A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if the image (formed) is real?

Answer:

24 cm

Explanation:

Given:

Mass of proton = 1.67 × 10⁻²⁷ Kg

kinetic energy = 2.5 × 10⁻¹³ J

magnetic field in the cyclotron, B = 0.75 T

Now,

Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]  = 2.5 × 10⁻¹³ J

where, v is the velocity of the electron

or

[tex]\frac{1}{2}\times1.67\times10^{-27}\times v^2[/tex]  = 2.5 × 10⁻¹³ J

or

v² = 2.99 × 10¹⁴

or

v = 1.73 × 10⁷ m/s

also,

centripetal force = magnetic force

or

[tex]\frac{mv^2}{r}[/tex]  = qvB

q is the charge of the electron

r is the radius of the dipole magnets

on substituting the respective values, we get

[tex]\frac{1.67\times10^{-27}\times1.73\times10^7}{r}[/tex]  = 1.6 × 10⁻¹⁹ × 0.75

or

r = 0.2408 m ≈ 24 cm

Hence, the correct answer is 24 cm

The radius of the dipole magnets in the cyclotron is mathematically given as

r=24 cm

Question Parameter(s):

A cyclotron is designed to accelerate protons (mass 1.67 x 10^-27 kg)

Up to kinetic energy of 2.5 x 10^-13 J.

Generally, the equation for the Kinetic energy is mathematically given as

K.E=0.5mv^2  

Therefore

0.5*1.67*10^{-27}*v^2  = 2.5 × 10^{-13} J

v = 1.73 × 10⁷ m/s

In conclusion

mv^2/r  = qvB

(1.67*10^{-27}*1.73*10^7)/r  = 1.6 × 10^{-19}* 0.75

r=24 cm

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Answer:

length = 2L, mass = M/2, and maximum angular displacement = 1 degree

Explanation:

We consider only small amplitude oscillations (like in this case), so that the angle θ is always small enough. Under these conditions recall that the equation of motion of the pendulum is:

[tex]\ddot{\theta}=\frac{g}{l}\theta[/tex]

And its solution is:

[tex]\theta=Asin(\omega t + \phi)[/tex]

Where [tex]\omega=\sqrt\frac{g}{l}[/tex] are the angular frequency of the oscillations, from which we determine their period:

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

Therefore the period of a pendulum will only depend on its length, not on its mass or angle, for angles small enough. So, the answer is the one with the greater length.

The period of a pendulum is only determined by its length and the acceleration due to gravity, and is independent of other factors such as mass and maximum displacement.

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. Therefore, none of the combinations of variables given will result in a greater period for the pendulum. The period is only determined by the length and the value of acceleration due to gravity.

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Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]h=\dfrac{v^2}{2g}[/tex]

[tex]h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}[/tex]                    

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.          

Answer:

The Answer is Letter A :)

Explanation:

When the ice skater sticks out her hands, she spins slower. Then she rotates really fast when she pulls her arms to her sides. This is an example of a fundamental law in physics called Conservation of Angular Momentum. This law relates two observable quantities: the speed of rotation and the shape.

I Hope It's Helpful

Hint The Brainliest :)

Answer:

Option (A)

Explanation:

Angular momentum is usually defined as a vector quantity that controls the rotational momentum of an object, body or a system. It is the product of the three quantities namely the mass, radius, and velocity of the rotating body.

The given question is based on the conservation of the angular momentum, where an ice skater when pulls in her arms during the time of spinning, the angular momentum remains conserved. It does not change.

Thus, the correct answer is option (A).

Answer:

a) 21.8 mts

b) 10.2 seconds

c) 4.28 m/s

Explanation:

Because the blimp is filled with helium, the acceleration won't be the gravity. We have to calculate the new acceleration:

[tex]a=\frac{Vf-Vo}{t}\\\\a=\frac{0-4.28}{`10.2}\\\\a=0.420 m/s^2[/tex]

in order to obtain the height we have to use the formulas of accelerated motion problems:

[tex]X=Vo*t+\frac{1}{2}*a*t^2\\\\X=4.28*(10.2)+\frac{1}{2}*(-0.420)*(10.2)^2\\X=21.8mts[/tex]

we can calculate the time with the same formula:

[tex]-21.8=0*t+\frac{1}{2}*(-0.420)*t^2\\solving\\t=10.2 seconds[/tex]

the velocity at the same height is given by:

[tex]Vf^2=Vo^2+2*a*x\\Vf=\sqrt{2*(-0.420)*(-21.8)} \\Vf=-4.28m/s[/tex]

the speed would be 4.28m/s because is a scalar value.

