Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
A football quarterback throws a football for a long pass. While in the motion of throwing, the quarterback moves the ball , starting from rest, and completes the motion in . Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
stopping potential becomes more and more negative why
Answer:
stopping potential is the negative potential applied to the circuit to stop the moving electrons so as to stop the flow of current
for high current high negative potential is applied
Two planes of charge with no thickness, A and B, are parallel and vertical. The electric field in region I to the left of plane A has magnitude 3σ/2????0 and points to the left. The electric field in the region to the right of B has magnitude 3σ/2????0 and points to the right. The electric field in the region between the two planes has magnitude σ/2????0 and points to the right. The surface charge density on planes A and B respectively is
Answer:
The surface charge density on planes A and B respectively is
[tex]\sigma__{A}} } = 2\sigma[/tex]
and
[tex]\sigma__{B}} = \sigma[/tex]
Explanation:
From the question we are told that
The electric field in region to the left of A is [tex]E_i = \frac{3 \sigma}{2 \epsilon_o}[/tex]
The direction of the electric field is left
The electric field in the region to the right of B is [tex]E_f = \frac{3 \sigma}{2 \epsilon_o}[/tex]
The direction of the electric field is right
The electric field in the region between the two planes is [tex]E_m = \frac{\sigma }{2 \epsilon_o }[/tex]
The direction of the electric field is right
Let the surface charge density on planes A and B be represented as [tex]\sigma__{A}} \ \ and \ \ \sigma__{B}} \ \ \ respectively[/tex]
From the question we see that
[tex]E_i = E_f[/tex]
Generally the electric to the right and to the left is due to the combined electric field generated by plane A and B so
[tex]E_i = E_f = \frac{3\sigma }{2\epsilon} = \frac{\sigma_A }{ 2 \epsilon_o } + \frac{\sigma_B }{ 2 \epsilon_o }[/tex]
=> [tex]\sigma__{A}} + \sigma__{B}} = 3 \sigma -- -(1)[/tex]
Generally the electric field at the middle of the plane A and B is due to the diffencence in electric field generated by plane A and B
i.e
[tex]\frac{\sigma }{2 \epsilon_o } = \frac{\sigma_A }{ 2 \epsilon_o } - \frac{\sigma_B }{ 2 \epsilon_o }[/tex]
=> [tex]\sigma__{A}} - \sigma__{B}} = \sigma[/tex]
=> [tex]\sigma__{A}} } = \sigma + \sigma__{B}[/tex]
From equation 1
[tex]\sigma + \sigma__{B}}+ \sigma__{B}} = 3 \sigma[/tex]
=> [tex]\sigma__{B}} = \sigma[/tex]
So
[tex]\sigma__{A}} } = \sigma + \sigma[/tex]
=> [tex]\sigma__{A}} } = 2\sigma[/tex]
which statement about force is incorrect
Answer:
What are the options?
Explanation:
______ is the total distance traveled divided by the total time of travel.
(5 Points)
speed
total speed
average speed
displacement
When an object falls, it is reacting to the force of gravity. true or false
Answer:
The answer is true, as gravity affects everything.
You want to move a spacecraft that is in 300 km circular orbit around Venus into another circular orbit around Venus of 600 km. Explain how you would accomplish this. Solve for the velocities that need to be applied.
Answer:
The final velocity will be half of the initial velocity of the spacecraft.
Explanation:
Angular momentum is conserved for the circular force motion and central force motion.
Considering
L = MVR = Constant
Where
M = Mass of the object
V = Velocity of the object
r = radius of circle
We know that
V = [tex]\frac{1}{R}[/tex]
So,
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{R_{1} }{R_{2} }[/tex]
As per the given data
[tex]R_{1}[/tex] = Initial Radius = 300 km
[tex]R_{2}[/tex] = Final Radius = 600 km
[tex]V_{1}[/tex] = Initial Velocity =
[tex]V_{2}[/tex] = Final Velocity =
Placing values in the formula
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{300 km}{600 km }[/tex]
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{1}{2}[/tex]
[tex]{V_{2}[/tex] = [tex]\frac{1}{2} V_{1}[/tex]
Hence, The final velocity will half of the initial velocity of the spacecraft.
Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?
1.When they are halfway to the bottom
2.When they reach the bottom of the fall
Answer:
2.When they reach the bottom of the fall
Explanation:
The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).
Thus, the correct option is "2" When they reach the bottom of the fall
A ball is thrown horizontally at 30 m/s from a height of 45m. How long is it in the air?
(i) How fast is it moving horizontally when it hits the ground?
Answer:
1.25s
1) 90.9m
Explanation:
From the question, we are given the following;
Speed v = 30m/s
Maximum height H = 45m
Required;
Time
Using the equation of motion S = ut + 1/2gt²
Substitute;
45= 30t + 1/2(9.8)t²
45 = 30t + 4.9t²
4.9t² + 30t - 45 = 0
Factorize;
t = -30 ±√30²-4(-45)(4.9)/2(4.9)
t = -30 ±√900+882/9.8
t = -30 ±42.21/9.8
t = -30 + 42.21/9.8
t = 12.21/9.8
t = 1.25secs
Hence the ball spent 1.25secs in air.
i) To get the horizontal distance, we will use the formula;
R = U√2H/g
R = 30√2(45)/9.8
R = 30√90/9.8
R = 30√9.18
R = 30(3.03)
R = 90.9m
Hence the horizontal distance is 90.9m
when is a bandage best suited to be used?
1. What is the total distance traveled?
A 3.0m
B 4.0m
C 5.0m
D 6.0m
Answer:
c
Explanation:
ASTRONOMY
Match the following phases of stellar evolution to its characteristics.
*please help me thx*
Answer:
1. Brown dwarf.
2. Red giant.
3. Black dwarf.
4. White dwarf.
5. Black hole.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
Basically, stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
Some of the examples of stars are; Canopus, Sun (closest to the Earth), Betelgeus, Antares, Vega etc.
1. Brown dwarf: does not undergo hydrogen fusion.
2. Red giant: has an outer shell of burning hydrogen and a core of helium.
3. Black dwarf: a white dwarf that has lost all its energy.
4. White dwarf: has a core of carbon that is not massive enough to fuse. A white dwarf also known as degenerate dwarf is a star or an electron degenerate matter and is as massive as the Sun but only about as large in size as planet Earth. Also, the maximum mass of a white dwarf is about 1.4 times the mass of the Sun.
5. Black hole: formed from the collapse of a white dwarf beyond 1.5 solar masses.
Answer:
other person is correct!!!
Explanation:
got 100 on test
A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.
Answer:
The value is [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
Explanation:
From the question we are told that
The force is [tex]F = 22.7 \ N[/tex]
The value of room temperature is [tex]T = 298 \ K[/tex]
Generally the rate at which its entropy changes as it stretches is mathematically represented as
[tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]
=> [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]
=> [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
1 What is the resistance of the inductive coil
takes 5A current across 240V, 50Hz supply
at 0.8 power factor?
A 48
B 4250
C 38.40
D 26.60
Answer:
38.40
Explanation:
This is the answer for this question
Can we use a hydrometer to
measure the density of milk?
Answer:
yes
Explanation:
i hope this helps not sure im right
Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT
Answer:
(A)
Explanation:
Answer:
275 x 10"N
Explanation:
the maximum normal force a pilot can withstand is about eight times his weight. What is the maximum radius of curvature that a jet planes pilot, pulling out of a vertical dive
Complete Question
the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?
Answer:
The value is [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]
Explanation:
From the question we are told that
The weight of the pilot is [tex]W = mg[/tex]
The maximum force a pilot can withstand is [tex]F_{max} = 7 W = 7 (mg)[/tex]
The speed is [tex]v = 250 \ m/s[/tex]
Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e
[tex]F_c = F_{max} - mg[/tex]
Here N is the normal force acting on the pilot
Now
[tex]F_c = \frac{m v^2 }{r}[/tex]
So
[tex]\frac{m v^2 }{r} = 7(mg) - mg[/tex]
=> [tex]r = \frac{v^2 }{6g}[/tex]
=> [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]
=> [tex]r = 1063 \ m[/tex]
A 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Answer:
4.5 × 10^6 Nm
Explanation:
Given that a 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Let's first calculate the area covered by the heel shoe by using the area of a circle.
Convert cm to m
0.6 / 100 = 0.006
A = πr^2
A = 22/7 × 0.006^2
A = 1.131 × 10^-4 m^2
The force exerted by the woman = mg
Force exerted = 52 × 9.8
Force exerted = 509.6 N
Pressure = force / Area
Pressure = 509.6 / 1.131 × 10^-4
Pressure = 4505853.3 N/m
Therefore, she exerts pressure of 4.5 × 10^6 Nm approximately on the floor.
A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s. The frictional force between the block and the floor is 10 N. The net work done on the block in 1 minutes is most nearly:
Answer:
5× 3 over 10 = 1.5
Explanation:
F = 1m/s × kg over n
The net work done on the block in 1 minutes is most nearly 1800 Joule.
What is work?
A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction. The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item.
Work has no direction and only magnitude. Work is a scalar quantity as a result.
Given that: A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s.
As the block is moving with uniform velocity; this force is equal to the frictional force in magnitude.
So, The net work done on the block in 1 minutes is =
force× displacement
= force × velocity × time
= 10 × 3 × 60 Joule
= 1800 Joule.
