculated using the capacitance and charge to find the voltage, which then helps determine the electric field strength.

To find the electric field strength inside a parallel-plate capacitor, you can use the relationship between the voltage (V), plate separation (d), and electric field (E). The formula is given by:

E = V / d

However, we are not given the voltage directly, but we have the charge (Q) and the plate area (A). The first step is to determine the capacitance (C) using the formula:

C = ε₀ (A / d)

Where ε₀ (epsilon naught) is the permittivity of free space (8.85 × 10⁻¹² F/m), A is the area of one plate, and d is the separation between the plates.

Given:

Plate area,

A = (3.00 cm × 3.00 cm)

= 9.00 cm²

= 9.00 × 10⁻⁴ m²

Plate separation,

d = 1.30 mm

= 1.30 × 10⁻³ m

Now calculate the capacitance:

C = ε₀ (A / d)

= (8.85 × 10⁻¹² F/m) (9.00 × 10⁻⁴ m² / 1.30 × 10⁻³ m)

≈ 6.13 × 10⁻¹⁵ F (Farads)

Next, use the charge (Q) and capacitance (C) to find the voltage (V):

Q = C × V

V = Q / C

Given charge,

Q = 0.708 nC

= 0.708 × 10⁻⁹ C

V = (0.708 × 10⁻⁹ C) / (6.13 × 10⁻¹⁵ F)

≈ 115.56 V

Finally, calculate the electric field strength:

E = V / d

E = 115.56 V / 1.30 × 10⁻³ m

≈ 88.89 kV/m

Thus, the electric field strength inside the capacitor is approximately 88.89 kV/m.

Answer:

explained below

Explanation:

solar system consist of sun, planets, asteroids, stars etc.

Milky Way is  a galaxy.

Combination of  many solar systems then forms galaxy.

Our solar system is a part of milky way galaxy.

If you consider marbles as planets and balls as stars.

so, we can say that in milky way is consist of 100-400 billions balls and 100 million marbles.

Final answer:

The tangential speed of the tack is 0.724 m/s, calculated using Vt = rω formula with the given values.

Explanation:

The tangential speed of the tack can be calculated using the formula:

Vt = rω

where Vt is the tangential speed, r is the distance from the rotation axis (0.357 m), and ω is the angular speed (2.03 times/s).

Substitute the values:

Vt = 0.357 m * 2.03 times/s = 0.724 m/s.

Answer:

Speed of the car after the collision is 1.2 m/s

Explanation:

It is given that,

Mass of first car, m₁ = 2 kg

Velocity of first car, v₁ = 6 m/s

Mass of second car, m₂ = 3 kg

Velocity of second car, v₂ = -2 m/s (it is travelling in opposite direction)

The two cars lock together and move along the road. This shows an inelastic collision. Let their common velocity is V. On applying the conservation of momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)V[/tex]

[tex]V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]V=\dfrac{2\ kg\times 6\ m/s+3\ kg\times (-2\ m/s)}{5\ kg}[/tex]

V = 1.2 m/s

So, after collision the speed of the cars is 1.2 m/s. Hence, this is the required solution.

Answer:

Percentage difference is 4.25 %.

Explanation:

Standard value of the density of wood, [tex]\rho_s=0.47\ g/cc[/tex]

Experimental value of the density of wood, [tex]\rho_e=0.45\ g/cc[/tex]

We need to find the percentage difference on the density of wood. It is given by :

[tex]\%=|\dfrac{\rho_s-\rho_e}{\rho_s}|\times 100[/tex]

[tex]\%=|\dfrac{0.47-0.45}{0.47}|\times 100[/tex]

Percentage difference in the density of the wood is 4.25 %. Hence, this is the required solution.

Answer:

18.73 m/s^2

Explanation:

f = 2662 rpm = 2662 / 60 rps

r = 6.725 cm = 0.06725 m

Acceleration, a = r w

a = r x 2 x pi x F

a = 0.06725 × 2 × 3.14 × 2662 / 60

a = 18.73 m/s^2

Answer:

y and length is directly relation

Explanation:

Given data

A single-slit diffraction pattern is formed on a distant scree

angles involved = small

to find out

what factor will the width of the central bright spot on the screen change

solution

we know that  for single slit screen formula is

mass ƛ /area = sin θ and y/L = sinθ

so we can say mass ƛ /area =  y/L

and y = mass length  ƛ / area       .................1

in equation 1 here we can see y and length is directly relation so we can say from equation 1 that  the width of the central bright spot on the screen change if the distance from the slit to the screen is doubled

C. I = 4.2A.

The magnetic field inside a solenoid is given by the equation:

B = μ₀NI/L

Clearing I for the equation above.

