A stone is tied to a string and whirled around in a circle at
aconstant speed. Is the string more likely to break when the
circleis horizontal or when it is vertical? Account for your
answer,assuming the constant speed is the same in each case.

Answer:

Part a)

[tex]v = 2.25 \times 10^6 m/s[/tex]

Part b)

[tex]r = 0.72 \times 10^{-10} m[/tex]

Explanation:

As we know that alpha particle remain at rest always so the energy of system of positron and alpha particle will remain constant

So we will have

[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]

here we know that

[tex]q_1 = 1.6 \times 10^{-19} C[/tex]

[tex]q_2 = 2(1.6 \times 10^{-19}) C[/tex]

also we have

[tex]r_1 = 2.00 \times 10^{-10} m[/tex]

[tex]r_2 = 1.00 \times 10^{-10} m[/tex]

[tex]v_1 = 3.00 \times 10^6 m/s[/tex]

[tex]m = 9.11 \times 10^{-31} kg[/tex]

now from above equation we have

[tex]2.304 \times 10^{-18} + 4.0995\times 10^{-18} = 4.608 \times 10^{-18} + \frac{1}{2}(9.11 \times 10^{-31})v^2[/tex]

[tex]v = 2.25 \times 10^6 m/s[/tex]

Part b)

Now when it come to rest then again by energy conservation we can say

[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]

now here final speed will be zero

[tex]2.304 \times 10^{-18} + 4.0995\times 10^{-18} = \frac{(9/times 10^9)(1.6\times 10^{-19})(2\times 1.6 \times 10^{-19})}{r_2}[/tex]

[tex]r = 0.72 \times 10^{-10} m[/tex]

Final answer:

To find the speed of the positron at a certain distance from the alpha particle and the separation distance when the positron comes to rest.

Explanation:

The speed of a positron at a certain distance from an α particle can be found using the conservation of energy. At 1.00 x 10^-10 m from the α particle, the positron's kinetic energy will be equal to the potential energy at that distance. Using this information, we can calculate the speed of the positron as it approaches the α particle.

To find the separation between the two when the positron comes to rest, we can set the final kinetic energy of the positron equal to zero and solve for the separation distance. This will give us the point at which the positron stops moving towards the α particle.

Answer:

1.81 x 10^-4 m/s

Explanation:

M = 98700 kg

m = 780 kg

d = 201 m

Let the speed of second asteroid is v.

The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.

[tex]\frac{GMm}{d^{2}}=\frac{mv^2}{d}[/tex]

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

Where, G be the universal gravitational constant.

G = 6.67 x 10^-11 Nm^2/kg^2

[tex]v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}[/tex]

v = 1.81 x 10^-4 m/s

Answer:

For 0.24 sec the person was in the air.

Explanation:

Given that,

Height = 1 m

Initial velocity = 3 m/s

We need to calculate the time

Using equation of motion

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Where, u = initial velocity

s = height

Put the value into the formula

[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]

[tex]4.9t^2+3t-1=0[/tex]

[tex]t = 0.24\ sec[/tex]

On neglecting the negative value of time

Hence, For 0.24 sec the person was in the air.

Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

[tex]K = \frac{1}{2}mv^{2}[/tex]

By substituting the values

[tex]1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}[/tex]

v = 4.9 m/s

Now convert metre per second into km / h

We know that

1 km = 1000 m

1 h = 3600 second

So, [tex]v = \left (\frac{4.9}{1000}   \right )\times \left ( \frac{3600}{1} \right )[/tex]

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

To find the speedometer reading, use the kinetic energy and mass to calculate the speed in meters per second and then convert it to kilometers per hour. The calculated speed is approximately 88 km/hr.

To determine the speedometer reading of the car in km/hr based on the kinetic energy given, we use the kinetic energy formula KE = ½ mv². We rearrange this formula to solve for speed (v): v = √(2KE/m).

Plugging in the provided kinetic energy of 1.2 x 10´ J and the car's mass of 1000 kg, we get:

v = √(2 × 1.2 x 10´ J / 1000 kg)

v = √(24 x 10´ J/kg)

v ≈ 24.49 m/s

To convert from m/s to km/hr, we multiply by 3.6:

v = 24.49 m/s × 3.6 km/hr per m/s ≈ 88.16 km/hr

Therefore, the speedometer would read approximately 88 km/hr.

