The Hale Telescope on Palomar Mountain in California has a mirror 200 in. (5.08 m) in diameter and it focuses visible light. A)Given that a large sunspot is about 10,000 mi in diameter, what is the most distant star on which this telescope could resolve a sunspot to see whether other stars have them? (Assume optimal viewing conditions, so that the resolution is diffraction limited.)

Answer:

[tex]E=1.56\times 10^{23}\ N/C[/tex]

Explanation:

Given that,

Electric force applied to the electron, [tex]F=25000\ N[/tex]

Charge on electron, [tex]q=1.6\times 10^{-19}\ C[/tex]

We need to find the electric force acting on the electron. The electric field is given by :

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

[tex]E=\dfrac{25000}{1.6\times 10^{-19}}[/tex]

[tex]E=1.56\times 10^{23}\ N/C[/tex]

So, the electric field acting on the electron is [tex]1.56\times 10^{23}\ N/C[/tex]. Hence, this is the required solution

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

1) a = 2.13 m/s²

2) F = 51.12 N

3) T_max = 851.62 N

4)a₁ = 6.08 m/s²

5) a₂ = 12.11 m/s²

This is based on the concept of frictional motion with coefficients of friction.

We are given;

Mass of smaller top crate; m₁ = 24 kg

Mass of larger bottom crate; m₂ = 86 kg

Coefficient of static friction; μ_s = 0.79

Coefficient of kinetic friction; μ_k = 0.62

For the top crate, the sum of forces will be expressed as;

F = m₁a₁  - - - (eq 1)

For the bottom crate, the sum of forces will be expressed as;

T - F = m₂a₂  - - - (eq 2)

Where F is frictional force exerted by small crate on big crate and T is tension.

1) We are told that the rope is pulled with a tension T = 234 N. And that the top crate will not slide. Thus, the acceleration of the small crate will be gotten from a combination of equation 1 and 2 to get the formula;

T = m₂a + m₁a

Making a which is the acceleration the subject gives;

a = T/(m₁+m₂)

Plugging in the relevant values gives;

a = 234/(24 + 86)

a = 2.13 m/s²

2) From eq(1) above, we see that the formula for the frictional force the small crate exerts on the big one is;

F = m₁a₁

Thus, Plugging in the relevant values gives;

F = 24 × 2.12

F = 51.12 N

3) maximum tension that the lower crate can be pulled at before the upper crate begins to slide will occur at the maximum friction. The formula for mx friction is;

(F₂)_max = μ_s•m₂•g

Using the concept of eq(1) We can say that;

F₂ = m₂a

m₂a = μ_s•m₂•g

m₂ will cancel out to give;

a = μ_s • g

From first answer earlier, we saw that;

T = m₂a + m₁a

T = (m₂ + m₁)a

Plugging in the relevant values gives;

T_max = (m₁ + m₂)μ_s*g

T_max = (24 + 86) × 0.79 × 9.8

T_max = 851.62 N

4) To get the acceleration of the upper crate, we will make use of the formula for kinetic friction which is:

F = μ_k*m₁g

From earlier, we saw that; F = m₁a₁

Thus;

m₁a₁ = μ_k*m₁g

m₁ will cancel out to get;

a₁ = μ_k*g

a₁ = 0.62 × 9.8

a₁ = 6.08 m/s²

5) the acceleration of the lower crate as the upper crate slides will be gotten by putting μ_k*m₁*g for m₁a₁ in eq(2) to get;

m₂a₂ = T - (μ_k*m₁*g)

Making a₂ the subject gives us;

a₂ = (T - μ_k•m₁•g)/m₂

Plugging in the relevant values;

a₂ = (1187 - (0.62 × 24 × 9.8))/86

a₂ = 12.11 m/s²

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Answer with Explanation:

The force that is exerted on a object is proportional to the rate of change of momentum. Mathematically

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

As we can see from the above equation the force that is exerted on the object is inversely proportional to the time in which this momentum change is brought meaning if the time interval in which the momentum change is brought about is larger smaller the force that acts on the object.

When the driver of a car crashes into a barrier the airbags in the car are deployed instantly as due to inertia of motion the driver continues to move in the original direction of motion thus hitting the air bag in the process. Since the air in the airbag is compressible it increases the time in which this momentum becomes zero, reducing the force that is exerted on the person, thus protecting the person.

Airbags reduce the force on a driver's head by increasing the time over which the head decelerates, thus applying the principle of impulse to protect during a car crash. They achieve this in tandem with seatbelts and car structures designed to crumple and absorb impact forces.