I = BL/μ₀N

With B = 1.0T, L = 3 x 10⁻²m, μ₀ = 4π x 10⁻⁷T.m/A and N = 5748turns

I = [(1.0T)(3 x 10⁻²m)]/[(4π x 10⁻⁷T.m/A)(5748turns)]

I = 4.15 ≅ 4.2A

The radius of the spinning chamber can be calculated using the centripetal force formula, F = m*v^2/r. The values for mass (m), velocity(v) and force(F) are given in the question, which can be substituted into the rearranged version of the formula for radius (r), r = m*v^2/F.

The subject question is based in physics, specifically the 'Circular Motion and Gravitation' topic, and is about finding the radius of a spinning chamber in an amusement park, with the information the force exerted and the speed of the chamber.

To find the radius, we would need to use the formula for the circular force, F = m*v^2/r, where F is the force, m is the mass, v the velocity and r the radius. Rearranging for r, the radius, we get r = m*v^2/F.

Substituting the values from the question, we have m = 84.4 kg (person's weight), v = 3.28 m/s (speed of the outer wall) and F = 488 N. Calculating these, we get r = (84.4 kg * (3.28 m/s)^2)/488 N.

This will give the radius of the chamber.

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The radius of the chamber can be determined using the concept of centripetal force. By substituting the given values into the formula for centripetal force (F = m * v² / r) and rearranging for r, the radius is found to be approximately 2.2 meters.

In order to solve the problem, one must understand the concept of centripetal force, which is described as a force that makes a body follow a curved path, with its direction orthogonal to the velocity of the body, towards the fixed point of the instantaneous center of curvature of the path.

In this scenario, the centripetal force is equal to the force pressing against the rider's back, which is 488N. This force can be calculated using the formula: F = m * v² / r, where F is the centripetal force, m is the mass of the rider, v is the speed of the outer wall, and r is the radius of the chamber.

Given that m = 84.4 kg, v = 3.28 m/s, and F = 488 N, by substituting these values into the formula and rearranging it we find that r = m * v² / F. Consequently, r = (84.4 kg * (3.28 m/s)²) / 488 N, which equals approximately 2.2 meters. Therefore, the radius of the chamber is approximately 2.2 meters.

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Answer:

option (b), (c), (g)

Explanation:

When the battery is disconnected, the charge on the plates of a capacitor remains same.

As the capacitance of the capacitor is directly proportional to the dielectric constant.

C = k C0

Now the charge remains same, So

Q = Q0

Q = k C0 x V0 / k

So, potential between the plates is V0 / k.

Energy, E = 1/2 x C X V^2 = 1/2 x k C0 x V0^2 / k^2 = E0 / k

So, energy becomes E0 / k.

Answer:

Part a)

[tex]\Delta U = 4\times 10^5eV[/tex]

Part b)

[tex]\Delta U = -4\times 10^5eV[/tex]

Explanation:

Part a)

Change in potential energy of a charge is given as

[tex]\Delta U = q\Delta V[/tex]

here we know that

[tex]q = e[/tex] for proton

also we have

[tex]\Delta V = 400 kV[/tex]

now we have

[tex]\Delta U = e(400 kV)[/tex]

[tex]\Delta U = 4\times 10^5eV[/tex]

Part b)

Change in potential energy of a charge is given as

[tex]\Delta U = q\Delta V[/tex]

here we know that

[tex]q = -e[/tex] for proton

also we have

[tex]\Delta V = 400 kV[/tex]

now we have

[tex]\Delta U = -e(400 kV)[/tex]

[tex]\Delta U = -4\times 10^5eV[/tex]

Answer:

False statement = There must be a non-zero net force acting on the object.  

Explanation:

An object is moving at a constant speed along a straight line. If the speed is constant then its velocity must be constant. We know that the rate of change of velocity is called acceleration of the object i.e.

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

a = 0

⇒ The acceleration of the object is zero.

The product of force and acceleration gives the magnitude of force acting on the object i.e.

F = m a = 0

⇒  The net force acting on the object must be zero.

So, the option (a) is not true. This is because the force acting on the object is zero. First option contradicts the fact.