Answer: Speed = 74.63 km/h

Explanation:

We are provided with the following information:

Distance traveled = 50 km

Time taken = 40 minutes = [tex]\frac{40}{60} \ hours[/tex]

Time taken = 0.67 hours

Since, we know that;

[tex]Speed = \frac{Distance}{Time}[/tex]

Speed = [tex]\frac{50}{0.67}[/tex]

Speed = 74.63 km/h

The average speed of the car is 74.63 km/h.

The average speed in km/h for a car that travels 50.0 km in 40.0 min can be calculated by converting the time from minutes to hours and then dividing the distance by the time.

Conversion: 40.0 min × (1 hr/60 min) = 0.67 hr

Average speed = distance/time = 50.0 km/0.67 hr = 74.63 km/h

Therefore, the average speed of the car is 74.63 km/h.

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Answer:

electron moving at 91.82 m/s

Explanation:

given data

distance d1 = 6 cm

distance d2 = 3 cm

to find out

how fast will the electron be moving

solution

we know potential energy formula that is

potential energy = [tex]k\frac{q1q2}{r}[/tex]

here k is 9× [tex]10^{9}[/tex] N-m²/C²

m mass of electron is 9.11 × [tex]10^{-31}[/tex] kg

and q = 1.60 × [tex]10^{-19}[/tex] C

we consider here initial potential energy = U1

U1 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.06^2}[/tex]

U1 = 3.84 × [tex]10^{-27}[/tex] J

and final potential energy = U2

U2 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.03}[/tex]

U2 = 7.68 × [tex]10^{-27}[/tex] J

and speed of electron = v

so we will apply here conservation of energy

0.5×m×v² = U2 - U1    ................1

so

0.5×9.11× [tex]10^{-31}[/tex] ×v² =  3.84 × [tex]10^{-27}[/tex]

v = 91.82 m/s

so electron moving at 91.82 m/s

Answer:

The speed of the electron is 91.86 m/s.

Explanation:

Given that,

Distance of electron from proton  = 6.00 cm

Distance of proton = 3.00 cm

We need to calculate the initial potential energy

[tex]U_{1}=\dfrac{kq_{1}q_{2}}{r}[/tex]

Put the value into the formula

[tex]U_{i}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{6.00\times10^{-2}}[/tex]

[tex]U_{i}=3.84\times10^{-27}\ J[/tex]

The final potential energy

[tex]U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{3.00\times10^{-2}}[/tex]

[tex]U_{f}=7.68\times10^{-27}\ J[/tex]

We need to calculate the speed of the electron

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=U_{f}-U_{i}[/tex]

Put the value into the formula

[tex]\dfrac{1}{2}\times9.1\times10^{-31}\times v^2=7.68\times10^{-27}-3.84\times10^{-27}[/tex]

[tex]v=\sqrt{\dfrac{2\times3.84\times10^{-27}}{9.1\times10^{-31}}}[/tex]

[tex]v=91.86\ m/s[/tex]

Hence, The speed of the electron is 91.86 m/s.

Answer:

V = 7.91 m/s(magnitude) @ 72.3°(direction)

Explanation:

We will have to do a parabolic movement analysis in order to determine the initial velocity. For this we will mainly use the following kinetic formulae:

[tex]1) X_{f}=X_{0}+V_{0}t+\frac{at^{2}}{2}\\\\2) V_{f}=V_{0}+at[/tex]

We will work backwards from the initial information: a final velocity of 11.5 m/s at a 78.2° angle. From this we can extract the final horizontal and vertical velocity:

[tex]V_{fx}=11.8\frac{m}{s}*Cos(78.2)=2.41\frac{m}{s}\\\\V_{fy}=11.8\frac{m}{s}*Sin(78.2)=11.55\frac{m}{s}[/tex]

We will use [tex]V_{fy}[/tex] with a negative value (since the diver is moving in a downwards direction). That way we can find the time elapsed between the highest point of the trayectory to instants before she contacts the water. At the maximum height of the movement, the velocity in the vertical direction is zero. Therefore we can rewrite equation 2) and solve for t:

[tex]-11.55\frac{m}{s}=0\frac{m}{s} - 9.8\frac{m}{s^{2}}t\\\\t= \frac{-11.55\frac{m}{s}}{- 9.8\frac{m}{s^{2}}}=1.17s[/tex]

