Explanation:
Resistors in circuits can either be connected in series or in parallel.
For a series connected resistors, the same current flows through the resistors while different voltages is passed across them. The formula for calculating effective resistances in a circuit is expressed as
[tex]R_T = R_1+R_2+R_3...+R_n[/tex] where R1, R2, R3 etc. are the resistances connected in series.
The effective resistances for series connected resistors is the sum of the individual resistors
For a parallel connected resistors, different current flows through the resistors while same voltage is passed across them. The formula for calculating effective resistances in a circuit is expressed as:
[tex]\frac{1}{R_T} = \frac{1}{R_1} +\frac{1}{R_2} +\frac{1}{R_3} +...\frac{1}{R_n}[/tex]
The effective resistances for parallel connected resistors is the sum of the reciprocal of the individual resistors
my heart strike him to dead.what figure of speech is that?
Answer:
Hyperbole
Explanation:
this is an extreme exaggeration or overstatement/ magnification
If I am driving down the highway going north at 50 miles per hour, and another car is driving south at 75 miles per hour. How fast is the car coming toward me?
Its an exam >.
I WILL GIVE BRAINLIEST
What happens to the energy in a closed system?
Answer:
can exchange energy with its surroundings through heat and work transfer. In other words, work and heat are the forms that energy can be transferred across the system boundary.
Find the distance along an arc on the surface of the earth that subtends a central angle of 1 minutes (1 minute = 1/60 degree). The radius of the earth is 3960 miles. Round to the thousandths. (3 decimal places)
Answer:
1.152 miles
Explanation:
Given: central angle = 1 minute = [tex](\frac{1}{60}) ^{o}[/tex]
radius of the earth = 3960 miles
The length of an arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r
where: [tex]\alpha[/tex] is the central angle, and r is the radius.
Thus,
Distance along the arc = [tex]\frac{\alpha }{360^{o} }[/tex] 2[tex]\pi[/tex]r
Distance along the arc = [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 2 x [tex]\frac{22}{7}[/tex] x 3960
= [tex]\frac{(\frac{1}{60}) ^{o} }{360^{o} }[/tex] x 24891.4286
= 1.1524
The required distance along an arc is 1.152 miles.
Plz help me fast WITH EXTRA POINTS AFTER SUBMITTING
Answer:
4 bobux
Explanation:
one bobux
two bobux
three bobux
four bobux
A 50 kg bicyclist starts his ride down the road with an acceleration of 1m/s2 in air with a density of 1.2 kg/m3. If his velocity at a given moment is 2m/s, how much force is he exerting? Assume the area of his body is 0.5m^2.
a. The bicyclist is exerting 1.1 N of force.
b. The bicyclist is exerting 49 N of force.
c. The bicyclist is exerting 50 N of force.
d. The bicyclist is exerting 51 N of force.
Answer:
b. The bicyclist is exerting 49 N of force
Explanation:
Given;
mass of bicyclist start, m = 50 kg
acceleration, a = 1 m/s²
density of air, ρ = 1.2 kg/m³
velocity, v = 2 m/s
Area of the bicyclist body, A = 0.5 m²
The drag force on the bicyclist is given by ;
Fd = 0.5CρAv²
where;
C is drag coefficient = 0.9 for bicycle
Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²
Fd = 1.1 N
The force of the bicyclist is given by;
F = ma
F = 50 x 1
F = 50 N
The effective force exerted by the bicyclist is given by;
Fe = F - Fd = 50 N - 1.1 N
Fe = 49 N
Therefore, the force exerted by the bicyclist is 49 N
How do compounds differ from mixtures such as lemonade
Answer:
A mixture is a combination of two or more substances in any proportion. This is different from a compound, which consists of substances in fixed proportions. ... The lemonade pictured above is a mixture because it doesn't have fixed proportions of ingredients.
Explanation:
Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at
Complete question is;
Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at
(x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)
Compute the length of the bungee cord at t = 7 seconds. (Round your answer to three decimal places.)
Answer:
Length of bungee cord = 85.734 ft
Explanation:
We are told that Adrian moves 9ft/sec. Thus, at 5.5 seconds, distance he moved is; 9 ft/sec × 5.5sec = 49.5 ft in the negative x (-x) direction. Therefore, the coordinate is (-49.5, 0).
Now, Allyson has moved 10ft/sec. Thus, at 7 seconds, distance he moved would be; 10 ft/sec x 7sec = 70 feet in the positive (+y) direction. Therefore, the coordinate is (0, 70).
Now, since they started from the origin, it means (0, 0) is a coordinate. Thus, we now have 3 coordinates which are; (0, 0), (0, 70) & (-49.5,0). These 3 coordinates would therefore combine to form a right triangle.
The hypotenuse is the distance between Allyson and Adrian.
Thus, from pythagoras theorem, we can find the distance between them which is same as the length of the cord.
Thus;
(-49.5)² + 70² = D².
D² = 2450.25 + 4900
D = √7350.25
D = 85.734 ft
A 126 N force is applied at an angle of 25.00 to a 8.50 kg block pressed against a rough vertical wall and the block slides down the wall at constant velocity. Calculate the coefficient of kinetic friction between the block and the wall.
Answer:
μk = 0.58
Explanation:
If the block is sliding down at constant speed, this means that no net force is acting upon it in the vertical direction.As the block is pressed on the wall, this means that it doesnt accelerate in the horizontal direction either, so no net force acts upon it in this direction also.In this direction, we have only two forces acting, equal and opposite each other, one is the normal force (exerted by the wall) and the other is the horizontal component of the applied force.If the applied force forms an angle of 25º with the wall (which is vertical), this means that we can get its projection along the horizontal direction, using simple trigonometry , as follows:[tex]F_{apph} = F_{app} * sin\theta = 126 N * sin 25 = 53.3 N[/tex]
⇒ [tex]F_{n} = - F_{apph} = -53.3 N[/tex]
In the vertical direction, we have three forces acting on the block: the weight pointing downward, the kinetic friction force (as we know that the block is sliding), and the vertical component of the applied force, in the same direction as the friction one.As we have already said, the sum of these forces must be 0.[tex]F_{g} + F_{appv} + F_{ff} = 0 (1)[/tex] where Fg is the weight of the block, Fappv is the vertical component of the applied force, and Fff is the kinetic friction force.Replacing these forces by their mathematical expressions, we have:[tex]F_{g} = m_{b} * g = 8.5 Kg * (-9.8 m/s2) = -83.3 N[/tex]
[tex]F_{appv} = F_{app}* cos\theta = 126 N * cos 25 = 114.2 N[/tex]
[tex]F_{ff} = \mu_{k}* F_{n} =\mu_{k} * (-53.3 N)[/tex]
Replacing in (1), and solving for μk, we finally get:μk = 0.58
Let k be the Boltzmann constant. If the thermodynamic state of gas at temperature T changes isothermally and reversibly to a state with three times the number of microstates as initially, the energy input to gas as heat is:______
a. Q = 0
b. Q = 3kT
c. Q = −3kT
d. kT ln 3
e. −kT ln 3
Answer:
d. kT ln 3
Explanation:
Given;
k as Boltzmann constant
Let initial initial microstate, = Фi
Let the final microstate, = Фf = 3Фi
at constant temperature, T
The energy input to gas as heat is given by;
Q = TΔS = Tk(lnФf - lnФi)
[tex]Q= kT*\frac{ln \phi _f}{ln \phi_i} \\\\Q= kT*\frac{3ln \phi _i}{ln \phi_i}\\\\Q= kT ln3[/tex]
Therefore, the energy input to gas as heat is kT ln 3
The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?
Answer:
a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
Explanation:
a) From Second Newton's Law, we form this equation of equilibrium:
[tex]\Sigma F = F_{E}-W = 0[/tex] (Eq. 1)
Where:
[tex]F_{E}[/tex] - Electrostatic force exerted on human, measured in Newton.
[tex]W[/tex] - Weight of the human, measured in Newton.
If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:
[tex]q\cdot E-m\cdot g = 0[/tex]
[tex]q = \frac{m\cdot g}{E}[/tex] (Eq. 2)
[tex]E[/tex] - Electric field, measured in Newtons per Coloumb.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravity acceleration, measured in meters per square second.
[tex]q[/tex] - Electric charge, measured in Coulomb.
As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that [tex]m = 60\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]E = -150\,\frac{N}{C}[/tex], the charge that a 60-kg human must have to overcome weight is:
[tex]q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }[/tex]
[tex]q = -3.923\,C[/tex]
The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:
[tex]F = \kappa \cdot \frac{q^{2}}{r^{2}}[/tex] (Eq. 3)
Where:
[tex]\kappa[/tex] - Electrostatic constant, measured in Newton-square meter per square Coulomb.
[tex]q[/tex] - Electric charge, measured in Coulomb.
[tex]r[/tex] - Distance between two people, measured in meters.
If we know that [tex]\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]q = -3.923\,C[/tex] and [tex]r = 100\,m[/tex], then the force of repulsion between two people is:
[tex]F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right][/tex]
[tex]F = 13.851\times 10^{6}\,N[/tex]
The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
Which is a belief held by sociologists who work from a social-conflict
perspective?
O A. The best approach for a study is from a micro-level orientation.
O B. Personal background has little impact on how individuals react
with one another.
C. Some social patterns are helpful, while others are harmful.
D. Data are irrelevant to the study of sociology.
SUBMIT
Answer:
C. Some social patterns are helpful, while others are harmful.
Explanation:
Hope this was helpful, Have an amazing,spooky Halloween!!
What is the car’s average velocity (in m/s) in interval between t=1.0s to t=1.5s?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula for velocity is;
Velocity (in m/s) = distance/time
The distance the car covered in the completed question is divided by the difference in the time interval
The difference in the time interval will be = 1.5s - 1.0s = 0.5s
NOTE: the distance must be in meters or be converted to meters
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 7.9 times the magnitude of the tangential acceleration. What is the angle
Answer:
The angle is 3.95 rad.
Explanation:
The angle can be calculated as follows:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]
Where:
[tex]\omega_{f}[/tex]: is the final angular speed
ω₀: is the initial angular speed = 0 (it starts from rest)
α: is the angular acceleration
θ: is the angle=?
The centripetal acceleration is:
[tex]a_{c} = \omega_{f}^{2}*r[/tex]
And the tangential acceleration is:
[tex] a_{T} = \alpha*r [/tex]
Since the magnitude of the centripetal acceleration is 7.9 times the magnitude of the tangential acceleration:
[tex]a_{c} = 7.9a_{T}[/tex]
[tex]\omega_{f}^{2}*r = 7.9*\alpha*r \rightarrow \alpha = \frac{\omega_{f}^{2}}{7.9}[/tex]
Now, the angle is:
[tex]\omega_{f}^{2} = 2(\frac{\omega_{f}^{2}}{7.9})\theta[/tex]
[tex] \theta = \frac{7.9}{2} = 3.95 rad [/tex]
Therefore, the angle is 3.95 rad.
I hope it helps you!
The angular distance traveled by the electric drill is 3.95 radians.
The given parameters;
initial angular speed, [tex]\omega_i[/tex] = 0centripetal acceleration, [tex]a_c[/tex] = 7.9aThe angular distance traveled by the electric drill is calculated as follows;
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]
The relationship between centripetal acceleration, tangential acceleration and angular speed is given as;
[tex]a_c = \omega ^2 r\\\\a = \alpha r\\\\a_c = 7.9a= 7.9\alpha r\\\\7.9\alpha r = \omega^2 r\\\\\alpha = \frac{\omega ^2}{7.9}[/tex]
Substitute the value of angular acceleration into the first equation;
[tex]\omega _f^2 = 0 + 2(\a (\frac{\omega _f^2}{7.9})\theta\\\\2\theta \omega_f^2 = 7.9\omega_f ^2\\\\\theta = \frac{7.9}{2} \\\\\theta = 3.95 \ rad[/tex]
Thus, the angular distance traveled by the electric drill is 3.95 radians.
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A small child weighs 60 N. If mommy left him sitting on top of the stairs, which are 12 m high, how much energy does the child have!
Please help ASAP
Answer:
6000 joules
Explanation:
I jus learned dis
Answer:6000j
Explanation:
Hope that helps
See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 86.4 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 1.33 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.
Answer:
(a) 24.56 N
(b) 142.28 N
Explanation:
(a)
The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.
The centripetal function is generated from either scenario by Equation:
⇒ [tex]Fc = \frac{mv^2}{r}[/tex]
On putting the values, we get
⇒ [tex]=\frac{12\times 1.33^2}{0.864}[/tex]
⇒ [tex]=24.56 \ N[/tex]
(b)
Use T to denote whatever arm stress we can get at the bottom including its circle:
⇒ [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]
⇒ [tex]T = mg + Fc[/tex]
⇒ [tex]=12\times 9.81+24.56[/tex]
⇒ [tex]=142.28 \ N[/tex]
Metals that have shine and luster?
Answer:
luster
Explanation:
1 (4 points) A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the magnitude of the net force on the ball
Answer:
100 N
Explanation:
The following data were obtained from the question:
Mass (m) = 2 Kg
Velocity (v) = 5 m/s
Radius (r) = 50 cm
Force (F) =.?
Next, we shall convert 50 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
50 cm = 50 cm × 1 m / 100 cm
50 cm = 0.5 m
Therefore, 50 cm is equivalent to 0.5 m.
Finally, we shall determine the magnitude of the net force on the ball by using the following formula:
F = mv²/r
Mass (m) = 2 Kg
Velocity (v) = 5 m/s
Radius (r) = 0.5 m
Force (F) =.?
F = mv²/r
F = 2 × 5²/ 0.5
F = 2 × 25/ 0.5
F = 50 / 0.5
F = 100 N
Therefore, the magnitude of the net force on the ball is 100 N.
The magnitude of the net force on the ball will be "100 N".
Force and speedAccording to the question,
Mass, m = 2 kg
Velocity, v = 5 m/s
Radius, r = 50 cm or,
= 50 × [tex]\frac{1}{100}[/tex]
= 0.5 m
We know the relation,
Force, F = [tex]\frac{mv^2}{r}[/tex]
By substituting the values, we get
= [tex]\frac{2\times (25)^2}{0.5}[/tex]
= [tex]\frac{2\times 25}{0.5}[/tex]
= [tex]\frac{50}{0.5}[/tex]
= 100 N
Thus the above response is appropriate.
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A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory with time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?
Answer:
Answer and steps in the pic
Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?
Answer:
12.5 m
Explanation:
The first thing we would do is to calculate the wavelength. To do this, we use the formula
v = fλ, where
v = wave speed
f = frequency
λ = wavelength
If we make wavelength the formula, we have
wavelength = speed / frequency
Now, we substitute the values we had been given and we have
wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m
half of this said wavelength will be
= 3.50 / 2
= 1.75 m
As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,
(10 / 1.75) = 5.7
So the separation will have to be 7 half wavelengths
= (7 * 1.75) = 12.5 m
Hollywood and video games often depict the bad guys being "blown away" when they’re shot by a bullet (i.e. once hit, their feet leave the ground and they fly backwards). Assuming that even if a handgun cartridge did generate enough momentum for the bullet to do this, why is it still nonsense on-screen?
Answer:
Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".
Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.
What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.
The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.