Answer: If a force on an object is aimed in the direction of the object’s velocity, the force does positive work (b).

Explanation:

Hi, the answer is oprion b. positive work.

If a force on an object is aimed in the direction of the object’s velocity, the force does positive work.

An object's kinetic energy will only change if the force acting on the object changes the object's speed. This will only happen if there is a component of the force in the direction that the object moves.

So, A force will do work only if the force has a component in the direction that the object moves.

The answer to the question is 'b. positive work.' If the force on an object is in the direction of the object’s velocity, positive work is done as it adds energy to the system.

If a force on an object is aimed in the direction of the object’s velocity, the force does positive work. Work is defined as the product of the force times the displacement times the cosine of the angle between them.

Since the angle between the force and the displacement is zero when the force is in the direction of velocity (cos 0 = 1), the work done is positive. The work done by a force in the direction of an object's motion adds energy to the system. 

An example of this would be pushing a lawn mower forward, where the force applied and the direction of the mower's motion are the same.

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Answer:

Because of that but also because of its magnetic properties.

Explanation:

But its abundance in water and fat, since Magnetic Resonance Imaging uses our own body's magnetic properties to produce the images, and for this is much better to use something that is abundant in our own body and has magnetic properties, that is, and hydrogen nucleus ( single proton), being useful since it behaves like a small magnet.

Answer:

(a) 6.56 km

(b) [tex]59.68^\circ[/tex] north of west

Explanation:

Given:

[tex]\vec{d}_1= 3.35\ km\ east = 3.35\ km\ \hat{i}\\\vec{d}_2= 9.31\ km\ north = 9.31\ km\ \hat{j}\\\vec{d}_3= 6.66\ km\ west = -6.61\ km\ \hat{i}\\\vec{d}_4= 3.65\ km\ south = -3.65\ km\ \hat{j}\\[/tex]

Let the resultant displacement vector be [tex]\vec{D}[/tex].

As the resultant is the vector sum of all the vectors.

[tex]\therefore \vec{D}=\vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} =(3.35\ km\ \hat{i})+(9.31\ km\ \hat{j})+(-6.61\ km\ \hat{i})+(-3.65\ km\ \hat{j})\\\Rightarrow \vec{D} =-3.31\ km\ \hat{i}+5.66\ km\ \hat{j}\\\textrm{Magnitude of the resultant displacement vector} = \sqrt{(-3.31)^2+(5.66)^2}\ km= 6.56\ km\\[/tex]

[tex]\textrm{Angle with the positive west} = \theta = \tan^{-1}(\dfrac{5.66}{3.31})= 59.68^\circ[/tex]

The magnitude of the resultant vector sum of all the displacements is 6.55 km, and its direction is 59.65° relative to due west.

To find the magnitude of the resultant vector (D = A + B + C + D), we first need to add the displacement vectors head-to-tail. Since each vector points in a cardinal direction, we can add their magnitudes algebraically after designating the direction for each: east and north are positive, while west and south are negative. The resultant in the east-west direction is 3.35 km (east) - 6.66 km (west) = -3.31 km (west). The resultant in the north-south direction is 9.31 km (north) - 3.65 km (south) = 5.66 km (north).

Now we use the Pythagorean theorem to find the magnitude of the resultant:

R = √((-3.31 km)² + (5.66 km)²) = √(10.96 km² + 32.05 km²) = √43.01 km² = 6.55 km

The direction is determined by the tangent of the angle relative to the west. Let φ be the angle, then

tan(φ) = opposite / adjacent = 5.66 km / 3.31 km

φ = arctan(5.66 / 3.31) = 59.65° (north of west)

Therefore, the resultant vector is 6.55 km in magnitude, at an angle of 59.65° relative to due west.