This t that we found represents the time from the maximum height to the surface of the water. Now we can use equartion 1) to determine the maximum height. We must remember a starting point of 3.9m above the water:

[tex]0 = h+3.9m+0\frac{m}{s}*t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\ \\h=\frac{9.8\frac{m}{s^{2}}1.17s^{2}}{2}-3.9m=2.9m\\\\[/tex]

Since we now know what is the maximum height we can determine the time that it took for the diver to reach that point using equation 1) and 2):

[tex]2.9m =V_{0y}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\0\frac{m}{s}= V_{0y}t-9.8\frac{m}{s^{2}}t[/tex]

We will solve for [tex]V_{ox}[/tex] in equation 2) and replace it in equation 1):

[tex]V_{0y} = 9.8\frac{m}{s^{2}}t\\\\2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}[/tex]

We solve for t (time from diving board to maximum height):

[tex]2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\2.9m=\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\5.8m=9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{ \frac{5.8m}{9.8\frac{m}{s^{2}}}}=0.77s[/tex]

Finally we can replace our t value in [tex]V_{0y}t=9.8\frac{m}{s^{2}}t[/tex] to obtain:

[tex]V_{0y}t=9.8\frac{m}{s^{2}}t=9.8\frac{m}{s^{2}}*0.77s=7.54\frac{m}{s}[/tex]

With this we have the information we need in order to determine the initial velocity of the diver. The magnitude will be calculated from the initial horizontal and vertical velocities (consider final and initial horizontal velocity to be equal in PARABOLIC MOVEMENTS):

[tex]V_{0}=\sqrt{V_{0x}^{2}+V_{0y}^{2}}=\sqrt{(2.41\frac{m}{s})^{2}+(7.54\frac{m}{s})^{2}}=7.91\frac{m}{s}[/tex]

And the direction is determined by an arctangent:

[tex]tan\theta=\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}}\\\\\theta=arctan(\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}})=72.3[/tex]

Final answer:

To find the diver's initial velocity, use equations involving vertical and horizontal components considering her final speed and displacement. The magnitude is 15.93 m/s, and the direction is horizontal.

Explanation:

To determine the diver's initial velocity, we first find the vertical component of her velocity using the equation vf=vi+at. Here, vf = 11.8 m/s, vi is the vertical component of the initial velocity, a = -9.81 m/s^2, and t = time in the air. Solving for vi, we get vi = 15.93 m/s. Then, to find the horizontal component of the initial velocity, we use the equation dx=vit+1/2at^2. Since the diver contacts the water at the highest point, her vertical displacement is 3.9m, and we can find the horizontal component of the initial velocity to be 0 m/s. Therefore, her initial velocity magnitude is 15.93 m/s, and her initial velocity direction is horizontal.

Answer:

Statement E is false.

Explanation:

Statement a says that the object would have moved a distance of 10 m.

At t = 0

x0  = 0

y0 = 0

At t= 2

x2 = 2 * 2^2 = 8

y2 = 3 * 2 = 6

[tex]d2 = \sqrt{x2^2 + y2^2}[/tex]

[tex]d2 = \sqrt{8^2 + 6^2} = 10 m[/tex]

Statement A is true.

Statement B says that the momentum of the object would be 4*i+1.5*j

To test this we need to know the speed.

The speed is the first derivative by time:

v = 4*t*i + 3*j

At t = 2 the speed is

v2 = 8*i + 3*j

The momentum is

P = m * v

P2 = 0.5 * (8*i + 3*j) = (4*i+1.5*j)

This statement is true.

Statement C says that the kinetic energy of the object would be 18 J.

The magnitude of the speed is

[tex]v = \sqrt{8^2 + 3^2} = 8.54 m/s[/tex]

The kinetic energy is

Ek = 1/2 * m * v^2

Ek = 1/2 * 0.5 * 8.54^2 = 18 J

Statement C is true

Statement D says that the force acts perpenducular to the direction of the unit vector j.

The acceleration is the second derivative respect of time

a = 4*i + 0*j

Since there is no acceleration in the direction of j we can conclude that the force is perpendicular to the direction of j.

Statement D is true.

Statement E says that the force acting on the object is of 4 N

f = m * a

f = 0.5 * 4 = 2 N

Statement E is false.

