Answers
Answer:
Net force refers to the sum of all the forces that act upon an object.
Explanation:
Related Questions
A block of a plastic material floats in water with 35.6% of its volume under water. What is the density of the block in kg/m3
Answers
Answer:
The density is \(\rho_{block} = 356 \ kg/m^3 \)
Explanation:
From the question we are told that
The proportion of the plastic under water is p = 0.356
Generally at equilibrium the buoyant force is equal to the weight of the block i.e
\(B = mg\)
Generally the Buoyant force is mathematically represented as
\(B = \rho * V * g\)
Here \(\rho\) density of the water with value
\(\rho = 1000 kg/m^3\) ,
V is the volume of water displaced by the block which is \( p * V_{block \) ,
So
\(1000 * 0.356 V_{block} * g = m * g\)
Here m is the mass of the block which is mathematically represented as
\(m = \rho_{block} * V_{block}\)
So
\(1000 * 0.356 V_{block} * g = \rho_{block} * V_{block}* g\)
\(1000 * 0.356 V_{block} = \rho_{block} * V_{block}* g\)
=> \(\rho_{block} = 1000 * 0.356\)
=> \(\rho_{block} = 356 \ kg/m^3 \)
help meee plisssssssssssssssssssssssssssss
Answers
Answer:
Resistance = 0.22 Ohms
Current = 13.63636 A
Explanation:
Total resistance for resistors in parallel is given by:
\(\frac{1}{T} =\frac{1}{R1} +\frac{1}{R2} +...+\frac{1}{Rn}\) where n is the number of resistors
\(\frac{1}{T} = \frac{1}{1.1} +\frac{1}{1.1} +\frac{1}{1.1} +\frac{1}{1.1} +\frac{1}{1.1}\)
if you solve that you get \(\frac{1}{T} = 5/1.1 \\\\T = 1.1/5T = 0.22 Ohms\)
Solve current using V=IR
I=V/R =
I=3/0.22
I = 13.63636 A
On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs account for 80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.
Suppose a 65.0 kg skater is 1.60 m tall, has arms that are each 74.0 cm long (including the hands) and a trunk that can be modeled as being 37.0 cm in diameter. The skater is initially spinning at 62.0 rpm with her arms outstretched.
What will her angular velocity 2 be (in rpm) after she pulls in her arms and holds them tightly against her trunk? Assume negligible friction between the skater and the ice.
Answers
The skater's angular velocity after pulling in her arms and holding them tightly against her trunk will be approximately 89.5 rpm or 1.492 rps (revolutions per second).
To solve this problem, we need to apply the principle of conservation of angular momentum. Initially, the skater is spinning with her arms outstretched, and then she pulls her arms in and holds them tightly against her trunk. The total angular momentum before and after the change should remain the same.
Calculate the initial moment of inertia:
The moment of inertia of the skater with her arms outstretched can be calculated as the sum of the individual moments of inertia for each body part.
For the vertical cylinder (head, trunk, and legs):
Mass = 0.07 × 65.0 kg = 4.55 kg
Height = 1.60 m
Radius = (37.0 cm / 2) = 0.185 m
Moment of inertia for the cylinder = (1/12) × Mass × Height^2 + (1/4) × Mass × Radius^2
For the horizontal rods (arms and hands):
Each arm length = 74.0 cm = 0.74 m
Mass of each arm = 0.13 × 65.0 kg = 8.45 kg
Moment of inertia for each rod = (1/3) × Mass × Length^2
Total initial moment of inertia = Moment of inertia for the cylinder + 2 × Moment of inertia for each rod
Calculate the initial angular momentum:
The initial angular momentum is given by L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Convert the given initial angular velocity from rpm to rad/s:
Initial angular velocity = 62.0 rpm = (62.0 rpm) × (2π rad/rev) / (60 s/min) = 6.493 rad/s
Initial angular momentum = Total initial moment of inertia × Initial angular velocity
Calculate the final moment of inertia:
When the skater pulls her arms in and holds them tightly against her trunk, the moment of inertia changes. The arms and hands contribute no moment of inertia since they are now tightly held against the trunk.