Answer:

Magnitude of change in momentum = 4.65 kg.m/s

Magnitude of impulse = 4.65 kg.m/s

Magnitude of the average force applied by the bat = 1550 N

Explanation:

Mass of the cricket ball, m = 0.155 kg

Initial velocity of the ball, u = 35.0 m/s

final velocity of the ball after hitting the bat, v = 65.0 m/s

Time of contact, t = 2.00 ms = 2.00 × 10⁻³ s

Now,

Magnitude of change in momentum = Final momentum - Initial momentum

or

Magnitude of change in momentum = ( m × v ) - ( m × u )

or

Magnitude of change in momentum = ( 0.155 × 65 ) - ( 0.155 × 35 )

or

Magnitude of change in momentum = 10.075 - 5.425 = 4.65 kg.m/s

Now, Magnitude of impulse = change in momentum

thus,

Magnitude of impulse = 4.65 kg.m/s

Now,

magnitude of the average force applied by the bat = [tex]\frac{\textup{Impulse}}{\textup{Time}}[/tex]

or

magnitude of the average force applied by the bat = [tex]\frac{\textup{4.65}}{\textup{3}\times\textup{10}^{-3}}[/tex]

or

Magnitude of the average force applied by the bat = 1550 N

Answer:

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction, [tex]\vec{F_{up}} = 3.50\times 10^{- 5} N[/tex]

Charge, Q = [tex] - 1.55\micro C = - 1.55\times 10^{- 6} C[/tex]

Now, we know by Coulomb's law,

[tex]F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}[/tex]

Also,

Electric field, [tex]E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}[/tex]

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

[tex]\vec{E} = \frac{\vec F_{up}}{Q}[/tex]

[tex]\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}[/tex]

[tex]\vec{E} = - 22.58 N/C[/tex]

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

[tex]|\vec{E}| = 22.58\ N/C[/tex]

The magnitude of the electric field is 22.58 × 10³ N/C and the direction is downward, which is opposite to the upward force applied.

The question is asking to find the magnitude and direction of an electric field that applies a force on a charged particle. The formula used to calculate the electric field (E) is given by E = F/q, where F is the force applied to the charge and q is the magnitude of the charge.

Given a force (F) of 3.50 × 10⁻⁵ N upward and a charge (q) of -1.55 μC (microcoulombs), which is equivalent to -1.55 × 10⁻⁶ C (coulombs), first, we convert the charge into coulombs by multiplying the microcoulombs by 10⁻⁶. Next, we substitute the values into the formula, yielding:

E = (3.50 × 10⁻⁵ N) / (-1.55 × 10⁻⁶ C), which simplifies to E = -22.58 × 10³ N/C. The negative sign indicates that the direction of the electric field is opposite to the direction of the force, so if the force is upward, the electric field is downward.

The magnitude of the electric field is thus 22.58 × 10³ N/C and the direction is downward

Answer:

8.1 m /s

Explanation:

Let the required velocity be v . This is a horizontal velocity so it will cover the horizontal distance of 12.3 m with this velocity without any acceleration .

Time taken t = distance / velocity

t = 12.3 /v

During this period ball also covers vertical distance with initial velocity zero and acceleration of g.

For vertical fall

initial velocity u = 0

Acceleration = g

Time = t

h = ut + 1/2 g t²

11.3 = 0 + .5 x 9.8 x (12.3 / v )²

v² = 65.6

v = 8.1 m /s

Answer:

a) Vx = -31.95 km/h       b) Vy = -45.07 km/h

c) t = 0.083 h                 d) d = 0.22 km

Explanation:

First we have to express these values as vectors:

ra = (2.5, 3.9) km      rb = (0,0)km

Va = (0, - 21) km/h    Vb = (31.95, 24.07) km/h

Now we can calculate relative velocity:

[tex]V_{A/B} = V_{A} - V_{B} = (0, -21) - (31.95, 24.07) = (-31.95, -45.07) km/h[/tex]

For parts (c) and (d) we need the position of A relative to B and the module of the position will be de distance.

[tex]r_{A/B} = (2.5, 3.9) + (-31.95, -45.07) * t[/tex]

[tex]d = |r_{A/B}| = \sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}[/tex]

In order to find out the minimum distance we have to derive and find t where it equals zero:

[tex]d' = \frac{-2*(2.5-31.95*t)*(-31.95)-2*(3.9-45.07*t)*(-45.07)}{2*\sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}} =0[/tex]

Solving for t we find:

t = 0.083 h

Replacing this value into equation for d:

d = 0.22 km

Answer:

The temperature at which the wing would be shorter is [tex]- 80.69^{\circ}C[/tex]

Solution:

The original length of the Aluminium wing, [tex]l_{w} = 25 m[/tex]

Temperature, T = [tex]21^{\circ}C[/tex]

Change in the wing's length, [tex]\Delta l_{w} = 0.06 m[/tex]