Answer:

(a) [tex]E_{ c} = 112.5 \mu J[/tex]

(b) [tex]E'_{ c} = 18 \mu J[/tex]

Solution:

According to the question:

Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]

Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]

(a)The Energy stored in the Capacitor is given by:

[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]

[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]

[tex]E_{ c} = 112.5 \mu J[/tex]

(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:

[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]

[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]

[tex]E'_{ c} = 18 \mu J[/tex]

(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]

(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]

The given parameters;

The energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]

When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;

[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]

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Answer:48 %

Explanation:

Given

outside temperature [tex]=25^{\circ}\approx 298 K[/tex]

Internal temperature[tex]=300^{\circ}\approx 573 K[/tex]

Maximum efficiency is carnot efficiency which is given by

[tex]\eta =1-\frac{T_L}{T_H}[/tex]

[tex]\eta =1-\frac{298}{573}[/tex]

[tex]\eta =\frac{275}{573}=0.4799[/tex]

i.e. [tex]47.99 %\approx 48 %[/tex]

No it is not possible to achieve this efficiency because carnot engine is the ideal engine so practically it is not possible but theoretically it is possible

Answer:

Truck will be at a distance of 720 m

Explanation:

Speed of the truck is given as 90 m/sec

Time for which truck travel t = 10 seconds

We have to find the distance that how much truck travel in 8 sec with speed of 90 m/sec

We know that distance is given by [tex]distance=velocity\times time[/tex]

So distance traveled by truck in 8 sec will be equal to [tex]=90\times 8 =720m[/tex]

So truck will be at a distance of 720 m

Answer:4.95 m/s

Explanation:

Given

Ball rises to a height of 1.25 m

Let u be the velocity of player while leaving ground .

Its final velocity will be zero as he reaches a height of 1.25 m

thus

[tex]u^2=2gh[/tex]

[tex]u=\sqrt{2\times 9.81\times 1.25}[/tex]

[tex]u=\sqrt{24.525}[/tex]

u=4.952 m/s

Final answer:

To reach a height of 1.25 meters, a basketball player must leave the ground with an initial upward velocity of 4.95 m/s, calculated using the kinematic equation for motion under constant acceleration due to gravity.

Explanation:

To find the required initial velocity, we use the kinematic equation which relates distance (d), initial velocity ([tex]v_i[/tex] ), final velocity ([tex]v_f[/tex]), acceleration (a), and time (t). We can ignore time since we are not asked for it and we know that the final velocity at the peak of the jump is 0 m/s because the ball will momentarily stop moving upwards before it starts falling down. We will use the acceleration due to gravity, which is approximately 9.81 m/s².

The equation is:

[tex]v_f^2 = v_i^2 + 2ad[/tex]

Setting [tex]v_f[/tex] to 0 m/s (since the ball stops at the peak), the equation simplifies to:

[tex]v_i[/tex] = √(2ad)

Plugging in the values we get:

[tex]v_i[/tex]  = √(2 * 9.81 m/s² * 1.25 m) = √(24.525 m²/s²)

[tex]v_i[/tex]  = 4.95 m/s

Therefore, the basketball player must leave the ground with an initial velocity of 4.95 m/s to reach a height of 1.25 meters.

Answer:

option D ) is correct.

Explanation:

When two charges Q₁ and Q₂ which are unequal in magnitude as well as in nature are at a distance of R then a force F is created between them which is given by the expression

 F = k ( Q₁ X Q₂) / R²

This force acts equally on both the charges Q₁ and Q₂ irrespective of the fact that they are unequal charges . In other words both Q₁ and Q₂ experience same force F whose value is given above.

In the given case too charges are unequal in magnitude and in nature . They will attract each other with equal force . So if A experiences force F , B too will experience the same force,

Hence option D ) is correct.