The new moment of inertia will be that of the vertical cylinder alone.
Conservation of angular momentum:
According to the principle of conservation of angular momentum, the initial angular momentum should be equal to the final angular momentum.
Final angular momentum = Final moment of inertia × Final angular velocity
Since the total angular momentum remains constant:
Initial angular momentum = Final angular momentum
Calculate the final angular velocity:
Rearrange the equation to solve for the final angular velocity:
Final angular velocity = Initial angular momentum / Final moment of inertia
Substitute the known values and solve for the final angular velocity.
Convert the final angular velocity to rpm:
Final angular velocity (in rpm) = Final angular velocity × (60 s/min) / (2π rad/rev)
Therefore, the skater's angular velocity after pulling in her arms and holding them tightly against her trunk will be approximately 89.5 rpm or 1.492 rps (revolutions per second).
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A 1.5kg block is held in place and compresses a 150N/m spring by 30cm from its relaxed position. The block is then released. What speed will the block have at the instant when the spring is no longer compressed?
Answers
Answer: simple harmonic motion
Simple harmonic motion. At the instant the spring is no longer compressed(equilibrium), all of our spring potential energy(kx^2/2) has been converted to kinetic energy(mv^2/2). All you have to do is find what your spring potential energy is when the spring is compressed using the spring constant(150N/m) and the distance it's compressed(30cm), use that as your kinetic energy, and solve for the velocity since you already know the mass.
What is the best description of the function of stars?
Stars are the recycling centers of the universe.
Stars are the light bulbs of the universe.
Stars are the batteries of the universe.
Stars are the motors of the universe.
Answers
Answer:
Stars are the light bulbs of the universe.
Answer:
Stars are the recycling centers of the universe
Explanation:
I just did on edge C
A student tries to unscrew a bolt with a spanner with 25 cm long. Thelargest force that he is able to apply is 100 N, while the minimummoment required to unscrew the nut is 30 N m. Can he unscrew the nut?
Answers
ANSWER
No, he cannot
EXPLANATION
First, let us draw a sketch of the problem:
For him to be able to unscrew the nut, the moment of the largest force he can apply must be greater than the minimum moment required to unscrew the nut.
To find the moment of the force, we apply the formula:
\(M=F\cdot d\)where F = force applied = 100 N;
d = distance between effort and bolt i.e. the length of the spanner = 25 cm = 0.25 m
Therefore, we have that the moment of the force is:
\(\begin{gathered} M=100\cdot0.25 \\ M=25Nm \end{gathered}\)Since the largest moment is 25Nm, and the minimum moment required to unscrew the nut is 30Nm, he cannot unscrew the nut.
The answer is NO.
-3 *10^-5 c Point charge placed at the origin of the coordinates Calculate the electric field at a point x=5m on the x axis
Answers
Answer: This is the best I could do.
Explanation:
Un camión viaja en una carretera recta con un valor de velocidad de 100km/h y frena con una aceleración constante de valor numérico de 2.27m/s². ¿Cuánto viajo antes de detenerse?
Answers
Con la cinematica podemos encontrar la distancia recorrida antes de detenerse es 169,96 m
Parámetros dados
velocidad del camión v= 100 km/h aceleración a= 2,27 m/s²Para encontrar
La distancia para detenerseEl sistema internacional de medidas (SI) es el sistema que deben tener todas las unidades del ejercicio, este es el primer paso si se tiene una mezcla de unidades
v= 100 km/h ( 1000 m/1 km) ( 1 h/3600 s) = 27,78 m/s
La cinemática estudia el movimiento de los cuerpos, estableciendo relaciones entre la posición, la velocidad y la aceleración
v² = v₀² – 2 a x
Donde v y v₀ son las velocidades inicial y final, a la aceleración del cuerpo y x la distancia recorrida
En este caso como el vehículo se detiene su velocidad final es cero
0= v₀² – 2 a x
x = \(\frac{v_o^2}{2a}\)
Calculemos
x = 27,78² / 2 2,27
x = 169,96 m
En conclusión usando la cinematica podemos encontrar la distancia recorrida antes de detenerse es
x= 169,96 m
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What is the magnitude of the resultant vector? Round your answer to the nearest tenth. m
Answers
The magnitude of the resultant vector to round the answer to the nearest tenth, we look at the digit in the hundredth's place. If this digit is 5 or greater, we round up. If it is less than 5, we round down.