Also, for Aluminium, at temperature between [tex]20^{\circ}C[/tex] to [tex]100^{\circ}C[/tex], the linear expansion coefficient, [tex]\alpha = 23.6\times 10^{- 6}/^{\circ}C[/tex]

Now, Change in length is given by:

[tex]\Delta l_{w} = l_{w}\alpha \Delta T[/tex]

[tex]0.06 = 25\times 23.6\times 10^{- 6}\(T - T')[/tex]

[tex]\frac{0.06}{5.9\times 10^{- 4}} = 21^{\circ}C _ T'[/tex]

[tex]T' = - 80.69^{\circ}C[/tex]

-Final answer:

To find the temperature at which the aluminum wing would be 6 cm shorter, use the thermal expansion formula. Given the initial length, final length, and coefficient of linear expansion for aluminum, the required temperature will be -285.8°C.

Explanation:

Thermal Expansion Formula: ΔL = αL0ΔT

To find the temperature at which the aluminum wing would be 6 cm shorter, we can use the thermal expansion formula. Given initial length L0 = 25m, final length = 25m - 0.06m = 24.94m, and coefficient of linear expansion for aluminum α = 22 × 10-6 °C. Rearranging the formula gives us the change in temperature ΔT = ΔL / (αL0). Substituting the values results in ΔT ≈ 306.8°C.

Now, ΔT = Temperature for 25meter length - Temperature for 24.94meter length.

Thus, Temperature for 24.94meter length = -285.8°C

Answer:

so speed = 165 m/s

Explanation:

given data

speed = (25.0 + A + B) urks/dort

A = 39

B = 18

1 urk = 58 m

to find out

Convert speed to meters per second

solution

we know speed =  (25.0 + A + B)

put A and B

speed =  (25.0 + 39+ 18) = 82 urk/dort

we know that hour is divided into 125 time units name dorts

so we can say

1 hour = 125 × dorts

and we know 1 hour = 3600 seconds

so

3600 = 125 × dorts

dorts = 28.8 seconds

and we have given

1 urk = 58 m

so 82 urk = 82 × 58 = 4756 m

so from speed

speed = 82 urk/dort

speed =  [tex]82 * \frac{58}{28.8}[/tex] m/s = 165.139

so speed = 165 m/s

Final answer:

The speed (82.0 urks/dort) converts to 165 meters per second when using the given conversion factors, with values A=39 and B=18 included.

Explanation:

To convert a speed from urks/dort to meters per second, we first need to express the speed in urks/dort in terms of meters and seconds, using the given conversions: 1 urk = 58.0 meters and 1 hour = 125 dorts. Given the values A=39 and B=18, the speed in urks/dort is 25.0 + 39 + 18 = 82.0 urks/dort.

Firstly, we convert urks to meters:

(82.0 urks/dort) × (58.0 meters/urk) = 4756 meters/dort

Now, we can convert dorts to seconds. Since 1 hour is divided into 125 dorts and there are 3600 seconds in an hour, there are 3600 seconds / 125 dorts = 28.8 seconds/dort.

Speed in meters per second (m/s) = 4756 meters/dort × (1 dort/28.8 seconds)

Speed in m/s = 165 m/s (rounded to three significant figures)

Therefore, the speed in meters per second is 165 m/s, rounded to three significant figures.

Final answer:

The magnitude of the electric field at the center of a square with four point charges (either all positive or three positive and one negative) each of magnitude 4.50 nC and sides of length 2.80 cm, is zero. This outcome is due to symmetry and the principle of superposition.

Explanation:

To find the magnitude of the electric field at the center of a square due to point charges located at its corners, we use the principle of superposition. This entails calculating the electric field contribution from each charge individually and then vectorially adding these contributions together.

For our case, where all charges have a magnitude of 4.50 nC and the square has sides of length 2.80 cm, the geometry of the problem simplifies the calculations significantly.

When all four charges are positive, they all contribute to the electric field in a symmetrically outward manner relative to the center. Thus, their contributions in terms of magnitude cancel out, making the net electric field at the center zero.

In the scenario where three charges are positive and one is negative, we approach the problem similarly. However, the negative charge introduces an incongruence in the symmetry. Its electric field contribution will 'attract' rather than 'repel' like the other three positive charges.

Despite this, due to the square's symmetry and equal magnitude of the charges, the net electric field at the center still cancels out, resulting in zero magnitude.

This conclusion is based on symmetry and the principle that equal point charges at equal distances in a square configuration result in a net zero electric field at the square's center, irrespective of the sign when the magnitudes are equal.