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

Answer:

153.6 kN

Explanation:

The elastic constant k of the block is

k = E * A/l

k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m

0.12% of the original length is:

0.0012 * 0.25 m = 0.0003  m

Hooke's law:

F = x * k

Where x is the change in length

F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)

The compressive load will generate a stress of

σ = F / A

F = σ * A

F = 80*10^6 * 0.048 * 0.04 = 153.6 kN

The smallest admisible load is 153.6 kN

In this exercise we have to use the knowledge of force to calculate that:

[tex]153.6 kN[/tex]

Then from the formula of the elastic constant we get that:

[tex]k = E * A/l\\k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m[/tex]

With that, now using Hooke's law we find that:

[tex]F = x * k\\F = 0.0003 * 729.6*10^6 = 218.88 kN[/tex]

The compressive load will generate a stress of

[tex]\sigma = F / A\\F = \sigma * A\\F = 80*10^6 * 0.048 * 0.04 = 153.6 kN[/tex]

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Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we know that

[tex]q_1 = +8.4\mu C[/tex]

[tex]q_2 = +5.6 \mu C[/tex]

force between two charges is given as

[tex]F = 0.66 N[/tex]

now we have

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

[tex]0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}[/tex]

[tex]r = 0.8 m[/tex]

so it is separated by 80 cm distance

Final answer:

The distance between the charges is approximately 20.73 cm.

Explanation:

To determine the distance between the charges, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is: F = k × (q1 × q2) / r², where F is the force, k is the Coulomb's constant (8.99 x 10⁹ Nm²/C²), q1 and q2 are the charges of the particles, and r is the distance between them.

In this case, we have the force (F) as 0.66 N, q1 as 8.4 uC, q2 as 5.6 uC, and we need to find the distance (r). Plugging these values into the formula, we get:

0.66 = (8.99 x 10⁹) × ((8.4 x 10⁻⁶) × (5.6 x 10⁻⁶)) / r²

Solving for r, we can rearrange the equation to get:

r² = ((8.99 x 10⁹) × ((8.4 x 10⁻⁶) × (5.6 x 10⁻⁶))) / 0.66

r² = 8.99 x 10⁹ x 8.4 x 10⁻⁶ x 5.6 x 10⁻⁶ / 0.66

r² = 8.99 x 8.4 x 5.6 / 0.66

r² = 429.643

r = √429.643

r ≈ 20.73 cm

Ans:

12500 N/C

Explanation:

Side of square,  a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

[tex]E = \frac{Kq}{r^2}[/tex]

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD = [tex]\sqrt{2}\times a[/tex]

where, a be the side of the square

So, Electric field at B due to charge at A

[tex]E_{A}=\frac{Kq}{a^2}[/tex]

[tex]E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]

EA = 6531.32 N/C

Electric field at B due to charge at C

[tex]E_{C}=\frac{Kq}{a^2}[/tex]

[tex]E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]

Ec = 6531.32 N/C

Electric field at B due to charge at D

[tex]E_{D}=\frac{Kq}{2a^2}[/tex]

[tex]E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}[/tex]

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

[tex]E=\sqrt{E_{x}^{2}+E_{y}^{2}}[/tex]

[tex]E=\sqrt{8840.5^{2}+8840.5^{2}}[/tex]

E = 12500 N/C

Explanation:

Answer:

[tex]1.250\times 10^4\ N/C.[/tex]

Explanation:

Given:

According to the Coulomb's law, the strength of the electric field at a point due to a charge [tex]q[/tex] at a point [tex]r[/tex] distance away is given by

[tex] E = \dfrac{kq}{r^2}[/tex]

where,

[tex]k[/tex] = Coulomb's constant = [tex]9\times 10^9\ Nm^2/C^2[/tex].

The direction of the electric field is along the line joining the point an d the charge.

The electric field at the point P due to charge at A is given by

[tex]E_A = \dfrac{kq}{a^2}[/tex]

Since, this electric field is along positive x axis direction, therefore,

[tex]\vec E_A = \dfrac{kq}{a^2}\ \hat i.[/tex]

[tex]\hat i[/tex] is the unit vector along the positive x-axis direction.

The electric field at the point P due to charge at B is given by

[tex]E_B = \dfrac{kq}{a^2}[/tex]

Since, this electric field is along negative y axis direction, therefore,

[tex]\vec E_B = \dfrac{kq}{a^2}\ (-\hat j).[/tex]

[tex]\hat j[/tex] is the unit vector along the positive y-axis direction.

The electric field at the point P due to charge at C is given by

[tex]E_C = \dfrac{kq}{r^2}[/tex]

where, [tex]r=\sqrt{a^2+a^2}=a\sqrt 2[/tex].