Airbags protect drivers during a crash by utilizing the principle of impulse, which is the product of the net force acting on an object and the time over which the force acts. The airbag increases the time over which the driver's head decelerates to a stop, in turn decreasing the force that acts on the head. By extending the stopping time from a fraction of a second (if hitting the dashboard) to a longer time (momentum is stopped by the airbag), the average force experienced by the driver's head is considerably lessened. This reduction in force reduces the likelihood of severe injuries. The airbag also has vents to deflate during the crash, preventing bounce-back injuries.

Further, the airbag works with seatbelts that may have variable tension to distribute forces across the body more evenly, thus reducing stress on any single part of the body. Modern vehicles include these safety features not just for comfort but for crucial safety improvements, allowing for a longer time of force application during a crash and therefore less force at any instant. The same concept applies to the car's structure with plastic components designed to crumple and prolong collision time, further reducing forces experienced by the passengers.

Answer:

Distance, d = 101.388 meters

Explanation:

It is given that,

The average velocity of plane, v = 3650 km/h = 1013.88 m/s

The time for which eye blinks, [tex]t=100\ ms=0.1\ s[/tex]

Let d is the distance covered by the jet. It can be calculated as :

[tex]d=v\times t[/tex]

[tex]d=1013.88\ m/s\times 0.1\ s[/tex]        

d = 101.388 meters

So, the distance covered by a fighter jet during a pilot's blink is 101.388 meters. Hence, this is the required solution.                                                                              

Final answer:

During a blink that lasts 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.

Explanation:

To calculate the distance a fighter jet travels during a pilot's blink, we can use the formula distance = velocity × time. The pilot's blink lasts 100 milliseconds (ms) which is 0.1 seconds because 1000 ms equals 1 second. The jet's velocity is given as 3650 kilometers per hour (km/h).

First, we'll need to convert the jet's velocity to meters per second (m/s) since the time of the blink is given in seconds. 1 km equals 1000 meters, and there are 3600 seconds in an hour, so:

Velocity in m/s = (Velocity in km/h) × (1000 m/km) / (3600 s/h) = (3650) × (1000) / (3600) = 1013.89 m/s

Now, we will multiply the velocity in m/s by the time in seconds to get the distance:

Distance = Velocity × Time = (1013.89 m/s) × (0.1 s) = 101.39 meters

So, during the blink of an eye, which is 0.1 seconds, a fighter jet traveling at an average velocity of 3650 km/h will cover approximately 101.39 meters.

Answer:

Explanation:

The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .

Distance traveled by it = .15 t

Distance traveled by hare = .15 t - .35 m

Time taken by hare to complete this distance

=  t - 5 x 60 s

Speed of hare

= Distance / time

( .15t-.35 ) / t - 300 , so

[tex]\frac{.15t-.35}{t-300} = 3.75[/tex]

t = 312.40

= 5 minutes 12.4 seconds

Distance of race

312.4 x speed of tortoise

= 312.4 x .15

= 46.85 m

(a) The length of the race is 91.1 meters.

(b) The race takes 605.0 seconds.

The tortoise moves at a speed of 0.15 m/s while the hare moves at a speed of 25 * 0.15 = 3.75 m/s.

The hare stops for 5 minutes, which is equal to 5 * 60 = 300 seconds.

In the 300 seconds that the hare was taking a nap, the tortoise was able to move a distance of 0.15 * 300 = 45 meters.

When the hare woke up, he started running at his top speed. He was able to catch up to the tortoise and surpass him by 35 cm. This means that the hare was able to run a distance of 45 + 0.35 = 45.35 meters.

Therefore, the length of the race is 45.35 meters.

The total time it took for the race to finish is 300 + 605 = 905 seconds.

So the answer is (a) 91.1 meters (b) 605.0 seconds

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The skateboarder's horizontal distance traveled after leaving the ramp can be calculated from her initial velocity and time in the air during projectile motion.

The skateboarder's motion can be analyzed using kinematic equations. When she leaves the ramp, she will follow a projectile motion trajectory. The horizontal distance she travels can be calculated using her initial vertical velocity and the time she is in the air.

The skateboarder touches down approximately 4.87 meters from the end of the 1.0-meter-high, 30° ramp after being launched at a speed of 7.6 m/s.

First, we'll find the horizontal and vertical components of the skateboarder's initial velocity. The angle of the ramp is 30°, and her speed is 7.6 m/s:

Horizontal component ([tex]V_x[/tex]) = 7.6 * cos(30°)

or, [tex]V_x[/tex] = 7.6 * 0.866

or, [tex]V_x[/tex] = 6.58 m/s

Vertical component ([tex]V_y[/tex]) = 7.6 * sin(30°)

or, [tex]V_y[/tex] = 7.6 * 0.5

or, [tex]V_y[/tex] = 3.8 m/s

solving for [tex]{V_f}_y[/tex] in the horizontal and vertical components of velocity as the skater leaves the ramp :

[tex]V_f}_y^2 = V_y^2 - 2* g * \Delta x[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{V_y^2 - 2 * g * \Delta x }[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|(3.8)^2 - 2 * 9.8 * 1.0|}[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|14.44 - 19.6|}[/tex]

or, [tex]{V_f}_y[/tex] = [tex]\sqrt{|-5.16|}[/tex]

or, [tex]{V_f}_y[/tex] = √(5.16)

or, [tex]{V_f}_y[/tex] = 2.27 m/s

In the x-axis, there is no acceleration so, [tex]a_x[/tex] = 0 m/s²

[tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex]+ 2 * [tex]a_x[/tex] * Δx

or, [tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex] + 2 * 0 * Δx

or, [tex]{V_f}_x^{2}[/tex] = [tex]V_x^2[/tex]

or, [tex]{V_f}_x[/tex] = [tex]V_x[/tex]

or, [tex]{V_f}_x[/tex] = 6.58 m/s

The skateboarder reaches a height of 1.0 m. We will use the following kinematic equation to find the time (T) it takes for her to fall this distance:

y = [tex]{V_f}_y[/tex] * t + 0.5 * (-g) * t²

Here, y = -1.0 m (since she’s falling), [tex]{V_f}_y[/tex] = 2.27 m/s, and g = 9.8 m/s². Plugging in the values:

-1.0 = 2.27 * T - 0.5 * 9.8 * T²

or, 4.9 * T² - 2.27 * T - 1.0 = 0

Solving this quadratic equation (4.9 * T² - 2.27 * T - 1.0 = 0) for T gives:

T ≈ 0.74 seconds

Next, we calculate the horizontal distance traveled using this time and the horizontal velocity:

Distance = [tex]{V_f}_x[/tex] * T = 6.58 m/s * 0.74 s

Distance = 4.87 meters

Therefore, the skateboarder touches down approximately 4.87 meters from the end of the ramp.

Answer:

0.0567 N

Explanation:

q1 = 3 micro coulomb

q2 = - 2 micro coulomb

OB = 50 cm

AB = 60 cm

By using Pythagoras theorem in triangle OAB

[tex]OA^{2}=OB^{2}+AB^{2}[/tex]

[tex]OA^{2}=50^{2}+60^{2}=6100[/tex]

OA = 78.1 cm

By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is

[tex]F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}[/tex]

F = 0.0885 N

The horizontal component of force is

= F CosФ = [tex]F\times \frac{OB}{OA}[/tex]

= 0.0885 x 50 / 78.1 = 0.0567 N

Answer and Explanation:

The motion of the viper is circular and it completes a semi- circle forming an arc and then retraces its path back.

Thus the force experienced by the bug is centripetal force thus centripetal acceleration and in the absence of this force the bug will get dislodged.

This centripetal force is mainly provided by the static friction between the blades and the bug.

When the wipers are moving with high velocity when turned on, larger centripetal force is required to keep the bug moving with the wiper on the arc  than at low setting.

Thus there are more chances for the bug to be dislodged at higher setting than at low setting.

Answer:

the greatest height at which the ball can reach is 19.62 m.

Explanation:

given,

time taken by the ball to reach the ground = 4 s

time at which ball reach at maximum height = 4/2 = 2 s

velocity of the ball at the top most point = 0 m/s

we know,

v = u + a t

0 = u + (-9.81 ) × 2

u = 19.62 m/s

maximum height achieved

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

[tex]s = 19.62\times 2 + \dfrac{1}{2}(-9.81)\times 2^2[/tex]

s = 19.62 m

hence, the greatest height at which the ball can reach is 19.62 m.

Explanation:

Charge, [tex]q=-30\ \mu C=-30\times 10^{-6}\ C[/tex]

Electric force, [tex]F=27\ mN=27\times 10^{-3}\ N[/tex]

We need to find the magnitude and direction of electric field at that location. The relation between the electric field and electric force is given by :

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

[tex]E=\dfrac{27\times 10^{-3}\ N}{-30\times 10^{-6}\ C}[/tex]

[tex]E=-900\ N/C[/tex]

For a negative charge, the direction of electric field is inward. The direction of electric force and electric field is same. So, the direction of electric field in this case is in upward direction. Hence, this is the required solution.

The magnitude of the electric field is 900 N/C, and the direction is downward. Therefore, the electric field at that location is  -900 N/C  (negative value indicating downward).

To determine the magnitude and direction of the electric field at the location where the point charge experiences an electrostatic force, we can use the relationship between the force  [tex]\mathbf{F}[/tex]  experienced by a charge  q  in an electric field [tex]\mathbf{E}[/tex] :

[tex]\mathbf{F} = q \mathbf{E}[/tex]

Given:

- The charge  q  is -30 µC (microcoulombs).

- The electrostatic force  [tex]\mathbf{F}[/tex]  is 27 mN (millinewtons) upward.

First, we convert the given quantities to standard SI units:

- Charge  [tex]q = -30 \, \mu \text{C} = -30 \times 10^{-6} \, \text{C}[/tex]

- Force  [tex]\mathbf{F} = 27 \, \text{mN} = 27 \times 10^{-3} \, \text{N}[/tex]

Next, we use the equation  [tex]\mathbf{F} = q \mathbf{E}[/tex] to solve for the electric field [tex]\mathbf{E}[/tex] :

[tex]\mathbf{E} = \frac{\mathbf{F}}{q}[/tex]

Substituting the given values:

[tex]\mathbf{E} = \frac{27 \times 10^{-3} \, \text{N}}{-30 \times 10^{-6} \, \text{C}} \\\\\mathbf{E} = \frac{27 \times 10^{-3}}{-30 \times 10^{-6}} \, \text{N/C} \\\\\mathbf{E} = \frac{27}{-30} \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -0.9 \times 10^{3} \, \text{N/C} \\\\\mathbf{E} = -900 \, \text{N/C}[/tex]

The negative sign indicates the direction of the electric field. Since the force on a negative charge is upward, the electric field must be directed downward (opposite to the direction of the force on a negative charge).

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m      

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

[tex]x = x_{0} + V_{0}t + \frac{1}{2}at^2[/tex]

[tex]1.66 = 0 + 0t + \frac{1}{2}9.8t^2[/tex]

[tex]1.66 = 4.9t^2[/tex]

[tex]\frac{1.66}{4.9}  = t^2[/tex]

[tex]\sqrt{0.339} = t\\ t = 0.582s[/tex]

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

Answer:

The total force exerted on the Y axis is: -52.07μC

Explanation:

This is an electrostatic problem, so we will use the formulas from the Coulomb's law:

[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]

We are interested only of the effect of the force on the Y axis. We can notice that the charge placed on the x=4cm will exers a force only on the Y axis so:

[tex]Fy1=9*10^9*\frac{6.05*10^{-9}*(-1.97)*10^{-9}}{(3.02*10^{-2})^2}\\[/tex]

Fy1=-117.61μC

For the charge placed on the origin we have to calculate the distance and the angle:

[tex]r=\sqrt{(4*10^{-2}m)^2 +(3.02*10^{-2}m)^2} \\r=5cm=0.05m[/tex]

we can find the angle with:

[tex]alpha = arctg(\frac{3.02cm}{4cm})=37^o[/tex]

The for the Force on Y axis is:

[tex]Fy2=9*10^9*\frac{6.05*10^{-9}*(5)*10^{-9}}{(3.02*10^{-2})^2}*sin(37^o)\\[/tex]

Fy2=65.54μC

The total force exerted on the Y axis is:

Fy=Fy1+Fy2=-52.07μC

Answer:

gravitational force between two objects is 8.37 N

Explanation:

given data

mass of object  = 85000 kg

distance between two object r  = 2.40 m

to find out

gravitational force between two objects

solution

we know that gravitational constant G is  6.67 × [tex]10^{-11}[/tex] m³/ s²-kg

so

gravitational force between two objects formula is

F  = [tex]G*\frac{m1m2}{r^2}[/tex]

here E is gravitational force and G is gravitational constant put here all value and m1 and m2 are mass of object

F  = [tex]6.67*10^{-11}*\frac{85000^2}{2.40^2}[/tex]

F = 8.37 N

gravitational force between two objects is 8.37 N

Answer:

a) [tex]R_s = 0.092[/tex]

b) [tex]R_p = 0.085[/tex]

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

[tex]R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2[/tex]

using snell's law to calculate θ t

[tex]sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}[/tex]

[tex]cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}[/tex]

a) [tex]R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2[/tex]

[tex]R_s = 0.092[/tex]

b) [tex]R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2[/tex]

[tex]R_p = 0.085[/tex]

Answer:

a) [tex]v_{3} =8.43 m/s[/tex]

b) [tex]v_{2}=2.15m/s[/tex]

c) ΔK=[tex]-28.18x10^4J[/tex]

d)ΔK=[tex]-10.33x10^4J[/tex]

Explanation:

From the exercise we know that there is a collision of a sports car and a truck.

So, the sport car is going to be our object number 1 and the truck object number 2.

[tex]m_{1}=1050kg\\v_{1}=-13m/s\\m_{2}=6320kg\\v_{2}=12m/s[/tex]

Since the two vehicles remain locked together after the collision the final mass is:

[tex]m_{3}=7370kg[/tex]

a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle

[tex]p_{1}=p_{2}[/tex]

[tex]m_{1}v_{1}+ m_{2}v_{2}=m_{3}v_{3}[/tex]

[tex]v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3}}=\frac{(1050kg)(-13m/s)+(6320kg)(12m/s)}{7370kg}[/tex]

[tex]v_{3}=8.43m/s[/tex]

b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before

[tex]m_{1}v_{1}+ m_{2}v_{2}=0[/tex]

[tex]v_{2}=\frac{-m_{1}v_{1}}{m_{2} }=\frac{-(1050kg)(-13m/s)}{(6320kg)}=2.15m/s[/tex]

c) To find the change in kinetic energy we need to do the following steps:

ΔK=[tex]k_{2}-k_{1}=\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]

ΔK=[tex]\frac{1}{2}(7370)(8.43)^{2}-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(12)^{2} )=-28.18x10^{4}J[/tex]

d) The change in kinetic energy where the two vehicles stopped in the collision is:

ΔK=[tex]k_{2}-k_{1}=0-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} )[/tex]

ΔK=[tex]-(\frac{1}{2}(1050)(-13)^{2}+\frac{1}{2}(6320)(2.15)^{2} )=-10.33x10^4J[/tex]

Answers:

a) [tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]

b) [tex]\rho_{liq}= 1.48 g/cm^{3}[/tex]

c) When we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.

Explanation:

a) According to Archimedes’ Principle:

A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.

In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium.  This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.

Hence:

[tex]W_(cylinder)=B[/tex] (1)

Where:

[tex]W_(cylinder)=m.g[/tex] is the weight of the wooden cylinder, where [tex]m[/tex] is its mass and [tex]g[/tex] gravity.

[tex]B[/tex] is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.

We can rewrite (1) as follows:

[tex]m_{cylinder}g=m_{water}g[/tex] (2)

On the other hand, we know density [tex]\rho[/tex] establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:

[tex]\rho=\frac{m}{V}[/tex] (3)

isolating the mass:

[tex]m=\rho V[/tex]    (4)

Now we can express (2) in terms of the density and the volume of cylinder and water:

[tex]\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g[/tex] (5)

In this case [tex]V_{water}[/tex] is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).

At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:

[tex]V_{cylinder}=\pi r^{2} h[/tex] (6)

Where [tex]r[/tex] is the radius and [tex]h=30 cm[/tex] the total height of the cylinder.

[tex]V_{water}=\pi r^{2} (h-h_{top})[/tex] (7)

Where [tex]h_{top}=13.5 cm[/tex] is the height of the top of the cylinder above the surface of water and [tex](h-h_{top})[/tex] is the height of the sumerged part of the cylinder.

Substituting (6) and (7) in (5):

[tex]\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g[/tex] (8)

Clearing [tex]\rho_{cylinder}[/tex]:

[tex]\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}[/tex] (9)

Simplifying;

[tex]\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}[/tex] (10)

Knowing [tex]\rho_{water}=1g/cm^{3}[/tex]:

[tex]\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})[/tex] (11)

[tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex] (12) This is the density of the wooden cylinder

b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ([tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]), but the density of the liquid [tex]\rho_{liq}[/tex] is unknown.

Applying again the Archimedes principle:

[tex]\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g[/tex] (13)

Isolating [tex]\rho_{liq}[/tex]:

[tex]\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}[/tex] (14)

Where:

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

[tex]V_{liq}=\pi r^{2} (h-h_{top})[/tex]

Then:

[tex]\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}[/tex] (15)

[tex]\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}[/tex] (16)

[tex]\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}[/tex] (17)

[tex]\rho_{liq}= 1.48 g/cm^{3}[/tex] (18) This is the density of the liquid

c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.

This is because in this case, when we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.

Answer:

Rate of heat loss=7992 kJ/h

Explanation:

First, we can consider the room as a closed system, so we can use the first law of the thermodynamic:

[tex]Qnet,in-Wnet,out = Esystem[/tex]

The incoming net heat is the incoming heat less the leaving heat, the leaving net work is the leaving work less the incoming work  and the system energy is only the change of the intern energy because we have a stationary system, so:

[tex](Qin-Qout)-(Wout-Win) =U2-U1[/tex]

From the problem we consider that there is no change of the temperature thus there is no change in the intern energy(U2=U1), moreover there is no incoming heat neither exit work, all the work is made to the system as a result the leaving heat is equal to the incoming work:

[tex](0-Qout)-(0-Win) =0[/tex]

[tex]-Qout+Win =0[/tex]

[tex]Qout=Win[/tex]

Substituting the known values we can get the rate of heat loss (exit heat):

[tex]Q_{out}=Q_{loss}=W_{ref}+W_{TV}+W_{res}+W_{fan}[/tex]

[tex]Q_{loss}=250 [W]+120 [W]+1800 [W]+50 [W][/tex]

[tex]Q_{loss}=2220 [W]=2220 [\frac{J}{s} ][/tex]

Converting to Joules/hours:

[tex]Q_{loss}=2220 [\frac{J}{s} ][\frac{3600 s}{1 h} ][/tex]

[tex]Q_{loss}=7992000 [\frac{J}{h} ][/tex]

Finally the rate of heat loss is:

[tex]Q_{loss}=7992 [\frac{kJ}{h} ][/tex]

Final answer:

The rate of heat loss from a room with several electrical appliances running and constant air temperature is 7992 kJ/h, calculated by summing the power of all appliances and converting to kJ/h.

Explanation:

To find the rate of heat loss from the room, we need to consider the energy input from all the electrical appliances and the fact that the air temperature in the room remains constant. The total power consumption of the appliances is the sum of the power of the refrigerator, TV, electric resistance heater, and fan, which is 250 W + 120 W + 1.8 kW + 50 W = 2.22 kW.

Since the temperature is constant, the rate of energy input equals the rate of heat loss. This input is the combined power consumption of the appliances and is entirely converted into heat (assuming 100% efficiency for simplicity).

To express this rate in kilojoules per hour (kJ/h), multiply the power in kilowatts by the number of hours and by 1000 to convert from kW to kJ since 1 kW = 1 kJ/s:

2.22 kW = 2.22 kJ/s

So, the rate of heat loss is:

2.22 kJ/s × 3600 s/h = 7992 kJ/h

Answer:

Explanation:

Given

Distance to grandmother's house=100 mi

it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph

Let Julie travel x mi with 30 mph and 100-x with 70 mph

[tex]\frac{x}{30}=\frac{100-x}{70}[/tex]

x=30 mi

Therefore

Julie's Average speed on the way to Grandmother's house[tex]=\frac{100}{\frac{50}{30}+\frac{50}{70}}[/tex]

=42 mph

On return trip

[tex]=\frac{100}{2\frac{30}{30}}=50 mph[/tex]

Final answer:

Julie's average speed to Grandmother's house is 42.02 mph. The average speed for the return trip cannot be calculated without the total time of travel or the division of time at each speed. Her average speed for the entire round trip is approximately 26.67 mph, and the average velocity is 0 mph since the starting and ending points are the same.

Explanation:

To calculate Julie's average speed on the way to Grandmother's house, we need to use the formula for average speed, which is total distance traveled divided by the total time taken. Since Julie drives half the distance at 30.0 mph and the other half at 70.0 mph for a total distance of 100 miles, we can calculate the time taken for each half. At 30 mph, for 50 miles, the time taken is ≈ 1.67 hours, and at 70 mph, for 50 miles, the time taken is ≈ 0.71 hours. The total time is ≈ 2.38 hours. Therefore, average speed is 100 miles ÷ 2.38 hours = 42.02 mph.

On the return trip, she drives half the time at 30.0 mph and half the time at 70.0 mph. However, without additional information, we cannot calculate the average speed for the return trip because we need the total time or the portion of time spent at each speed. The calculation is different from the first trip because this time it is dependent on time, not distance. For her entire trip, if Julie returned home 7 hours and 30 minutes after she left, and assuming the same 100-mile distance back, her average speed for the entire trip is the total distance (200 miles) divided by the total time (7.5 hours), which is ≈ 26.67 mph. However, since she returns to the starting point, her displacement is zero, and thus her average velocity for the entire trip is 0 mph, similar to the example in Figure 2.10.

Answer:

260 g

Explanation:

Given:

Let n be the number of such silicon dioxide that would give the desired mass of sand.

According to the question, the surface area of all the sand particles will be equal to the surface area of the cube.

[tex]\therefore n\times \textrm{Area of a sand particle}=\textrm{Surface area of a cube }\\\Rightarrow n\times 4\pi r^2= 6a^2\\\Rightarrow n = \dfrac{6a^2}{4\pi r^2}[/tex]

Let the total mass of sand required be M.

[tex]\textrm{Total mass of sand} = \textrm{Mass of all the required sand particle}\\\Rightarrow M = n\times \textrm{mass of one sand particle}\\\Rightarrow M = n\times \textrm{Density of sand particle}\times \textrm{Volume of a sand particle}\\\Rightarrow M = n\times\rho \times \dfrac{4}{3}\pi r^3\\[/tex]

[tex]\Rightarrow M = \dfrac{6a^2}{4\pi r^2}\times\rho \times \dfrac{4}{3}\pi r^3\\\Rightarrow M = 2\times\rho \times a^2r\\\Rightarrow M = 2\times2600 \times (1)^2\times 5\times 10^{-5}\\\Rightarrow M =0.260\ kg\\\Rightarrow M =260\ g\\[/tex]

Hence, the total mass of the sand required will be equal to the 260 g.

Answer:

a) The taxi is 107 s in motion

b) The average velocity is 26.2 m/s

Explanation:

First, the car travels with an acceleration of 2.00 m/s². The equations for position and velocity that apply for the car are:

x = x0 + v0 t + 1/2 a t²

v = v0 + a t

where

x = position at time t

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = speed

Let´s calculate how much distance and for how long the taxi travels until it reaches a speed of 29.0 m/s:

Using the equation for velocity:

v = v0 + a t

v - v0 / a = t

(29.0 m/s - 0 m/s) / 2 m/s² = t

t = 14.5 s

Then, in the equation for position:

x = x0 + v0 t + 1/2 a t²

x = 0 + 0 + 1/2 * 2.00 m/s² * (14.5 s)²

x = 210 m

Then, the vehicle travels at constant speed for 87 s. The distance traveled will be:

x = v * t

x = 29.0 m/s * 87.0 s = 2.52 x 10³ m

Lastly the car stops (v = 0) in 5 s. In this case, the car has a constant negative acceleration:

Using the equation for velocity:

v = v0 + a t

if v=0 in 5 s, then:

0 = 29.0 m/s + a * 5.00 s

a = -29.0 m/s / 5.00 s  

a = -5.80 m/s²

Using now the equation for the position, we can calculate how far has the taxi traveled until it came to stop:

x = x0 + v0 t + 1/2 a t²

x = 0 + 29.0 m/s * 5.00 s -1/2 * 5.80 m/s² * (5.00s)²

x = 72.5 m

a) The taxi has been in motion for:

Total time = 14.5 s + 87.0 s + 5.00s = 107 s

Note that we have always used x0 = 0, then, we have calculated the displacement for each part of the trip.

Adding all the displacements, we will get the total displacement:

Total displacement = 210 m + 2.52 x 10³ m + 72.5 m = 2.80 x 10³ m

Average speed = total displacement / total time

Average speed = 2.80 x 10³ m / 107 s = 26.2 m/s

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

[tex]1\ m=\frac{1}{1609.344}\ miles[/tex]

1 hour = 60×60 seconds

[tex]1\ s=\frac{1}{3600}\ hours[/tex]

[tex]28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h[/tex]

Top speed of the cheetah is 64.2 mi/h

Equation of motion

[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2[/tex]

Acceleration of the cheetah is 8.68 m/s²

2)

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s[/tex]

It takes a cheetah 3.31 seconds to reach its top speed.

3)

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m[/tex]

It travels 47.5 m in that time

4) When s = 120 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s[/tex]

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

The cheetah's top speed of 28.7 m/s is approximately 64.2 mi/h. It takes a cheetah 3.31 seconds to reach its top speed, traveling a distance of 47.4 meters during this acceleration. To reach a stationary rabbit 120 meters away, it would take the cheetah a total of 5.84 seconds.

The question involves converting speeds from meters per second to miles per hour, finding the time taken to achieve a certain speed, calculating the distance traveled in that time, and determining the time required to reach a target.

Answer:

A) x and y components of the electric field  (Ep) at the origin.

Epx = -1620.5 N/C

Epy = -1620.5 N/C

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

   Ep= 2291.7 N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence  

1nC= 10⁻⁹C

Data

K= 9x10⁹N*m²/C²

q₁ = q₂= +6.5nC=+6.5 *10⁻⁹C

d₁=d₂=0.190m

Graphic attached

The attached graph shows the field due to the charges:

Ep₁ : Electric Field at point P  (x=0, y=0) due to charge q₁. As the charge q₁ is positive (q₁+) ,the field leaves the charge.

Ep₂: Electric Field at point  P (x=0, y=0) due to charge q₂. As the charge q₂  is positive (q₂+) ,the field leaves the charge

Ep: Total field at point P due to charges q₁ and q₂.

Because q₁ = q₂ and d₁ = d₂, then, the magnitude of Ep₁ is equal to the magnitude of Ep₂

Ep₁ = Ep₂ = k*q/d² = 9*10⁹*6.5*10⁻⁹/0.190m² = 1620.5 N/C

Look at the attached graphic :

Epx = Ep₁= -1620.5 N/C

Epy = Ep₂= -1620.5 N/C

A) x and y components of the electric field  (Ep) at the origin.

Ep = (1620.5(-i)+1620.5(-j)) N/C

B) Magnitude of the electric field  (Ep) at the origin.

[tex]E_{p} =\sqrt{1620.5^{2}+1620.5^{2} } = 2291.7 \frac{N}{C}[/tex]

Answer: 0.506 m

Explanation: To solve this we use the relationship for the harmonic in the string which are given by the following expression:

λ=2*L/3

λ=2*0.76 m/3= 0.506 m

Answer:

7.468 kN

Explanation:

Here the force of 7468 Newton is given.

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

The number is 7468.0

Here, the only solution where the number of significant figures is kilo

1 kilonewton = 1000 Newton

[tex]1\ Newton=\frac{1}{1000}\ kilonewton[/tex]

[tex]\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton[/tex]

So 7468 N = 7.468 kN

Answer:

3.28 m

3.28 s

Explanation:

We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.

Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

V0 = 4 m/s

a = -2.45 m/s^2 (because the acceleration is down slope)

Then:

X(t) = 4*t - 1.22*t^2

And the equation for speed is:

V(t) = V0 + a * t

V(t) = 4 - 2.45 * t

If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:

0 = 4 - 2.45 * t

4 = 2.45 * t

t = 1.63 s

Replacing that time on the position equation:

X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m

To find the time it will take to return we equate the position equation to zero:

0 = 4 * t - 1.22 * t^2

Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:

0 = t * (4 - 1.22*t)

t1 = 0

0 = 4 - 1.22*t2

1.22 * t2 = 4

t2 = 3.28 s

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound.  The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

[tex]d=v\times t[/tex]

[tex]d=340\ m/s\times 5\ s[/tex]  

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.    

Answer:

13.65 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 9.5 m

a = Acceleration = 9.81 m/s²

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as-u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 9.5-0^2}\\\Rightarrow v=13.65\ m/s[/tex]

The flower pot is moving at a speed of 13.65 m/s when it crashes the sidewalk.

Answer:

(a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Explanation:

Given that,

A bird watcher meanders through the woods, walking 1.46 km due east, 0.123 km due south, and 4.24 km in a direction 52.4° north of west.

(a). We need to calculate the displacement

Using Pythagorean theorem

[tex]D=\sqrt{(OA)^2+(AB)^2}[/tex]

[tex]D=\sqrt{(1.46-4.24\sin52.4)^2+(0.123-4.24\cos52.4)^2}[/tex]

[tex]D=3.11\ km[/tex]

(b). We need to calculate the average velocity

Using formula of average velocity

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = displacement

T = time

Put the value into the formula

[tex]v_{avg}=\dfrac{3.11}{1.285}[/tex]

[tex]v_{avg}=2.42\ km/hr[/tex]

Hence, (a). The displacement is 3.11 m.

(b). The average velocity is 2.42 km/hr.

Answer:

The correct answer is option 'a' 'The momentum is always conserved while as the kinetic energy may be conserved'

Explanation:

The conservation of momentum is a basic principle in nature which is always valid in an collision between 'n' number of objects if there are no external forces on the system. It is valid for both the cases weather the collision is head on or glancing or weather the object is elastic or inelastic.

The energy is only conserved in a collision that occurs on a friction less surface and the objects are purely elastic. Since in the given question it is mentioned that only the surface is friction less and no information is provided regarding the nature of the objects weather they are elastic or not hence we cannot conclusively come to any conclusion regarding the conservation of kinetic energy as the objects may be inelastic.

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h  east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is  

hypotenuse/[tex]\sqrt{2}[/tex]

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/[tex]\sqrt{2}[/tex] km/h= 63,6396 km/h.  

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east,  here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

[tex]\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h [/tex],

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= [tex]sin^{-1}'frac{63,6396}{629,5851}[/tex]=5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

626,6396 km/h south + 63,6396km/h  east.

Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get

Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get  

Xdetour=[/tex]\sqrt{2*  11,6672x^{2} }[/tex]  = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows  see fig 2