It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this, it is still nonsense on screen in Hollywood and video.
Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:
τs = -κ θ.
Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.
Required:
a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.
b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device?
Complete Question
Consider a 50-turn circular loop with a radius of 1.55 cm in a 0.35-T magnetic field. This coil is going to be used in a galvanometer that reads 45 μA for a full-scale deflection. Such devices use spiral springs which obey an angular form of Hooke's law, where the restoring torque is:
τs = -κ θ.
Here κ is the torque constant and θ is the angular displacement, in radians, of the spiral spring from equilibrium, where the magnetic field and the normal to the loop are parallel.
Required:
a. Calculate the maximum torque, in newton meters, on the loop when the full-scale current flows in it.
b. What is the torque constant of the spring, in newton meters per radian, that must be used in this device? Assume the full scale deflection is 60° from the spring's equilibrium position
Answer:
a
[tex]\tau_{m} = 5.95 *10^{-7} \ N \cdot m[/tex]
b
[tex]\beta = 2.83 *10^{-7} \ N \cdot m / rad [/tex]
Explanation:
From the question we are told that
The number of turns is N = 50
The radius is r = 1.55 cm = 0.0155 m
The magnetic field is B = 0.35 T
The induced current is [tex]I = 45 \mu A = 45 *10^{-6} \ A[/tex]
Generally the area of loop is mathematically represented as
[tex]A = \pi r^2[/tex]
=> [tex]A =3.142 * 0.0155^2[/tex]
=> [tex]A =0.000755\ m^2[/tex]
Generally the maximum torque is mathematically represented as
[tex]\tau_{m} = N * B * I * A[/tex]
=> [tex]\tau_{m} = 50 * 0.35 * 45 *10^{-6} * 0.000755[/tex]
=> [tex]\tau_{m} = 5.95 *10^{-7} \ N \cdot m[/tex]
Generally the torque 60° from the spring's equilibrium position is mathematically represented as
[tex]\tau = N * B * I * A * sin (60)[/tex]
=> [tex]\tau = 50 * 0.35 * 45 *10^{-6} * 0.000755 * sin (60)[/tex]
=> [tex]\tau = 2.973 *10^{-7} \ N \cdot m [/tex]
Generally the toque constant of the spring is mathematically represented as
[tex]\beta = \frac{\tau}{60}[/tex]
=> [tex]\beta = \frac{\tau}{\frac{\pi}{3}}[/tex]
=> [tex]\beta = \frac{2.973 *10^{-7}}{\frac{\pi}{3}}[/tex]
=> [tex]\beta = 2.83 *10^{-7} \ N \cdot m / rad [/tex]
A cart with an unknown mass is at rest on one side of a track. A student must find the mass of the cart by using Newton’s second law. The student attaches a force probe to the cart and pulls it while keeping the force constant. A motion detector rests on the opposite end of the track to record the acceleration of the cart as it is pulled. The student uses the measured force and acceleration values and determines that the cart’s mass is 0.4kg . When placed on a balance, the cart’s mass is found to be 0.5kg . Which of the following could explain the difference in mass?
Answer choices:
A) The track was not level and was tilted slightly downward.
B) The student did not pull the cart with a force parallel to the track.
C) The wheels contain bearings that were rough and caused a significant amount of friction.
D) The motion sensor setting was incorrect. The student set it up so that motion away from the sensor would be the negative direction.
Answer: The correct answer is A) The track was not level and was tilted slightly downward.
Explanation: This is because of the two values: 0.4 kg and 0.5 kg. I won't go into much detail but due to this difference of mass, we know that the track was not level.
"The track was not level and was tilted slightly downward" could explain the difference in the mass.
Mostly because the university student or learners calculates a mass of just over the spring quantity, the vehicle speed seems to have been higher than there would have had to be.Option B, as well as Option C, are wrong because the acceleration would've been smaller in each of these 2 circumstances, so that computed mass would've been larger.
Thus Option A is appropriate.