Answer:

The surface charge density on the plate, [tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]

Explanation:

It is given that,

Area of parallel plate capacitor, [tex]A=2\times 10^{-3}\ m^2[/tex]

Separation between the plates, [tex]d=5\times 10^{-2}\ mm[/tex]

Electric field between the plates, [tex]E=8.5\times 10^{6}\ V/m[/tex]

We need to find the surface charge density on the plates. The formula for electric field is given by :

[tex]E=\dfrac{\sigma}{\epsilon}[/tex]

Where

[tex]\sigma[/tex] = surface charge density

[tex]\sigma=E\times \epsilon[/tex]

[tex]\sigma=8.5\times 10^{6}\ V/m\times 8.85\times 10^{-12}\ C^2/Nm^2[/tex]

[tex]\sigma=0.000075\ C/m^2[/tex]

[tex]\sigma=7.5\times 10^{-5}\ C/m^2[/tex]

Hence, this is the required solution.

Final answer:

The surface charge density on the plates of a parallel-plate air-filled capacitor 7.5225 x [tex]10^{-5}[/tex] [tex]C/m^2[/tex].

Explanation:

To determine the surface charge density on the plates of the parallel-plate air-filled capacitor, we can use the relationship between the electric field (E), the permittivity of free space
[tex](\sigma\))[/tex]. The electric field is defined as
[tex](E = \frac{\sigma}{\varepsilon_0}).[/tex]

Given that the electric field (E) is
[tex](8.5 \times 10^6 V/m)[/tex] and the permittivity of free space [tex]((\varepsilon_0))[/tex] is
[tex](8.85 \times 10^{-12} C^2/N \cdot m^2)[/tex], we can rearrange the formula to solve for the surface charge density [tex]((\sigma)):[/tex]
[tex](\sigma = E \cdot \varepsilon_0 = (8.5 \times 10^6 V/m) \times (8.85 \times 10^{-12} C^2/N \cdot m^2) = 7.5225 \times 10^{-5} C/m^2).[/tex]

Thus, the surface charge density on the plates is
[tex](7.5225 \times 10^{-5} C/m^2).[/tex]

Answer:

impulse = 6.09 kg m/s

Force = 6090 N

Explanation:

As we know that the impulse is defined as the change in momentum of the ball

so here we will have

[tex]\Delta P = mv_f - mv_i[/tex]

now we know that

[tex]v_f = 25 m/s[/tex]

initial speed is given as

[tex]v_i = -17 m/s[/tex]

now impulse is given as

[tex]\Delta P = 0.145(25 - (-17))[/tex]

[tex]\Delta P = 6.09 kg m/s[/tex]

Now we also know that average force is defined as the rate of change in momentum

[tex]F = \frac{\Delta P}{\Delta t}[/tex]

so we have

[tex]F = \frac{6.09}{1 \times 10^{-3}}[/tex]

[tex]F = 6.09 \times 10^3 N[/tex]

The magnitude of the impulse delivered by the bat to the ball is 6.09 kg*m/s. The magnitude of the average force exerted by the bat on the ball during a contact time of 1.0 ms is 6090 N.

The question relates to the concept of impulse and average force in a collision scenario. Impulse is given by the change in momentum of the object, and it can also be calculated as the average force (F) multiplied by the time duration (t) of the impact.

Considering the baseball scenario where the ball has a mass of 0.145 kg, enters the strike zone at 17.0 m/s and leaves in the opposite direction at a speed of 25.0 m/s, we can calculate the magnitude of the impulse delivered by the bat as follows:

To find the magnitude of the average force exerted by the bat on the ball, we use the time duration of the contact:

So, the magnitude of the average force exerted by the bat on the ball is 6090 N.

Answer:

The number of oscillation is 36.

Explanation:

Given that

Mass = 280 g

Spring constant = 3.3 N/m

Damping constant [tex]b=8.4\times10^{-3}\ Kg/s[/tex]

We need to check the system is under-damped, critical damped and over damped by comparing b with [tex]2m\omega_{0}[/tex]

[tex]2m\omega_{0}=2m\sqrt{\dfrac{k}{m}}[/tex]

[tex]2\sqrt{km}=2\times\sqrt{3.3\times280\times10^{-3}}=1.92kg/s[/tex]

Here, b<<[tex]2m\omega_{0}[/tex]

So, the motion is under-damped and will oscillate

[tex]\omega=\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}[/tex]

The number of oscillation before the amplitude decays to [tex]\dfrac{1}{e}[/tex] of its original value

[tex]A exp(\dfrac{-b}{2m}t)=A exp(-1)[/tex]

[tex]\dfrac{b}{2m}t=1[/tex]

[tex]t = \dfrac{2m}{b}[/tex]

[tex]t=\dfrac{2\times280\times10^{-3}}{8.4\times10^{-3}}[/tex]

[tex]t=66.67\ s[/tex]

We need to calculate the time period of one oscillation

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

[tex]T=\dfrac{2\times3.14}{\sqrt{\omega_{0}^2-\dfrac{b^2}{4m^2}}}[/tex]

[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{k}{m}-\dfrac{b^2}{4m^2}}}[/tex]

[tex]T=\dfrac{2\times3.14}{\sqrt{\dfrac{3.3}{280\times10^{-3}}-\dfrac{(8.4\times10^{-3})^2}{4\times(280\times10^{-3})^2}}}[/tex]

[tex]T=1.83\ sec[/tex]

The number of oscillation is

[tex]n=\dfrac{t}{T}[/tex]

[tex]n=\dfrac{66.67}{1.83}[/tex]

[tex]n=36[/tex]

Hence, The number of oscillation is 36.

Answer:

[tex]E = 9.66\times 10^{-6} N/C[/tex]

direction is Horizontal

Explanation:

As we know that the string is horizontal here

so the tension force in the string is due to electrostatic force on it

now we will have

[tex]F = qE[/tex]

so here the force is tension force on it

[tex]F = 6.57 \times 10^{-2} N[/tex]

[tex] Q = 6.80 \times 10^3 C[/tex]

now we have

[tex]6.57 \times 10^{-2} = (6.80 \times 10^3)E[/tex]

[tex]E = 9.66\times 10^{-6} N/C[/tex]

direction is Horizontal

The magnitude of the electric field on the charged sphere in this scenario is approximately 1.17 x 10^-5 N/C. The direction of the electric field is horizontal, which is the same direction as the tension in the thread.

To start, we can use the equilibrium condition where the tension in the thread is equal to the force due to the electric field and gravity on the sphere. The formula to calculate the electric force is F = qE, and the gravitational force is F = mg, where F is the force, q is the electric charge, E is the electric field, m is the mass of the object, and g is the gravity constant.

Tension - electric force - gravitational force equals zero: T - F_electric - F_gravity = 0. We fill in the previous formulas: T - qE -mg = 0. This can be rearranged to E = (T + mg) / q.

In this case, the sphere's mass m is 0.018 kg, the tension T is 6.57 x 10^-2 N, and the sphere's charge q is 6.80 x 10^3 C, and we use g = 9.81 m/s². So, E = ((6.57 x 10^-2) + (0.018 * 9.81)) / 6.80 x 10^3.

This leads to an electric field magnitude of approximately 1.17 x 10^-5 N/C. The direction of the electric field is the same as the direction of the tension, which is horizontal due to the thread being horizontal.

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Answer:

Speed of the cars after the collision is 3.34 m/s.

Explanation:

It is given that,

Mass of one car, m₁ = 1500 kg

Velocity of this car, v₁ = + 30 m/s ( in east )

Mass of other car, m₂ = 3000 kg

Velocity of other car, v₂ = - 20 m/s (in south)

The two cars stick together after the collision. It is a case of inelastic collision. Let v is the speed of cars after collision. It can be calculated using the conservation of linear momentum as :

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

[tex]v=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}[/tex]

[tex]v=\dfrac{1500\ kg\times 30\ m/s+3000\ kg\times (-20\ m/s)}{1500\ kg+3000\ kg}[/tex]

v = -3.34 m/s

So, the speed of the cars after the collision is 3.34 m/s. Hence, this is the required solution.

Answer:

10.83 m/s

Explanation:

Case 1:

m = mass attached to spring = 2 kg

x = compression of the spring = 40 cm = 0.40 m

k = spring constant

v₀ = maximum speed = 10 m/s

Using conservation of energy

maximum spring potential energy = maximum kinetic energy

(0.5) k x² = (0.5) m v₀²

k (0.40)² = (2) (10)²

k = 1250 N/m

Case 2 :

m = mass attached to spring = 2 kg

A = amplitude = 50 cm = 0.50 m

x = compression of the spring at half amplitude = A/2 = 50/2 = 25 cm = 0.25 m

k = spring constant = 1250 N/m

v = speed = ?

Using conservation of energy

maximum spring potential energy = spring potential energy + kinetic energy

(0.5) k A² = (0.5) k x² + (0.5) m v²

(1250) (0.50)² = (1250) (0.25)² + (2) v²

v = 10.83 m/s