An electric field of 790,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -3.00 uC at this spot? (14C = 10 6C) Give your answer in Sl unit rounded to two decimal places.

Answer:

Its initial velocity will be greater than final velocity so option (b) will be correct option  

Explanation:

As we throw the any object vertically the motion of the object will be opposes by the gravity.

And as the velocity of object is opposes by gravity, the final velocity goes on decreasing and finally it becomes zero.

So the initial velocity is always greater than final velocity when the object is thrown vertically upward.

So option (b) will be the correct option  

Answer:

electric flux is 280  Nm²/C  

so correct option is D 280  Nm²/C

Explanation:

radius r = 0.50 m

angle = 30 degree

field strength = 713 N/C

to find out

the electric flux through the surface

solution

we find here electric flux by given formula that is

electric flux = field strength × area× cos∅   .......1

here area = πr² = π(0.50)²

put here all value in equation  1

electric flux = field strength × area× cos∅  

electric flux = 713 × π(0.50)² × cos60

we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk

so we use cos60

electric flux = 280  Nm²/C

so correct option is D 280  Nm²/C

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

[tex]V=E\times d[/tex]

[tex]V=3000\ N/C\times 0.7\ m[/tex]

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

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

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:

The strength of the field is 22.84 N/C.

Explanation:

Given that,

Speed [tex]v= 7.5\times10^{5}\ m/s[/tex]

Distance = 7.0 cm

We need to calculate the acceleration

Using equation of motion

[tex]v^2-u^2=2as[/tex]

Put the value in the equation

[tex]0-(7.5\times10^{5})^2=2\times a\times7.0\times10^{-2}[/tex]

[tex]a =-\dfrac{(7.5\times10^{5})^2}{2\times7.0\times10^{-2}}[/tex]

[tex]a =-4.017\times10^{12}\ m/s^2[/tex]

We need to calculate the strength of the field

Using newton's second law and electric force

[tex]F = ma = qE[/tex]

[tex]-qE=-ma[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

Put the value into the formula

[tex]E=\dfrac{9.1\times10^{-31}\times4.017\times10^{12}}{1.6\times10^{-19}}[/tex]

[tex]E=22.84\ N/C[/tex]

Hence, The strength of the field is 22.84 N/C.

Final answer:

The strength of the electric field that can stop an electron moving at 7.5 x 10^5 m/s within a distance of 7 cm can be calculated using the work-energy principle, where the work done by the electric field is equal to the change in kinetic energy of the electron.

Explanation:

The strength of the electric field required to bring an electron moving to the right at 7.5 x 105 m/s to rest in a distance of 7.0 cm can be found using the work-energy principle and knowing the force exerted by an electric field on a charge. The work done by the electric field is equal to the change in kinetic energy of the electron, which is initially ½ mv2 and becomes zero when the electron is at rest.

To calculate the strength of the field, we can use:

Work (W) = Electric field (E) x Charge (e) x Distance (d)

½ mv2 = E * e * d

where m is the mass of the electron, v is its initial velocity, e is the elementary charge (approximately 1.6 x 10−12 C), and d is the distance (7.0 cm or 0.07 m). We solve for E to find the strength of the electric field

Answer:

Her velocity at the bottom of the hill is 15.61m/s and she travel up the 30° slope 24.85m

Explanation:

For simplicity purpose, we can analyze the section of the snowboarder's travel in the hill, in the horizontal surface and in the slope separately.

In the hill, we will say that the x-axis is parallel to the hill, and y-axis is perpendicular. Using geometry, we can see that the angle of the snowboarder's weight force from the y-axis is 15°. The velocity of the snowboarder will increase in the direction parallel to the hill, in a constant acceleration motion:

[tex]F_x: W_x = ma\\W*sin(15) = ma\\mg*sin(15) = ma\\a = g*sin(15) = 9.81m/s^2 * sin(15) = 2.54m/s^2[/tex]

With the acceleration, we can use the equations for constant acceleration motion:

[tex]v_f^2 - v_o^2 = 2a*d\\v_f^2 - (0m/s)^2=2*2.54m/s^2*48m\\vf = \sqrt{2*2.54m/s^2*48m}=15.61 m/s[/tex]

This would be her velocity at the bottom of the hill.

As there is no friction, she would reach the bottom of the slope with this velocity.

In the slope, the line of reasoning is similar as in the hill, with the difference that the acceleration will oppose velocity.

[tex]F_x: -W_x = ma\\-W*sin(30) = ma\\-mg*sin(30) = ma\\a = -g*sin(30) = -9.81m/s^2 * sin(30) = -4.905m/s^2[/tex]

[tex]v_f^2-v_o^2=2a*d\\d=\frac{(v_f^2-v_o^2)}{2a}=\frac{((0m/s)^2-(15.61m/s)^2)}{2(-4.905m/s^2)} = 24.85m[/tex]

The speed of the snowboarder at the bottom of the hill is 15.62 m/s.

The distance the snowboarder travel up the 30° slope is 24.9 m.

The acceleration of the snowboarder is calculated as follows;

[tex]W sin\theta - F_f = ma\\\\mgsin\theta - 0 = ma\\\\mg sin(\theta) = ma\\\\a = g(sin\theta)\\\\a = 9.8 \times sin(15)\\\\a = 2.54 \ m/s^2[/tex]

The speed of the snowboarder at the bottom of the hill is calculated as follows;

[tex]v^2 = u^2 + 2ah\\\\v^2 = 0 + 2ah\\\\v = \sqrt{2ah} \\\\v = \sqrt{2(2.54)(48)} \\\\v = 15.62 \ m/s[/tex]

The acceleration of the snowboarder up 30° slope is calculated;

[tex]-Wsin(\theta)- F_f = ma\\\\-Wsin(\theta) -0 = ma\\\\-mgsin(\theta)= ma\\\\-gsin(\theta) = a\\\\a = -9.8 \times sin(30)\\\\a = -4.9 \ m/s^2[/tex]

The distance the snowboarder travel up the 30° slope is calculated as follows;

[tex]v^2 = u^2 - 2ah\\\\-2ah = v^2- u^2\\\\-2ah = v^2 -0\\\\-2ah = v^2\\\\h = \frac{v^2}{-2a} \\\\h = \frac{(15.62)^2}{-2(-4.9)} \\\\h = 24.9 \ m[/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:

[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:

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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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:

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.

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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:

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:

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:

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 magnitude of resultant vector and direction are 1.843 m and 77.47° east of north.

Explanation:

Given that,

Magnitude of displacement due to east = 2.8 km

Magnitude of displacement due to north = 2.8 km

Magnitude of displacement due to west = 2.4 km

Magnitude of displacement due to south = 1 km

We need to calculate the resultant of the displacement

[tex]D = d_{1}+d_{2}+d_{3}+d_{4}[/tex]

[tex]D=2.8\hat{i}+2.8\hat{j}-2.4\hat{i}-1\hat{j}[/tex]

[tex]D=0.4\hat{i}+1.8\hat{j}[/tex]

The magnitude of the resultant vector

[tex]D=\sqrt{(0.4)^2+(1.8)^2}[/tex]

[tex]D=1.843\ m[/tex]

We need to calculate the direction

Using formula of direction

[tex]\tan\theta=\dfrac{j}{i}[/tex]

Put the value into the formula

[tex]\tan\theta=\dfrac{1.8}{0.4}[/tex]

[tex]\theta=\tan^{-1}4.5[/tex]

[tex]\theta=77.47^{\circ}[/tex]

Hence, The magnitude of resultant vector and direction are 1.843 m and 77.47° east of north.

Answer:

a) [tex]F_a=0.152 \mu N[/tex]

b) [tex]F_b=203.182 \mu N[/tex]

Explanation:

The center of mass of an homogeneous sphere is its center, therefore you can use Newton's universal law of gravitation to find both questions.

[tex]F_g=G\frac{m_1m_2}{d}[/tex]

[tex]G=6.674*10^{-11} NmKg^{-2}[/tex]

a) d = 19m

[tex]F_a = G\frac{8.4*10^{4}*9.8}{19^2}[/tex]

[tex]F_a=0.152 \mu N[/tex]

b) d = 0.52

[tex]F_b = G\frac{8.4*10^{4}*9.8}{0.52^2}[/tex]

[tex]F_b=203.182 \mu N[/tex]

Answer:

(a) GF = 1.522 x (10 ^ -7)  N

(b) GF = 2.032 x (10 ^ -4)  N

Explanation:

The magnitude of the gravitational force follows this equation :

GF = (G x m1 x m2) / (d ^ 2)

Where G is the gravitational constant universal.

G = 6.674 x (10 ^ -11).{[N.(m^ 2)] / (Kg ^ 2)}

m1 is the mass from the first body

m2 is the mass from the second body

And d is the distance between each center of mass

m2 is a particle so m2 it is a center of mass itself

The center of mass from the sphere is in it center because the sphere has uniform density

For (a) d = 19 m

GF = {6.674 x (10 ^ -11).{[N.(m ^ 2)] / (Kg ^ 2)} x 8.4 x (10 ^ 4) Kg x 9.8 Kg} / [(19 m)^ 2]

GF = 1.522 x (10 ^ -7)  N

For (b) d = 0.52 m

GF = 2.032 x (10 ^ -4)  N

Notice that we have got all the data in congruent units

Also notice that the force in (b) is bigger than the force in (a) because the distance is shorter

Answer:

Explanation:

The car is moving with uniform velocity . Hence there is no acceleration in the car .It indicates that net force on the car is zero . Since force in backward direction is exerted by the wind,  to make net force zero , the forward push by the car must be equal to backward force by the wind. In other words

Forward force by car = backward force by wind = 5000 N.

Answer:

[tex]g'=3.71\ m/s^2[/tex]

Explanation:

Given that,

Time period of a pendulum on the earth's surface, T₁ = 1.2 s

Time period of the same pendulum on Mercury, T₂ = 1.95 s

The time period of the pendulum is given by :

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

On earth :

[tex]T_1=2\pi \sqrt{\dfrac{l}{g}}[/tex]

[tex]1.2=2\pi \sqrt{\dfrac{l}{9.8}}[/tex].............(1)

Let g' is the acceleration due to gravity on Mercury. So,

[tex]1.95=2\pi \sqrt{\dfrac{l}{g'}}[/tex]............(2)

From equation (1) and (2) :

[tex]\dfrac{1.2}{1.95}=\sqrt{\dfrac{g'}{9.8}}[/tex]

[tex]g'=(\dfrac{1.2}{1.95})^2\times 9.8[/tex]

[tex]g'=3.71\ m/s^2[/tex]

So, the acceleration due to gravity on the mercury is [tex]3.71\ m/s^2[/tex]. Hence, this is the required solution.

To determine how many yards are in a mile, knowing that a mile equals 5280 feet and a yard contains 3 feet, divide the total feet in a mile by the feet in a yard, resulting in 1760 yards in a mile.

To find how many yards are there in a mile, given that a mile is 5280 feet long and a yard contains 3 feet, we can divide the total number of feet in a mile by the number of feet in a yard. Using the formula for conversion, we calculate:

Yards in a mile = Total feet in a mile ÷ Feet in a yard

By substituting the given values:

Yards in a mile = 5280 ft ÷ 3 ft

Yards in a mile = 1760

This calculation clearly shows that there are 1760 yards in a mile. This example emphasizes the importance of understanding unit conversions in mathematics, allowing us to easily switch between units of measurement.

Answer:

[tex]F_{max}=2.95*10^{5}N[/tex]

[tex]\Delta l=0.25cm[/tex]

Explanation:

E=1.50x10^10 N/m2     Young's modulus of bone

σmax=1.50x10^8 N/m2       Max stress tolerated by the bone

Relation between stress and Force:

[tex]\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}[/tex]

[tex]F_{max}=\sigma_{max}*\pi*d^{2}/4}=1.50*10^{8}*\pi*(2.5*10^{-2})^{2}=2.95*10^{5}N[/tex]

Relation between stress and strain:

Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:

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

[tex]\epsilon=\frac{\Delta l}{l}[/tex]      

We solve these equations to find the bone compression:

[tex]\Delta l=l*\frac{\sigma}{E}=25*\frac{1.50*10^{8}}{1.50*10^{10}}=0.25cm[/tex]      

Final answer:

To find the maximum force that can be exerted on the femur bone in the leg, we need to calculate the stress using Young's modulus. The maximum force is approximately 7.35 * 10^7 N. To find the amount by which the bone shortens, we need to calculate the strain using Young's modulus and the stress.

Explanation:

To find the maximum force that can be exerted on the femur bone in the leg, we need to calculate the stress. The stress is equal to the force divided by the cross-sectional area of the bone. The cross-sectional area can be determined using the diameter of the bone.

(a) The diameter of the bone is 2.50 cm, which is equal to 0.025 m. The cross-sectional area can be calculated using the formula for the area of a circle: π * (radius)^2. In this case, the radius is half of the diameter, so the area is approximately 0.4909 m2. To find the maximum force, we can use the formula for stress: stress = force / area. Rearranging the formula, we have: force = stress * area. Plugging in the values, we get: force = (1.50 * 108 N/m2) * (0.4909 m2). The maximum force that can be exerted on the femur bone is approximately 7.35 * 107 N.

(b) To find the amount by which the bone shortens, we need to calculate the strain. The strain is equal to the change in length divided by the original length of the bone. The change in length can be determined using the formula for strain: strain = change in length / original length. Rearranging the formula, we have: change in length = strain * original length. We can calculate the strain using the stress and Young's modulus: strain = stress / Young's modulus. Plugging in the values, we get: strain = (1.50 * 108 N/m2) / (1.50 * 1010 N/m2). The change in length can be calculated using the formula: change in length = strain * original length. Plugging in the values, we get: change in length = (1.50 * 108 N/m2) / (1.50 * 1010 N/m2) * 25.0 cm.

Answer:

a)Vg=13.42m/s :Speed with which the ball hits the ground

b) t₁= 2.74s : Time the ball remains in the air

c)h=9.19m: Maximum height reached by the ball

Explanation:

We apply the kinematic equations of parabolic motion:

a) Vg= Vo

Vg:speed with which the ball hits the ground

Vo: initial speed

Initial Speed ​​Calculation

[tex]v_{o} =\sqrt{v_{ox}^{2} +v_{oy} ^{2}  }[/tex]

[tex]v_{o} =\sqrt{16^{2} +12^{2}  }[/tex]

Vo=13.42m/s

Vg=13.42m/s

b)Calculation of the time the ball remains in the air

t₁=2*t₂

t₁;time the ball remains in the air

t₂ time when the ball reaches the maximum height

Vf=Vo-g*t₂ : When the ball reaches the maximum height Vf = 0

0=13.42-9.8*t₂

9.8*t₂=13.42

t₂=13.42 ÷9.8

t₂=1.37s

t₁=2*1.37s

t₁= 2.74s

c)Calculation of the maximum height reached by the ball

When the ball reaches the maximum height Vf = 0

Vf²=V₀²-2*g*h

0= V₀²-2*g*h

2*g*h= V₀²

h= V₀² ÷  2*g

h= 13.42² ÷2*9.8

h=9.19m

Answer:

Explanation:

We shall write velocities in vector form

Ship A travels in the direction of 32 °west of north with velocity 28 mph

V₁ = - 28 Sin 32 i + 28 Cos 32 j

Ship B travels in the direction of 24 ° east  of north with velocity 35 mph

V₂ = 35 Sin 24 i + 35 Cos 24 j

Their relative velocity

= V₁ -V₂ =  - 28 Sin 32 i + 28 Cos 32 j -  (35 Sin 24 i + 35 Cos 24 j )

-14.83 i - 14.23 i + 23.74 j - 31.97 j

= - 29.06 i - 8.23 j

Distance between them = relative velocity x time

- 29.06 i - 8.23 j x 2

= - 58.12 i - 16.46 j

magnitude²  =( 58.12 ) ² + ( 16.46)² = 60.40²

magnitude = 60.40 km

Speed of ship A as seen by ship B

= Relative velocity of A wrt B

= - 28 Sin 32 i + 28 Cos 32 j -  (35 Sin 24 i + 35 Cos 24 j )

 =  - 29.06 i - 8.23 j

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]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time      ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at²     ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

so

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

Answer:

1868.5 m

Explanation:

For AB :

u = 0 m/s

a = + 1.68 m/s^2

t = 14.2 s

Let the distance is s1 and the velocity at B is v.