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a. In one short sentence, explain why we call the force of gravity an attractive force.
b. Does a force of gravity exist between any two objects
Answer:
Explanation:
(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.
(b) Yes, it does and the formula for force of gravity between any two object is
F = G[tex]\frac{m1m2}{r}[/tex]
where m1 and m2 are masses of the first and second object respectively
r is the distance between the center of the two masses
G is the gravitational constant
Compare the amount of thermal energy required to MELT a solid with the amount of thermal energy released when the same liquid becomes a solid.
Two bodies, A and B, have equal kinetic energies. The mass of A is nine times that of B. The ratio of the momentum of A to that of B is:_______
a. 1:9
b. 1:3
c. 1:1
d. 3:1
e. 9:1
A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?
Answer:
The displacement of the canoe is 1.46 m
Explanation:
Given that,
Mass of canoe = 61 kg
Mass of man = 86.5 kg
Length = 4 m
Let the the displacement of the canoe is x'
We need to calculate the displacement of the man
Using formula of displacement
[tex]x=x_{2}-x_{1}[/tex]
Put the value into the formula
[tex]x=4-(0.75+0.75)[/tex]
[tex]x=2.5\ m[/tex]
We need to calculate the displacement of the canoe
Using conservation of momentum
[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]
[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]
[tex]86.5\times2.5=(61+86.5)\times x'[/tex]
[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]
[tex]x'=1.46\ m[/tex]
Hence, The displacement of the canoe is 1.46 m
A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational field strength of Planet X?
A. The gravitational field strength of Planet X is mg.
B. The gravitational field strength of Planet X is Wx/m.
C. The gravitational field strength of Planet X is 9.8 N/kg.
D. The gravitational field strength of Planet X is mWx.
Answer: B. The gravitational field strength of Planet X is Wx/m.
Explanation:
Weight is a force, and as we know by the second Newton's law:
F = m*a
Force equals mass times acceleration.
Then if the weight is:
Wx, and the mass is m, we have the equation:
Wx = m*a
Where in this case, a is the gravitational field strength.
Then, isolating a in that equation we get:
Wx/m = a
Then the correct option is:
B. The gravitational field strength of Planet X is Wx/m.
for an emitted wavelength of 500 nanometers and a redshift of 0.4 what will be the observed wavelength g
Answer:
The observed wavelength is [tex] \lambda = 700nm[/tex] (color - Red)
Explanation:
From the question we are told that
The wavelength of the emitter is [tex]\lambda_ e = 500 nm = 500 *10^{-9} \ m[/tex]
The redshift is R = 0.4
Generally red shift is mathematically represented as
[tex]R = \frac{ \lambda - \lambda_e }{\lambda_e}[/tex]
=> [tex]0.4 = \frac{ \lambda - 500 *10^{-9} }{500 *10^{-9} }[/tex]
=> [tex] \lambda - 500*10^{-9} = 200*10^{-9} [/tex]
=> [tex] \lambda = 700 *10^{-9}[/tex]
=> [tex] \lambda = 700nm[/tex]
use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is
Complete question:
use the hubble's law to determine the distance to a quasar receding at 75% the speed of light. The speed of light is 300,000 km/sec. assume Hubble's constant is 2.2 x 10⁻⁵ km/s/Lyr
Answer:
The distance to the quasar is 1.02 x 10¹⁰ Lyr
Explanation:
Given;
speed of light, v = 300, 000 km/sec
Hubble's constant, H₀ = 2.2 x 10⁻⁵ km/s/Lyr
percentage of the quasar recession = 75% of speed of light
Hubble's Law is given by;
[tex]v =H_od\\\\d = \frac{v}{H_o}\\\\d= \frac{(0.75*300,000)}{2.2*10^{-5}}.Lyr\\\\d = 1.02*10^{10} \ Lyr[/tex]
Therefore, the distance to the quasar is 1.02 x 10¹⁰ Lyr