In the study of physics, we use vectors to represent quantities that have both direction and magnitude. It is often the case that we want to add two or more vectors together to obtain a single vector that represents the net result of these additions. The process of adding two or more vectors together is known as vector addition.The magnitude of the resultant vector is the length of the line that represents it on a scale drawing.
When we add two or more vectors together, the resultant vector is the vector that represents the net result of these additions. To find the magnitude of the resultant vector, we use the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.
In the case of vector addition, the hypotenuse is the resultant vector, and the other two sides are the component vectors. If we have two vectors a and b, the magnitude of the resultant vector is given by the following equation:|R| = √(ax2 + bx2)where R is the resultant vector, a and b are the component vectors, and x is the angle between the vectors.
For example, if the answer is 12.345, we would round it to 12.3.
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Three children are riding on the edge of a merry-go-round that is a disk of mass 94 kg, radius 1.1 m, and is spinning at 25 rpm. The children have masses of 21.2 kg, 28 kg, and 34.2 kg. IF THE CHILD WHO HAS THE MASS OF 28 kg moves to the centre of merry go ground what is the new magnitude of angular velocity in rpm?
Answers
We are given the following information
Mass of merry-go-round = mm = 94 kg
Radius of merry-go-round = rm = 1.1 m
Angular velocity of merry-go-round = 25 rpm
Masses of three children = 21.2 kg, 28 kg, and 34.2 kg
If the child who has a mass of 28 kg moves to the center of merry go ground what is the new magnitude of angular velocity in rpm?
The initial momentum of the merry-go-round must be equal to the final momentum of the merry-go-round.
\(L_i=L_f\)The momentum is the product of the moment of inertia and the angular velocity, so the equation becomes
\(\begin{gathered} I_i\cdot\omega_i=I_f\cdot\omega_f \\ \omega_f=\frac{I_i\cdot\omega_i}{I_f} \end{gathered}\)Where the initial moment of inertia is the sum of the initial moment of inertia of merry-go-round and all the children.
\(\begin{gathered} I_i=I_m+I_1+I_2+I_3 \\ I_i=\frac{m_mr^2}{2}+m_1r^2+m_2r^2_{}+m_3r^2_{} \end{gathered}\)The final moment of inertia is the sum of the final moment of inertia of merry-go-round and all the children except the child with the mass of 28 kg .
\(\begin{gathered} I_f=I_m+I_1+I_3 \\ I_f=\frac{m_mr^2}{2}+m_1r^2+m_3r^2_{} \end{gathered}\)First, convert the angular velocity of merry-go-round from rpm to rad/s
\(25\; \frac{r}{\min}\times\frac{2\pi}{\sec}\times\frac{1\min}{60\sec}=\frac{5}{6}\pi\; \; \frac{rad}{s}\)So, the new magnitude of angular velocity is
\(\begin{gathered} \omega_f=\frac{I_i\cdot\omega_i}{I_f} \\ \omega_f=\frac{(\frac{m_mr^2}{2}+m_1r^2+m_2r^2_{}+m_3r^2_{})\cdot w_i}{\frac{m_mr^2}{2}+m_1r^2+m_3r^2_{}} \\ \omega_f=\frac{(\frac{94\cdot1.1^2}{2}+21.2\cdot1.1^2+28\cdot1.1^2+34.2\cdot1.1^2)\cdot\frac{5}{6}\pi}{\frac{94\cdot1.1^2}{2}+21.2\cdot1.1^2+34.2\cdot1.1^2} \\ w_f=3.33\; \; \frac{\text{rad}}{s} \end{gathered}\)Finally, convert the final angular velocity to rpm
\(3.33\; \frac{\text{rad}}{\text{sec}}\times\frac{1\; \text{rev}}{2\pi\; \text{rad}}\times\frac{60\sec}{1\min}=31.80\; \text{rpm}\)Therefore, the new magnitude of angular velocity is 31.80 rpm
How can scientist best confirm and validate the result of an experiment so they can publish their findings?
A. By removing the control from the experiment
B. By adding more variables to the experiment
C. By decreasing the number of trials
D. By creating a replicable experiment
Answers
\(Question\)
How can scientist best confirm and validate the result of an experiment so they can publish their findings?
Answer:
Hi, there!
D. By creating a replicable experiment Is The correct Answer!
Hope this Helps!!
-xXxAnimexXx- ♚♛♕♔ッ✨♚
Answer:. By creating a replicable experiment
Explanation:
the long-term average weather for a region is called
Answers
Answer:
Climate.
Explanation:
Weather can be defined as the atmospheric conditions of a particular area over a short period of time.
The elements of weather include precipitation, wind, temperature, atmospheric pressure, relative humidity, cloud, and wind speed.
The long-term average weather for a region is called climate.
This ultimately implies that, when the average atmospheric conditions prevailing in a specific region persists for a very long period of time, it is known as climate.
A ball is traveling at a constant speed of 4 m/s in a circle with a radius of 0.8 m. What is the centripetal acceleration of the ball?
Answers
According to the question the centripetal acceleration of the ball is 20 m/s².
What is centripetal acceleration?Centripetal acceleration is the acceleration that a body experiences when it is moving in a curved path. It is always directed towards the center of the curve, and its magnitude is equal to the square of the body's velocity divided by the radius of the curve. It is also known as the radial acceleration, since it is directed along the radius of the curve.
The centripetal acceleration of an object in a circular path is given by the equation:
\(a_c\) = v²/r
where a_c is the centripetal acceleration, v is the speed of the object, and r is the radius of the circular path.
In this case, the speed of the ball is 4 m/s, and the radius of the circular path is 0.8 m. Plugging these values into the equation, we get:
\(a_c\) = 4²/0.8 = 16/0.8 = 20 m/s²
Therefore, the centripetal acceleration of the ball is 20 m/s².
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When the south pole of a magnet approaches this iron nail from the left, what
happens to its domains?
A. All the arrows align to point down.
OB. All the arrows align to point right.
C. All the arrows align to point up.
OD. All the arrows align to point left.
t
Answers
In order to answer the question of where the arrows of the magnetic domains point, one must consider the properties of the magnets' poles: A) The arrows line up and point upward.
What comprises a magnet?Iron or cobalt is a common component of permanent magnets. An alloy composed of aluminum is called alnico. Involves a systematic process minerals are capable of making powerful electromagnets. They are widely utilized in industrial and consumer electronics applications.
Why was magnet named?The word "lithos magnes" from classical Greek is where the name first appeared. According to Pliny's explanation in his "Naturalis Historia," the term is derived from the myth of the Greek shepherd Magnes on Mount Ida, whose iron staff and the nails in his shoes were drawn to the magnetite stones.
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Answer:
POINTS LEFT
Explanation:
just did it
A body of mass 100g is fixed to a wheel and rotates in a circular path of 500mm diameter Calculate The moment of inertia nertial when the speed of the wheel from creases from 450rpm to 750rpm
Answers
The moment of inertia is 0.00625 kg·m^2 for both the initial and final speeds
To calculate the moment of inertia when the speed of the wheel increases, we need to know the distribution of mass in the body. Assuming the body is a thin ring with the mass concentrated on the rim, we can use the formula for the moment of inertia of a thin ring:
I = m * \(r^2\)
where I is the moment of inertia, m is the mass, and r is the radius.
Given:
Mass of the body (m) = 100 g = 0.1 kg
Diameter of the circular path = 500 mm
Radius (r) = Diameter / 2 = 500 mm / 2 = 250 mm = 0.25 m
Calculate the initial moment of inertia:
Initial speed = 450 rpm
Initial angular velocity (ω1) = 450 rpm * 2π / 60 = 47.12 rad/s
Using the formula for moment of inertia:
I1 = m * r^2
I1 = 0.1 kg * (0.25 m)^2
I1 = 0.00625 kg·m^2
Calculate the final moment of inertia:
Final speed = 750 rpm
Final angular velocity (ω2) = 750 rpm * 2π / 60 = 78.54 rad/s
Using the formula for moment of inertia:
I2 = \(m * r^2\)
I2 = \(0.1 kg * (0.25 m)^2\)
I2 = \(0.00625 kgm^2\)
Therefore, the moment of inertia remains the same as the mass and radius do not change. The moment of inertia is 0.00625 kg·m^2 for both the initial and final speeds.
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Please helpppppppppppppppppppp
Answers
Answer:
Explanation:
what do you really need help with
3
3
If a jogger runs a 10 kilometer race in 60 minutes, what is
her average speed?
A
10 km/hr
B
5 km/hr
С 6 km/hr
D
1.66 km/hr
Answers
If a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr. Details about average speed can be found below.
How to calculate average speed?Average speed can be calculated by dividing the distance moved by a body by the time taken. That is;
Average speed = Distance/Time
According to this question, a jogger runs a 10 kilometer race in 60 minutes. The average speed is calculated as follows:
Average speed = 10km/1hr
Average speed = 10km/hr.
Therefore, if a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr.
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Object A is negatively charged. Object A and Object B
attract. Object B and Object C repel. Object C and Object
D repel. What type of charge does Object B, Object C, and
Object D possess?
Answers
Answer:
Malrpr00qpq9owoowopwiaahaulaqkkkala9asoLHahababajjajalls
Explanation:
hhoootyiñlf7ogffyiklmhf
Two blocks (with masses of 2.0 kg and 4.0 kg) are on a bench tied together with string. They are being pulled to the right with a force of 30N. They each experience a friction force between the block and the bench.
(Refer to image)
The 2 kg block experiences a friction force with a friction coefficient of 0.30 and the 4 kg experiences a friction with a friction coefficient of 0.20.
Assume that g (the acceleration due to gravity) is 10.0 m/s/s.
Find the magnitude of the friction forces. Find the magnitude of the acceleration of the blocks. Use these answers to help you find the answer to the question:
What is tension in the string connecting the two blocks? (Submit just this answer in Newtons)
Answers
Answer:
T = 34/3 N
Explanation:
Magnitude of the friction force on 2kg block = 0.3x10x2 = 6N
Magnitude of the friction force on 4kg block = 0.2x10x4 = 8N
Magnitude of the acceleration of the blocks
F = ma
30 - 8 - 6 = (2+4)a
a = 8/3 m s^-2
Tension in the string connecting the two blocks
Consider the 2kg block,
T - f = ma
T - 6 = 2(8/3)
T = 34/3 N
Help me please
What is the shape of the velocity vs time graph for an object going at constant speed?
A- A curve that rises gradually to the right
B- A curve that decreases gradually to the right
C- A straight line that rises gradually to the right
D- A strait line that decreases gradually to the right
E- A straight line that is level and neither increases or decreases going to the right
Answers
Answer: a or b can u pls give me brainlest
Explanation:A straight line is a curve with constant slope. Since slope is acceleration on a velocity-time graph, each of the objects represented on this graph is moving with a constant acceleration.
Can anyone help with the attached question
Answers
The wave front, also known as the wavefront, is an imaginary surface that represents equivalent points of a wave that vibrate in sync. The correct option is D ; 2.0 .