Since, this electric field is along the direction, which is making an angle of [tex]45^\circ[/tex] below the positive x-axis direction, therefore, the direction of this electric field is given by [tex]\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)[/tex].

[tex]\vec E_C = \dfrac{kq}{2a^2}\ (\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j))\\=\dfrac{kq}{2a^2}\ (\dfrac{1}{\sqrt 2}\hat i-\dfrac{1}{\sqrt 2}\hat j)\\=\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\[/tex]

Thus, the total electric field at the point P is given by

[tex]\vec E = \vec E_A+\vec E_B +\vec E_C\\=\dfrac{kq}{a^2}\hat i+\dfrac{kq}{a^2}(-\hat j)+\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\=\left ( \dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )\hat i+\left (\dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )(-\hat j)\\=\dfrac{kq}{a^2}\left [\left ( 1+\dfrac{1}{2\sqrt 2} \right )\hat i+\left (1+\dfrac{1}{2\sqrt 2}  \right )(-\hat j)\right ]\\=\dfrac{kq}{a^2}(1.353\hat i-1.353\hat j)[/tex]

[tex]=\dfrac{(9\times 10^9)\times (4.25\times 10^{-6})}{2.42^2}\times (1.353\hat i-1.353\hat j)\\=(8.837\times 10^3\hat i-8.837\times 10^3\hat j)\ N/C.[/tex]

The magnitude of the electric field at the given point due to all the three charges is given by

[tex]E=\sqrt{(8.837\times 10^3)^2+(-8.837\times 10^3)^2}=1.250\times 10^4\ N/C.[/tex]

Answer:

All of the above

Explanation:

The correct answer is option E (All of the above)

All the options are alternative source of energy.

The option given are not the traditional way (energy production from coal) of extracting energy as the loss of natural resources does not take place in these source of energy.

energy extracted from wind,  solar light , hydrogen ,etc are the source of energy  the alternative source of production of energy because do not exploit the natural resources, reduce the carbon emission and energy produced by them is clean energy.

Answer:1.44 s,10.17 m

Explanation:

Given

two balls are separated by a distance of 41.1 m

One person drops the ball from a height of 41.1 and another launches a ball with velocity of 41.1 m exactly at the same time.

Let the ball launches from ground travels a distance of x m in t sec

For Person on window

[tex]41.1-x=ut+\frac{1}{2}gt^2[/tex]

[tex]41.1-x=0+\frac{1}{2}\times 9.81\times  t^2--------1[/tex]

For person at ground

[tex]x=v_ft-\frac{1}{2}gt^2---------2[/tex]

add (1) & (2)

[tex]41.1=v_ft-\frac{1}{2}gt^2+\frac{1}{2}gt^2[/tex]

[tex]41.1=v_ft[/tex]

and [tex]v_f[/tex] is given by

[tex]v_f=\sqrt{2\times 9.81\times 41.1}=28.39 m/s[/tex]

[tex]t=\frac{41.1}{28.39}=1.44 s[/tex]

Substitute value of t in equation 1

[tex]41.1-x=0+\frac{1}{2}\times 9.81\times  1.44^2[/tex]

41.1-x=10.171 m

Thus the two ball meet at distance of 10.17 m below the window.

Answer:

The answer is 2.3 eV

Explanation:

As energy should be conserved, we have to understand that the energy gap between the ground state and excited state is equal to the energy of the emitted photons.

The wavelength of the emitter photons is

[tex]\lambda = 539\ nm = 539 \times 10^{-9}\ m[/tex],

then their energy is

[tex]E = \frac{h c}{\lambda} = 36.855 \times 10^{-20}\ J[/tex]

where h is the Planck constant,

[tex]h = 6.626 \times 10^{-34}\ Js[/tex]

and c is the speed of light,

[tex]c = 2.998 \times 10^{8}\ m/s[/tex].

If we want our answer in electron-volts instead of joules...

[tex]1 J = 6.242 \times 10^{18}\ eV[/tex]

then

[tex]E = 36.855 \times 10^{-20} J = 2.3\ eV[/tex].

The energy gap between the ground state and excited state in the laser material is approximately 3.69 x 10^-19 Joules. This was calculated using the energy equation of photons, considering the given wavelength of 539 nm.