Use first equation of motion

v = u + at

v = 0 + 1.68 x 14.2 = 23.856 m/s

Use third equation of motion

[tex]v^{2}=u^{2}+2as_{1}[/tex]

[tex]23.856^{2}=0^{2}+2\times1.68\times s_{1}[/tex]

s1 = 169.38 m

For BC:

Let the distance is s2.

s2 = v x t

s2 = 23.856 x 68 = 1622.21 m

For CD:

u = 23.856 m/s

a = - 3.7 m/s^2

v = 0

Let the distance is s3.

Use third equation of motion

[tex]v^{2}=u^{2}+2as_{3}[/tex]

[tex]0^{2}=23.856^{2}-2\times 3.7 \times s_{3}[/tex]

s3 = 76.91 m

The total distance traveled is

s = s1 + s2 + s3

s = 169.38 + 1622.21 + 76.91 = 1868.5 m

Thus, the total distance traveled is 1868.5 m.  

Answer:

Explanation:

Energy stored in a capacitor

= 1/2 CV²

C is capacitance and V is potential of the capacitor .

When capacitor is charged to 24 V ,

E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J

When it is charged to 12 volt

E₂ = 1/2 CV²

.5 X 2.4 X 12 X12

= 172.8 J

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:

(a). The speed of the electron is [tex]3.68\times10^{4}\ m/s[/tex]

(b). The distance traveled by the electron is [tex]4.53\times10^{-5}\ m[/tex]

Explanation:

Given that,

Initial velocity = 50 km/s

Electric field  = 50 N/C

Time = 1.5 ns

(a). We need to calculate the speed of the electron 1.5 n s after entering this region

Using newton's second law

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

Using formula of electric force

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

from equation (I) and (II)

[tex]-qE= ma[/tex]

[tex]a = \dfrac{-qE}{m}[/tex]

(a). We need to calculate the speed of the electron

Using equation of motion

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

Put the value of a in the equation of motion

[tex]v = 50\times10^{3}-\dfrac{1.6\times10^{-19}\times50}{9.1\times10^{-31}}\times1.5\times10^{-9}[/tex]

[tex]v=36813.18\ m/s[/tex]

[tex]v =3.68\times10^{4}\ m/s[/tex]

(b). We need to calculate the distance traveled by the electron

Using formula of distance

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

Put the value in the equation

[tex]s = 3.68\times10^{4}\times1.5\times10^{-9}-\dfrac{1}{2}\times\dfrac{1.6\times10^{-19}\times50}{9.1\times10^{-31}}\times(1.5\times10^{-9})^2[/tex]

[tex]s=0.0000453\ m[/tex]

[tex]s=4.53\times10^{-5}\ m[/tex]

Hence, (a). The speed of the electron is [tex]3.68\times10^{4}\ m/s[/tex]

(b). The distance traveled by the electron is [tex]4.53\times10^{-5}\ m[/tex]

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.

To locate the emergency landed plane, we add the vectors representing the plane's two separate legs, breaking them into components using trigonometry. Summing these vectors gives us the direct path for the rescue crew in terms of both distance and bearing from the airport to the plane.

To assist in this emergency landing scenario, we need to compute the resulting position vector by analyzing the two separate motions of the plane. The first motion has the plane fly 170 km at 68° east of north, and the second has it flying 230 km at 48° south of east. By representing these movements as vectors and adding them, we find the direct path the rescue crew should take.

This vector addition can be done graphically or by using trigonometry to break each leg of the plane's journey into its horizontal (east-west) and vertical (north-south) components. After determining the components, we can find the direct distance and bearing from the airport to the plane's location. The past examples and explanations equip us with strategies to calculate the required velocity of the plane relative to the ground and the direction the pilot must head by accounting for the known wind velocities, when necessary.

In summary, to find the direction and distance for the rescue crew, we add the vectors representing the plane's path, utilizing trigonometry to solve the components and then applying vector sum principles to find the result. This procedure allows us to efficiently direct the rescue efforts.