What is a wavefront in physics?A wave front is described as a surface where the wave's phase remains constant. At any one time, all particles of the medium are moving in the same direction along a certain wave front. Two kinds of wave fronts are very essential. They are plane and spherical wave fronts, respectively. A wavefront is a surface on which the wave disturbance is in the same phase at all places. For example, the ripples of water created when a stone is tossed into a pond.
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A car initially traveling at a speed of 15.0 m/s accelerates uniformly to a speed of 20.0 m/s over a distance of 40.0 meters. What is the magnitude of the car's acceleration?
Answers:
1.1 m/s^2
2.0 m/s^2
2.2 m/s^2
9 m/s^2
Answers
Answer:
\(\boxed {\boxed {\sf 2.2 \ m/s^2}}\)
Explanation:
We are asked to solve for the magnitude of the car's acceleration.
We are given the initial speed, final speed, and distance, so we will use the following kinematic equation.
\({v_f}^2={v_i}^2+2ad\)
The car is initially traveling at 15.0 meters per second and accelerates to 20.0 meters per second over a distance of 40.0 meters. Therefore,
\(v_f\)= 20.0 m/s\(v_i\)= 15.0 m/s d= 40.0 mSubstitute the values into the formula.
\((20.0 \ m/s)^2= (15.0 \ m/s)^2 + 2 a (40.0 \ m)\)
Solve the exponents.
(20.0 m/s)² = 20.0 m/s * 20.0 m/s = 400.0 m²/s² (15.0 m/s)² = 15.0 m/s * 15.0 m/s = 225.0 m²/s²\(400.0 \ m^2/s^2 = 225.0 \ m^2/s^2 + 2 a(40.0 \ m)\)
Subtract 225.0 m²/s² from both sides of the equation.
\(400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 225.0 \ m^2/s^2 -225 \ m^2/s^2 +2a(40.0 \ m)\)
\(400.0 \ m^2/s^2 - 225.0 m^2/s^2 = 2a(40.0 \ m)\)
\(175 \ m^2/s^2 = 2a(40.0 \ m)\)
Multiply on the right side of the equation.
\(175 \ m^2/s^2 =80.0 \ m *a\)
Divide both sides by 80.0 meters to isolate the variable a.
\(\frac {175 \ m^2/s^2}{80.0 \ m}= \frac{80.0 \ m *a}{80.0 \ m}\)
\(\frac {175 \ m^2/s^2}{80.0 \ m}=a\)
\(2.1875 \ m/s^2 =a\)
Round to the tenths place. The 8 in the hundredth place tells us to round the 1 up to a 2.
\(2.2 \ m/s^2=a\)
The magnitude of the car's acceleration is 2.2 meters per second squared.
Consider an atom. Which contributes most to the size of the atom?
Answers
Answer:
Protons contribute the most to the size of the atom
Answer:
Protons
Explanation:
Protons contribute more to both the mass and size of an atom. Protons contribute more to an atom's mass while electrons contribute more to its size.
Is the acceleration change or constnt?
Answers
Acceleration means going faster
How much power is used if a force of 35n is used to push a box a distance of 10 meters in 5 seconds ?
Answers
Answer:
pratyush the first boy984
Which correctly describes latent heat?
A. The heat of molecules that are under pressure
B. The heat held inside of ice crystals colder than -2°C
C. The heat absorbed or lost by a substance while it's changing state
D. The heat used to change the temperature of a liquid
Answers
Option C. The heat absorbed or lost by a substance while it's changing state correctly describes latent heat
Latent heat is the heat absorbed or lost by a substance while it is changing state.
The latent heat is a type of heat that is transferred during phase change, i.e., while a substance undergoes a change of state.
For example, when ice melts into liquid water, or when liquid water evaporates into water vapor, heat is absorbed from the surroundings.
Latent heat is not associated with a temperature change; rather, it's associated with a change of state.
For instance, the temperature of water remains at 100°C while boiling.
When water is boiling, the latent heat of vaporization is absorbed and utilized to break the hydrogen bonds holding water molecules together to change water from the liquid phase to the gaseous phase.
When the water is boiling, adding more heat won't increase the water's temperature, instead, the extra heat will be absorbed to change the phase of water molecules.
Therefore, the correct answer to the given question is option C: The heat absorbed or lost by a substance while it is changing state.
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6. If an object accelerates at 3m/s/s, how long does it take for the object to travel at a speed of 12 m/s.
Answers
Answer:
4 seconds
Explanation:
Assuming that the object started from rest,
v = at
--> t = v/a = (12 m/s) / (3 m/s^2)
= 4 seconds
A cannonball is fired horizontally from a 10 m high cliff at 20 m/s. How long will the cannonball be in the air? How far away will the cannonball strike the ground?
Answers
Answe ¡Buenos días! –
#3 ¡Buenas tardes! –
#4 ¡Bienvenid
A carousel is (more or less) a disk of mass, 15,000 kg, with a radius of 6.14. What torque must be applied to create an angular acceleration of 0.0500 rad/s^2?round to 3 significant figures
(Plssss help me im suffering from severe brainrot)
Answers
To calculate the torque required to create an angular acceleration, we can use the formula:
Torque = Moment of Inertia × Angular Acceleration
The moment of inertia of a disk can be calculated using the formula:
Moment of Inertia = (1/2) × Mass × Radius^2
Given:
Mass = 15,000 kg
Radius = 6.14 m
Angular Acceleration = 0.0500 rad/s^2
First, calculate the moment of inertia:
Moment of Inertia = (1/2) × Mass × Radius^2
Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2
Next, calculate the torque:
Torque = Moment of Inertia × Angular Acceleration
Torque = Moment of Inertia × 0.0500 rad/s^2
Now, let's plug in the values and calculate:
Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2
Moment of Inertia ≈ 283,594.13 kg·m^2
Torque = 283,594.13 kg·m^2 × 0.0500 rad/s^2
Torque ≈ 14,179.71 N·m
Rounding to three significant figures, the torque required to create an angular acceleration of 0.0500 rad/s^2 is approximately 14,180 N·m.
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The objective lens and the eyepiece of a microscope are 16.5 cm apart. The objective lens has a magnification of 64× and the eyepiece has a magnification of 10 ×. Assume that the image of the objective lies very close to the focal point of the eyepiece.
Part A) Calculate the overall magnification of the microscope.
Part B) Calculate the focal length of the eyepiece.
Part C) Calculate the focal length of the objective.
Answers
M = M_objective x M_eyepiece
M = 64 x 10
M = 640
Therefore, the overall magnification of the microscope is 640×.
Part B) The magnification of the eyepiece is given by:
M_eyepiece = f_objective / (f_objective + d)
where f_objective is the focal length of the objective lens, and d is the distance between the objective lens and the eyepiece. Since the image of the objective lies very close to the focal point of the eyepiece, we can assume that d is equal to the focal length of the eyepiece, which we will denote as f_eyepiece. Substituting these values into the equation above, we get:
10 = f_objective / (f_objective + f_eyepiece)
We can rearrange this equation to solve for f_eyepiece:
f_eyepiece = f_objective / 10 - 1
The magnification of the objective lens is given by:
M_objective = f_eyepiece / (f_eyepiece - d)
Substituting the value of d, we get:
M_objective = f_eyepiece / (f_eyepiece - f_eyepiece)
M_objective = 1
Therefore, the magnification of the objective lens is 1×. We can use this value to solve for f_objective:
M = M_objective x M_eyepiece
640 = 1 x 10 x (f_objective / (f_objective + f_eyepiece))
Substituting the expression for f_eyepiece that we derived earlier, we get:
640 = 10 x f_objective / (f_objective / 9)
640 = 90
Solving for f_objective, we get:
f_objective = 14.06 cm
Therefore, the focal length of the objective lens is 14.06 cm.