The energy gap between the ground state and the excited state in a laser material can be calculated using the energy equation of photons, E=hc/λ, where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength (539 x 10^-9 m in this case). Upon calculation, we find the energy to be about 3.69 x 10^-19 Joules. Hence, the energy gap between the ground state and the excited state is 3.69 x 10^-19 Joules.

This calculation entails quantum physics because it is based on the principle that when an electron in an atom moves from an excited state to a ground state, it emits energy in the form of a photon. This emitted photon has a certain wavelength and frequency, in this case, 539 nm.

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Answer:

It will have an acceleration of [tex]0.71 m/s^{2}[/tex].

Explanation:

Newton’s second law is defined as:  

[tex]F = ma[/tex]   (1)  

Equation 1 can be rewritten for the case of the acceleration:

[tex]a = \frac{F}{m}[/tex]

The force and the acceleration are directly proportional (if one increase the other will increase too).

[tex]a = \frac{5.0 N}{7.0 Kg}[/tex]

But 1 N is equivalent to [tex]1 Kg.m/s^{2}[/tex]

[tex]a = \frac{5.0 Kg.m/s^{2}}{7.0 Kg}[/tex]

[tex]a = 0.71 m/s^{2}[/tex]    

So a body will have an acceleration as a consequence of the action of a force.  

Answer:

x = 2.044 m

Explanation:

given data

initial vertical component of velocity = Vy = 2sin18

initial horizontal component of velocity = Vx = 2cos18

distance from the ground yo = 5m

ground distance y = 0

from equation of motion

[tex]y = yo+ V_y t +\frac{1}{2}gt^2[/tex]

[tex]0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2[/tex]

solving for t

t = 1.075 sec

for horizontal motion

[tex]x = V_x t[/tex]

x = 2cos18*1.075

x = 2.044 m

The motion of sand is due to the movement of conveyor belt. The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.

The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.

The second equation of the motion for distance can be given as,

[tex]y=ut+\dfrac{2}{2}gt^2[/tex]

Here, [tex]u[/tex] is the initial body, [tex]g[/tex] is the acceleration of the body due to gravity and [tex]t[/tex] is the time taken by it.

Given information-

The conveyor is tilted at an angle of 18° above the horizontal.

The Sand is moved without slipping at the rate of 2 m/s.

The sand is collected in a big drum 5 m below the end of the conveyor belt.

The horizontal component of the velocity is given as,

[tex]v_y=2\cos 18[/tex]

The vertical component of the velocity is given as,

[tex]v_y=2\sin18[/tex]

Put the value in the above equation as,

[tex]y-y_0=v_yt+\dfrac{1}{2}gt^2[/tex]

[tex]0-5=2\sin18 (t)+\dfrac{1}{2}\times9.8\tiems t^2\\t=1.075\rm sec[/tex]

The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is,

[tex]d=v_xt\\d=2\cos18\times1.075\\d=2.044\rm m[/tex]

Thus, the horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.

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Answer : The velocity of the student relative to the bus is 1 m/s to the east.

Explanation :

Relative speed : It is defined as the speed of a moving object when compared to the another moving object.

There are two condition of relative speed.

(1) It two bodies are moving in same direction then the relative speed will be the difference of two bodies.

[tex]\text{Relative speed}=|V_2-V_1|[/tex]

(2) It two bodies are moving in opposite direction then the relative speed will be the sum of two bodies.

[tex]\text{Relative speed}=V_1+V_2|[/tex]

As per question,

The speed of bus, [tex]V_1[/tex] = 10 m/s to east

The speed of student, [tex]V_2[/tex] = 9 m/s to east

In this problem, the two bodies are moving in same direction then the relative speed will be the difference of two bodies.

[tex]\text{Relative speed}=|V_2-V_1|[/tex]

[tex]\text{Relative speed}=|9-10|m/s[/tex]

[tex]\text{Relative speed}=1m/s[/tex]

Therefore, the velocity of the student relative to the bus is 1 m/s to the east.

Explanation:

Given that, you are taking a picture of a giraffe that is standing far away from you. The image is just too small, so you swap the 60-mm-focal-length lens in your camera for a 960 mm telephoto lens such that,

[tex]m_1=960\ mm[/tex]

[tex]m_2=60\ mm[/tex]

We need to find the the factor with which the size of the image increases. It can be calculated by taking ratios of both [tex]m_1\ and\ m_2[/tex]

So, [tex]\dfrac{m_1}{m_2}=\dfrac{960}{60}[/tex]

[tex]\dfrac{m_1}{m_2}=16[/tex]

So, the size of image increases by a factor of 16. Hence, this is the required solution.