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, [tex]q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=2.4\ \mu C=2.4\times 10^{-6}\ C[/tex]

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

[tex]U=k\dfrac{q_1q_2}{r}[/tex]

[tex]r=k\dfrac{q_1q_2}{U}[/tex]

[tex]r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}[/tex]

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

The 2.40 μC point charge must be placed approximately 0.34 meters away from the -7.80 μC point charge for their electric potential energy to be -0.500 J.

To find the distance between a -7.80 μC point charge and a 2.40 μC point charge such that their electric potential energy is -0.500 J, we use the formula for electric potential energy:

U = k × (q₁ × q₂) / r

where U is the electric potential energy, k is Coulomb's constant (8.99 × 10⁹ Nm²/C²), q₁ and q₂ are the point charges, and r is the distance between them.

Given:

q₁ = -7.80 μC = -7.80 × 10⁻⁶ C

q₂ = 2.40 μC = 2.40 × 10⁻⁶ C

U = -0.500 J

Rearranging the formula to solve for r, we get:

r = k × (q₁ × q₂) / U

Substituting the given values:

r = (8.99 × 10⁹ Nm²/C²) × (-7.80 × 10⁻⁶ C × 2.40 × 10⁻⁶ C) / -0.500 J

r ≈ 0.34 meters

Therefore, the 2.40 μC point charge must be placed approximately 0.34 meters away from the -7.80 μC point charge for their electric potential energy to be -0.500 J.

Answers:

a) [tex]t=0.311 s[/tex]

b) [tex]a=86.847 m/s^{2}[/tex]

c) [tex]y=1.736 m[/tex]

d) [tex]V=17.369 m/s[/tex]

Explanation:

For this situation we will use the following equations:

[tex]y=y_{o}+V_{o}t+\frac{1}{2}at^{2}[/tex] (1)  

[tex]V=V_{o} + at[/tex] (2)  

Where:  

[tex]y[/tex] is the height of the model rocket at a given time

[tex]y_{o}=0[/tex] is the initial height of the model rocket

[tex]V_{o}=0[/tex] is the initial velocity of the model rocket since it started from rest

[tex]V[/tex] is the velocity of the rocket at a given height and time

[tex]t[/tex] is the time it takes to the model rocket to reach a certain height

[tex]a[/tex] is the constant acceleration due gravity and the rocket's thrust

The average velocity of a body moving at a constant acceleration is:

[tex]V=\frac{V_{1}+V_{2}}{2}[/tex] (3)

For this rocket is:

[tex]V=\frac{27 m/s}{2}=13.5 m/s[/tex] (4)

Time is determined by:

[tex]t=\frac{y}{V}[/tex] (5)

[tex]t=\frac{4.2 m}{13.5 m/s}[/tex] (6)

Hence:

[tex]t=0.311 s[/tex] (7)

Using equation (1), with initial height and velocity equal to zero:

[tex]y=\frac{1}{2}at^{2}[/tex] (8)  

We will use [tex]y=4.2 m[/tex] :

[tex]4.2 m=\frac{1}{2}a(0.311)^{2}[/tex] (9)  

Finding [tex]a[/tex]:

[tex]a=86.847 m/s^{2}[/tex] (10)  

Using again [tex]y=\frac{1}{2}at^{2}[/tex] but for [tex]t=0.2 s[/tex]:

[tex]y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}[/tex] (11)

[tex]y=1.736 m[/tex] (12)

We will use equation (2) remembering the rocket startted from rest:

[tex]V= at[/tex] (13)  

[tex]V= (86.847 m/s^{2})(0.2 s)[/tex] (14)  

Finally:

[tex]V=17.369 m/s[/tex] (15)  

Answer:

T = (5 ×3) + (5 × 9.8 ) N    =   64 N

so option D is correct

Explanation:

given data

mass = 5 kg

acceleration = 3 m/s²

to find out

tension

solution

we know acceleration on bucket is upward so

we say

T - mg = ma

here T is tension , m is mass and g is acceleration due to gravity = 9.8 m/s² and a is acceleration so

T = ma + mg

put here value

T = (5 ×3) + (5 × 9.8 ) N = 64 N

so option D is correct

Answer:

Explanation:

We just need to sum the forces, we can do this easily in their Cartesian form.

Knowing the magnitude and angle with the positive x axis, we can find Cartesian representation of the vectors using the formula

[tex]\vec{A}= |\vec{A}| \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]

where [tex]|\vec{A}|[/tex] its the magnitude of the vector and θ the angle with the positive x axis.

So, for our forces we got:

[tex]\vec{F}_1 \ = \ 66 \ lbf \ * \ ( \ cos (30\°)\ , \ sin(30\°) \ )[/tex]

[tex]\vec{F}_2 \ = \ 110 \ lbf \ * \ ( \ cos (45\°)\ , \ sin(45\°) \ )[/tex]

[tex]\vec{F}_3 \ = \ 138 \ lbf \ * \ ( \ cos (120\°)\ , \ sin(120\°) \ )[/tex]

this will give us:

[tex]\vec{F}_1 \ = ( \ 57.157 \ lbf \ , \ 33 \ lbf \ )[/tex]

[tex]\vec{F}_2 \ = ( \ 77.782 \ lbf \ , \ 77.782 \ lbf \ )[/tex]

[tex]\vec{F}_3 \ = ( \ - 69 \ lbf \ , \ 119.511 \ lbf \ )[/tex]

Now, we just sum the forces:

[tex]\vec{F}_{net} \ = \ \vec{F}_1 \ + \ \vec{F}_2 \ + \ \vec{F}_3[/tex]

[tex]\vec{F}_{net} \ = ( \ 57.157 \ lbf \ , \ 33 \ lbf \ ) + ( \ 77.782 \ lbf \ , \ 77.782 \ lbf \ ) + (\ - 69 \ lbf \ , \ 119.511 \ lbf \ )[/tex]

[tex]\vec{F}_{net} \ = ( \ 57.157 \ lbf \ + \ 77.782 \ lbf \ - \ 69 \ lbf \, \ 33 \ lbf \ + \ 77.782 \ lbf \ + \ 119.511 \ lbf \ )[/tex]

[tex]\vec{F}_{net} \ = ( \ 65.939 \ lbf \, \ 230.293 lbf \ )[/tex]

This is the net force, to obtain the magnitude, we just need to find the length of the vector, using the Pythagorean formula:

[tex]|\vec{F}_{net}| = \sqrt{(F_{net_x})^2+(F_{net_y})^2}[/tex]

[tex]|\vec{F}_{net}| = \sqrt{(65.939 \ lbf)^2+(230.293 lbf)^2}[/tex]

[tex]|\vec{F}_{net}| = \ 239.547 \ lbf[/tex]

To obtain the angle with the positive x-axis we can use the formula:

[tex]\theta \ = \ arctan( \frac{F_y}{F_y})[/tex]

[tex]\theta \ = \ arctan( \frac{230.293 lbf}{65.939 \ lbf})[/tex]

[tex]\theta \ = \ arctan( 3.492)[/tex]

[tex]\theta \ = \ 74.02[/tex]

So, the answer its

[tex]magnitude = \ 239.547 \ lbf[/tex]

[tex]angle_{ (to the x axis)} = \ 74.02 [/tex]

Rounding up:

[tex]magnitude = \ 239.5 \ lbf[/tex]

[tex]angle_{ (to the x axis)} = \ 74.0 [/tex]

Answer:

[tex]\tau = 6.35 s[/tex]

A = 3.2 cm

Explanation:

given,

Amplitude of the vibration of the top of lamppost = 6.4 cm

after 8.8 s the amplitude is 1.6 cm

time constant for damping of oscillation = ?

Amplitude at 4.4 second= ?

using formula

[tex]A = A_0e^{-\dfrac{T}{\tau}}[/tex]

[tex]1.6 = 6.4\times e^{-\dfrac{8.8}{\tau}}[/tex]

taking ln both side

[tex]ln (1.6) = ln(6.4)-\dfrac{8.8}{\tau}[/tex]

[tex]\tau = 6.35 s[/tex]

[tex]A = A_0e^{-\dfrac{T}{\tau}}[/tex]

[tex]A =6.4\times e^{-\dfrac{4.4}{6.35}}[/tex]

A = 3.2 cm

The amplitude of the oscillation 4.4 s after the quake stopped is:

A= 3.2cm

This refers to the maximum length which an object can attain when it oscillates or vibrates.

Given that the amplitude from the top of the lamppost is given as:

6.4cm

We need to find the time constant for the damping of the oscillation.

We use the formula

[tex]A= Aoe[/tex]^-T/t

We would input the values

1.6 = 6.4 x [tex]e[/tex]^-8.8/t

Taking Ln on both sides

Ln(1.6) = ln(6.4)^-8.8/t

t=6.35₈

A= [tex]Aoe[/tex]^T/t

A= 3.2cm

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Answer:

10945.9 kg/m^3

Explanation:

mass of cylinder, m = 1 kg

Height of cylinder, h = 55 mm = 0.055 m

Diameter of cylinder = 46 mm

Radius of cylinder, r = 23 mm = 0.023 m

The formula of the volume of the cylinder is given by

[tex]V = \pi r^{2}h[/tex]

V = 3.14 x 0.023 x 0.023 x 0.055

V = 9.136 x 10^-5 m^3

Density is defined as mass per unit volume .

[tex]Density = \frac{1}{9.136 \times 10^{-5}}[/tex]

Density = 10945.9 kg/m^3

Answer:

-2370000 N force acts on the charge particle    

Explanation:

We have given electric field E = 790000 N/C

Charge [tex]q=-3\mu C=-3\times 10^{-6}C[/tex]

We know that force on any charge particle due to electric field is given by

[tex]F=qE[/tex], here q ia charge and E is electric field

So force [tex]F=-3\times 10^{-6}\times 790000=-2370000N[/tex]

So -2370000 N force acts on the charge particle  

Answer:

The work done by gravity is zero.

Explanation:

Given that,

Gravitational force = 2000 N

Circumference = 80000 km

We need to calculate the work done by gravity

Using formula of work done

[tex]W=F\cdot d\cos\theta[/tex]

Here, [tex]\cos\theta[/tex] = 0

Because, for a circular motion is always perpendicular to the force.

Where, F = force

d = distance

Put the value into the formula

[tex]W=2000\times80000\times0[/tex]

[tex]W=0[/tex]

Hence, The work done by gravity is zero.

The work done on the satellite by gravity is 160,000,000,000 joules.

The work done on a satellite by gravity can be calculated using the formula: work = force x distance. In this case, the force is the gravitational force of 2000 N and the distance is the circumference of the orbit, which is 80000 km. To convert km to meters, we multiply by 1000, so the distance is 80000 x 1000 = 80,000,000 meters. Plugging these values into the formula, the work done on the satellite by gravity is 2000 x 80,000,000 = 160,000,000,000 J (joules).

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Final answer:

The value of q2, with its sign, can be found using Coulomb's Law. By plugging in the given values for q1, the distance, and the force experienced, we can calculate the value of q2 as -2.25C. The negative sign indicates that q2 is a negative charge.

Explanation:

In order to find the value of q2, we can use Coulomb's Law, which 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. Here, we are given the charge of particle 1 (q1 = +3.3C), the distance between the particles (d = 0.24m), and the force experienced by particle 1 (F = 4.1N). Let's denote the charge of particle 2 as q2.

Using Coulomb's Law, we can write:

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

Plugging in the given values, we have:

4.1N = (9 x 10^9 N m^2/C^2)(3.3C * q2) / (0.24m)^2

Simplifying the equation, we can solve for q2:

q2 = (4.1N * (0.24m)^2) / (9 x 10^9 N m^2/C^2 * 3.3C)

Calculating this equation gives us the value of q2 as +2.25C. Since the force experienced by particle 1 is attractive, with a positive charge (+3.3C), the value of q2 must be negative to create an attractive force. Therefore, the value of q2 is -2.25C.

Answer:

The correct answer is option'B': Change in entropy

Explanation:

We know from the second law of thermodynamics for any spontaneous process the total entropy of the system and it's surroundings will increase.

Meaning that any unaided process will  move in a direction in which the entropy of the system will increase.It is because the system will always want to increase it's randomness

Answer:

Height of the tree, h = 20.72 meters

Explanation:

Given that,

The sun is 21° above the horizontal, [tex]\theta=21^{\circ}[/tex]

Length of the shadow, d = 54 m

Let h is the height of the tree. It can be calculated using trigonometry as :

[tex]tan\theta=\dfrac{perpendicular}{base}[/tex]

Here, perpendicular is h and base is 54 meters.

[tex]tan(21)=\dfrac{h}{54}[/tex]

[tex]h=tan(21)\times 54[/tex]

h = 20.72 meters

So, the height of the tree is 20.72 meters. Hence, this is the required solution.

Answer:

Time taken, [tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

[tex]T\ cos\theta-mg=0[/tex]

[tex]T=\dfrac{mg}{cos\theta}[/tex]

Sum of forces in x direction,

[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]

[tex]mg\ tan\theta=\dfrac{mv^2}{r}[/tex].............(1)

Also, [tex]r=l\ sin\theta[/tex]

Equation (1) becomes :

[tex]mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}[/tex]

[tex]v=\sqrt{gl\ tan\theta.sin\theta}[/tex]...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

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

Put the value of T from equation (2) to the above expression:

[tex]T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]

[tex]T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]

On solving above equation :

[tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]

Hence, this is the required solution.

The time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].

As it is given to us that the small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone.

Now since the ball is moving in circular as well as vertical motion, therefore, the sum of vertical forces at any given moment of time can be written as,

[tex]\rm T\ cos \theta = mg\\\\T = \dfrac{mg}{Cos\theta}[/tex]

Also, the sum of the forces in the x-direction,

[tex]\rm T\ sin\theta= \dfrac{mv^2}{r}[/tex]

Substitute the value of T,

[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{r}[/tex]

[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{l\ sin\theta}[/tex]

[tex]v=\sqrt{gl\ tan\theta \cdot sin\theta}[/tex]

We know that the time is taken by the ball to circulate around the axis can be given by the formula,

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

Substitute the value of v,

[tex]T = \dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta\cdot sin\theta}}[/tex]

[tex]T=2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex]

Hence, the time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].

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Answer:

a) r = 6.81 cm : radius of the sphere

b) A = 582.78 cm² : surface area of the sphere

c) V = 1322.91 cm³ :  volume of the sphere

Explanation:

Formula to calculate the surface area of the sphere:

A = 4×π×r² Formula (1)

Formula to calculate the volume of the sphere:

V = (4/3)×π×r³ Formula (2)

Problem development

a)

d = 5.36 in

1 in = 2.54 cm

[tex]d = 5.36 in * \frac{2.54cm}{1in} = 13.614 cm[/tex]

r = d/2

Where:

r: sphere radius (cm)

d: sphere diameter (cm)

r = 13.614/2 = 6.81 cm

b)

We replace in formula (1)

A = 4×π×(6.81)² = 582.78 cm²

c)

We replace in formula (2)

V = (4/3)×π×(6.81)³ = 1322.91 cm³

Answer:

The ball hit the pyramid 18.52m down the pyramid face.

Explanation:

As you can see in the image below, the will hit the pyramid at at point where the distance travelled vertically divided by the distance travelled horizontally is equal to tan(50), since it's at this moment where the path of the ball will coincide with the walls of the pyramid.

The horizontal vellocity of the ball will remain constant at a value of 7m/s along the whole journey. This is because there is no horizontal acceleration that can affect the horizontal velocity. On the contrary, the vertical velocity will start at 0m/s and will increase because of gravity.

The distance travelled horizontally will be:

[tex]x = v_x*t = 7t[/tex]

The distance travelled vertically will be:

[tex]y = \frac{1}{2}gt^2+v_o_yt+y_o | v_o_y = 0m/s, y_o=0m\\ y = \frac{1}{2}gt^2[/tex]

So, then:

[tex]\frac{y}{x} = tan(50)\\\frac{\frac{1}{2}gt^2}{v_xt} =tan(50)\\\frac{1}{2}gt = v_xtan(50)\\t= \frac{2v_xtan(50)}{g} = \frac{2*7m/s*tan(50)}{9.81m/s^2}=1.7s[/tex]

At time = 1.7s:

[tex]x = v_x*t = 7t = 7m/s*1.7s = 11.9m[/tex]

[tex]y=\frac{1}{2}gt^2 =\frac{1}{2}*9.81m/s^2*(1.7s)^2=14.19m [/tex]

Using Pythagorean theorem, we can find the distance:

[tex]d = \sqrt{x^2+y^2}=\sqrt{(11.9m)^2+(14.19m)^2} = 18.52m[/tex]

Answer:

The net electric charge is [tex]-1.352\times10^{-12}\ C[/tex]

Explanation:

Given that,

No of electron [tex]n=8.45\times10^{6}[/tex]

We need to calculate the net electric charge in coulombs

Using formula of net electric charge

Net electric charge = number of electron X charge of one electron

[tex]Q=ne[/tex]

Put the value into the formula

[tex]Q=8.45\times10^{6}\times(-1.6\times10^{-19})[/tex]

[tex]Q=-1.352\times10^{-12}\ C[/tex]

Hence, The net electric charge is [tex]-1.352\times10^{-12}\ C[/tex]