Learn more about work here:
https://brainly.com/question/18094932
#SPJ2
2. 290 grams of water is to be heated from 24.0% to 100.0°C to make a cup of
tem, how much heat must be added? The specific heat of water is 4.18 J/g•C
Answer:
92127.2 j
Explanation:
Given that :
Mass of water = 290 g
Specific heat of water (C) = 4.18 j/g.C
Initial Temperature (t1) = 24°C
Final temperature (t2) = 100°C
Using the relation :
Q = mCdt
dt = change in temperature =( 100 - 24)° = 76°C
Q = quantity of heat ; C = specific heat capacity;
m = mass of substance
Q = 290 g * 4.18 j/g.C * 76°C
Q = 92127.2 j
a 60N at an angle of 30°from
horizontal
Explanation:
Force, F = 60 N
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find horizontal and vertical component of the force.
Horizontal component,
[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]
Vertical component,
[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]
So, the horizontal and vertical component are 51.96 N and 30 N respectively.
Do clouds have respiration
Answer:
yes it also does have but not in the exact form but it does 30 percent of respiration to produce rain
(c) The plates are moved farther apart with each plate maintaining the same net charge. In a coherent paragraph-length response, apply concepts of work and energy to explain how the electric potential difference between the plates changes, if at all, when the plates are moved farther apart.
Answer:
Energy is applied on the charge to do work.
Explanation:
Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..
A television camera is positioned 4 km from the base of a rocket launching pad. In order to keep the rocket in focus as it takes off, the camera must be programmed with the distance to the rocket and how that distance is changing. If we assume that the rocket rises vertically at a speed of 200 km/hr, how fast is the distance from the camera to the rocket changing when the rocket has risen 3 km
Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.
The intensity of sound is measured on the decibel scale, dB. The equation dB=10 log I represents the decibel level, where I is the ratio of the sound to the human hearing threshold. A noise is 150,000 times greater than the human hearing threshold. Which shows a valid step in the process of finding the decibel level of the noise?
a. 150,000 = 10 log I
b. 15,000 = log I
c. dB = 10 log 150,000
d. 10dB = log 150,000
e. 10/dB= log 150,000
Answer:
The correct option is c
Explanation:
From the question we are told that
The ratio of the noise to human hearing threshold is [tex]I = 150 000[/tex]
Generally from the equation given we have that
dB = 10 log I
So
dB = 10 log 150000
The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct
Given the equation for calculating the intensity of sound which is measured in decibel expressed as:
dB=10 log I
where;
I is the ratio of the sound to the human hearing threshold
Given that noise is 150,000 times greater than the human hearing threshold.
Substitute I = 150,000 into the expression above;
dB = 10 log 150,000
Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000
Learn more here: https://brainly.com/question/14468501
A small hole P is made in a piece of cardboard. The hole is illuminated by a torch as shown in Fig. 16.3. The pencil of light coming out of the hole falls on a mirror.
5th question
Answer:
According to law of reflection i know in case of plane mirror the angle of reflection is equal to the angle of incidence. So,if you placed the eye on point A the hole can be seen.
Explanation:
A jet pilot puts an aircraft with a constant speed into a vertical loop. (a) Which is greater, the normal force exerted on the seat by the pilot at the bottom of the loop or that at the top of the loop
Answer:
A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.
Explanation:
Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).
When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:
?F(c) = ma(c) = n - mg
n = mg + ma(c)
= m[g + a(c)]
Since centripetal acceleration equals v² / r, the equalization enhances:
n = m[g + (v² / r)]
what is force?answer in one line.
Two spherical balls are placed so that their centers are 3.61 m apart. The
force between them is 1.65 x 10-7 N. If the mass of the smaller ball is 81 kg,
what is the mass in kilograms of the other ball?
Answer:
The mass of the other ball is 397.775 kg.
Explanation:
Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.
The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.
The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.
In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:
[tex]F=G\frac{M1*M2}{d^{2} }[/tex]
where:
F = It is the module of the force exerted between both bodies, and its direction is found on the axis that joins both bodies. G = It is the constant of Universal Gravitation, whose value is 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]In this case:
F= 1.65*10⁻⁷ NG= 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]M1= 81 kgM2= ?d= 3.61 mReplacing:
[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]
Solving for M2:
[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]
M2= 397.775 kg
The mass of the other ball is 397.775 kg.
A 2.00 kg copper pot is placed on a glass shelf and then one side of the shelf falls so that it becomes angled at 35.0° to the horizontal. The coefficient of static friction between copper and glass is 0.680 and the coefficient of kinetic friction is 0.530. Will the pot slide, and if so, what is its acceleration?
Answer:
Explanation: about 2 Mph when it falls and sorry if I’m wrong