Answer:

Electric force, [tex]F=8.16\times 10^{-15}\ N[/tex]

Explanation:

A protein molecule in an electrophoresis gel has a negative charge. Electric field, E = 1700 N/C

There are 30 excess electrons. The charge on 30 electrons is, q = 30e

[tex]q=30\times 1.6\times 10^{-19}\ C=4.8\times 10^{-18}\ C[/tex]

The electric force in terms of electric field is given by :

[tex]F=q\times E[/tex]

[tex]F=4.8\times 10^{-18}\ C\times 1700\ N/C[/tex]

[tex]F=8.16\times 10^{-15}\ N[/tex]

So, the magnitude of the electric force on a protein is [tex]8.16\times 10^{-15}\ N[/tex]. Hence, this is the required solution.

Answer:

8.16 x 10^-15 N

Explanation:

Number of excess electrons = 30

Electric field, E = 1700 N/C

Charge of one electron, e = 1.6 x 10^-19 C

Total charge, q = charge of one electron x number of electrons

q = 30 x 1.6 x 10^-19 = 48 x 10^-19 C

The relation between the electric field and the force is given by

F = q E  

F = 48 x 10^-19 x 1700

F = 8.16 x 10^-15 N

Answer: 208.3 s

Explanation:

Hi!

You need to calculate the velocity of the boat relative to the shore, in each trip.

Relative velocities are transformed according to:

[tex]V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\[/tex]

Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s

In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s

In upstream trip we have:

[tex]V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms[/tex]

In dowstream we have:

[tex]V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms[/tex]

In both cases the distance travelled is 1000m. Then the time it takes the round trip is:

[tex]T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s[/tex]

Answer:

Explanation:

We shall apply Pascal's law here to find the solution .According to this law if we apply a pressure on a particular area over an in-compressible liquid it gets transferred to other point inside or over the surface of the liquid without getting diminished.

in the present case force is applied on a car to crush it which is enormous in magnitude. This force is 20000 N, which is also called output force  Let the surface on which car is placed be A₂. We apply input force F₁ at other point over a small area be it A₁.

According to Pascal law ,

20000/ A₂ = F₁ / A₁

F₁ = 20000 X (A₁ / A₂)

= 20000 X 1/8

= 2500 N.

To find the uncertainty in the area of a circle with a radius containing two significant figures, use the formula A = πr² and round the result to two significant figures, taking into account the rules of rounding and the implications for uncertainty.

The question pertains to the calculation of the uncertainty in the area of a circle when provided with the radius which has a certain number of significant figures. Given the radius (r) of a circle as 4.5×104 cm, which contains two significant figures, the area (A) can be calculated using the formula A = πr². However, since π is a constant with an infinite number of significant digits and the radius has only two, the answer must be limited to two significant figures to reflect the precision of the given measurement. Consequently, the area will be calculated and then rounded to two significant figures to give an approximate value of the uncertainty.

Assuming the calculated area is A, we must consider the rounding rules to determine the uncertainty. When we round to two significant figures, this may imply an uncertainty, as the final digit in the rounded number can vary up or down by one. This is an important concept to understand, as rounding can lead to a loss of information and increase the relative uncertainty of the final answer.

Answer:D-velocity

Explanation:

Given

Constant linear acceleration i.e. acceleration is constant

a=constant

and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]

Thus velocity is changing uniformly to give constant acceleration

While displacement and distance are function of time thus they are not changing uniformly.

During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.

Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.

Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.

Therefore,

During constant linear acceleration, the following quantity changes uniformly: velocity.

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Answer:

Answer:

energy

Explanation:

The capacity to do work is called energy.

There are several types of energy. Some of the types of energy are given below:

1. Kinetic energy

2. Potential energy

3. Solar energy

4. thermal energy

5. Sound energy

6. Nuclear energy, etc.

The SI unit of energy is Joule.

Some other units of energy are

erg, eV, etc.

1 J = 10^7 erg

1 eV = 1.6 x 10^-19 J

The commercial unit of energy is kilowatt hour.

1 Kilo watt hour = 3.6 x 10^6 J

Explanation: