Suppose that the math man, a super here that fights crime with math and physics, can decelerate the rate of gravity. During a recent fight with a diamond thief, Math man fell from the top of a 500 meter building. The equation D=t^2*.25. How long us he falling from the top of the building to the ground?
Answer:
44.7 seconds
Explanation:
D = 500 m
500 = .25 t^2
500/.25 = t^2
2000 = t^2
t = 44.7 seconds
*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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Why do batteries discharge more quickly in cold weather?
give me answer
who will give me answer in easy language i will give him 5 star ⭐⭐⭐
Answer:
lowering the ambient temperature causes chemical reaction to proceed more slowly,so a battery used in a low temperature produces less current than that at high temperature . As cold batteries run down they quickly reach the point where they cannot deliver enough current to keep up the demand.. hence discharge more quickly in cold weather..
Write a properly formatted hypothesis statement to answer this question: How does the amount of salt added to ice affect the rate at which the ice will melt?
Specify how you plan to change the independent variable by using terms such as increase or decrease. Also, specify how the dependent variable will change in response by using terms such as increase, decrease, or stays the same.
Criteria pts
Correct placement of IV 5
Correct placement of DV 5
If, then format 5
IV indicates either "increases" or "decreases" 5
DV indicates either "increases", "decreases", or "stays the same" 5
The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
The independent variable which is can be changed by increasing the rate of salt added to the equation.
The dependent variable which is ice will change or melt in response as it will decrease if the rate of the salt added increases.
What is the effect of salt on the melting temperature of ice?Salt does not really lower the temperature of an ice cubes, it is known to just lowers their freezing point, that is lowers their melting point.
Note that if salt is around, ice cubes are known to be colder to be solid, and they tend to melt at a temperature that is said to be lower than the freezing point of pure water.
If the ionic compound salt is known to be added, it tends to lowers the freezing point of the water, which implies that the ice on the ground is not able to freeze that layer of water at all.
Therefore, The hypothesis will be:
H₀ = The amount of salt added to ice will not affect the rate at which the ice will melt.
H₁ = The amount of salt added to ice will affect the rate at which the ice will melt.
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Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
A. it gets stuck in the atmosphere
B. it is magnetically attracted
C. it is attached by a rope to the Earth
D. it falls at the same rate the Earth curves
The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.
Newton's Cannonball:Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.
Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.
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Your car rides on springs, so it will have a natural frequency of oscillation. (Figure 1) shows data for the amplitude of motion of a car driven at different frequencies. The car is driven at 29 mph over a washboard road with bumps spaced 12 feet apart; the resulting ride is quite bouncy.
1)Determine the frequency of the oscillation, caused by the bumps. 1 mile is 5280 feet.
2)Should the driver speed up or slow down for a smoother ride?
Answer:
See below
Explanation:
29 mile/hr * 5280 f/mile / 3600 s/hr = 42.53 ft/ sec
42.53 ft / sec / 12 feet = 3.54 cycles / sec = 3.54 Hz
for frequency I suppose....where is the referred to figure?
Physics occurs all the time but often goes unnoticed. Here is your chance to reflect on physics in action. Other than the examples used in this lesson, think of a time where you witnessed the conservation of angular momentum. Describe the objects that had angular momentum and how angular momentum was conserved. You may also create an example if you cannot recall one in your personal experience.
Angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
What is principle of conservation of angular momentum?The principle of conservation of angular momentum states that the total angular momentum acting on an object is constant, provided there is no external torque acting on the object.
Angular momentum of a system is conserved as long as there is no net external torque acting on the system.
Examples of conservation of angular momentumthe spinning ice skatersomeone spinning in an office chaira child spinning on roller coasterThus, angular momentum is conserved in the above examples such as the ice skater, the torque or the rotating effect of the force is almost equal to zero because there is negligible friction between the skates and the ice.
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how to calculate the half time
The formular for calculating half life is:
T1/2=0.693/Π
A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of the ball in T/3 seconds ?
Answer:the
8/9 h
Explanation:
Height = 1/2 a T^2 now change to T/3
now height = 1/2 a (T/3)^2 = 1/9 1/2 a T^2 <===== it is 1/9 of the way down or 8/9 h
quations (3.4) and (3.5) are equivalent expressions for Lagrange's equations.
Exercise 3.1 Using the Nielsen form, determine the equation of motion
for a mass m connected to a spring of constant k.
Exercise 3.2 Using the Nielsen form, determine the equations of motion
for a planet in orbit around the Sun. (Answer: mr - mro² = - GMm and
mrö +2mr00.)
3.2 Hamilton's principle
echanical system composed of N particles can be described by n = 3N
inande
Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the
Explanation:
Can we take water instead of clock oil in Milikan oil drop experiment. Explain.
Answer:
No you can't cuz,if you put water instead of clock oil in Millikan oil drop your experiment will fail and it won't turn out the way you wanted it to be
If the gravitational potential energy of an object 10 m above the ground is 50 J, what is its Ep, if it moves to 30 m above the ground?
Answer:
150 J
Explanation:
Moving 3 times higher will increase the P E x 3 = 150 J
(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C
If a charge of −4.72 µC is placed at this point, what are the magnitude and direction of the force on it? Magnitude _______N
Direction?
- toward the left
- upward
-downward
- toward the right
The electric field strength at a point 1.00 cm to the left of the middle is -2.0 x 10⁷ N/C.
The magnitude of the force is 94.4 N and direction of the force on it towards the right.
Electric field strengthThe electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first charge
E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second charge
E2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third charge
E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
Net electric fieldE = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C
E = -2.0 x 10⁷ N/C
Force on the charge −4.72 µCF = Eq
F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = 94.4 N
Thus, the direction of the force will be towards the right.
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A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?
A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then
(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s
(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.
c) The speed of the 0.950- kg puck after the collision is 2.82 m/s
d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.
Given:Mass of ice puck, m₁ = 0.350 kg
Mass of another puck, m₂ = 0.950 kg
Velocity of ice puck, v₁ = 5.22 m/s
Velocity of another puck, v₂ = 0 m/s
[tex]v^{'}_1= ?[/tex]
[tex]v^{'}_2= ?[/tex]
[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]
[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]
[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]
Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.
[tex]v^{'}_2 = v^{'} + v_1[/tex]
[tex]= -2.40+5.22[/tex]
[tex]= 2.82 m/s[/tex]
Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.
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a 30kg crate is pulled up a ramp 15m long and 2m high by a constant force of 100n. the crate starts from resta nd has a velocity of 2m/s when it reaches the ramp. what is the frictional force between the crate and the ramp. use the principal of conservation energy.
The frictional force between the crate and the ramp is determined as 35.2 N.
Energy lost to frictional forceApply the principle of conservation energy to calculate the change in the energy of the crate.
Change in energy of the crate = energy loss to friction
P.Ei - K.Ef = E
mgh - ¹/₂mv² = E
where;
m is mass of the crateh is vertical height traveled by the cratev is the final velocity of the crate(30)(9.8)(2) - (0.5)(30)(2²) = E
528 J = E
Frictional force between the crate and the rampE = Fd
where;
F is the frictional forced is the distance traveled by the crateF = E/d
F = (528)/(15)
F = 35.2 N
Thus, the frictional force between the crate and the ramp is determined as 35.2 N.
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A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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8- A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, how many gal does it contain ?
A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, then it contains 0.599132136584 gal .
A bottle weighs 25 pounds ,
we know that 1 Gallon = 8.34 Lb
so, 25 lb = 2.995660682922 gal
and if the bottle weighs 5 pounds then ,
5 lb = 0.599132136584 gal
hence, 2.4 gal less.
Define pound.
a. a unit of weight equivalent to l6 ounces avoirdupois (453.59237 grams), the fundamental unit of weight in the FPS system;
b. a unit of weight equal to 12 ounces troy or 12 ounces apothecaries' (373.2418 grams).
It is an imperial unit of mass or weight measurement.
Define gallons.
In both imperial and US customary units, the gallon is a unit of volume.
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Let's begin with a simple calculation of the weight of air using density. Find the mass of air and its weight in a living room that has a 4.3m×5.4m floor and a ceiling 2.9 m high. What are the mass and weight of an equal volume of water?
a. What volume of water would have a mass equal to the mass of air in the room?
1. The mass of the air, given the data is 85.88 Kg
2. The weight of the air is 841.624 N
3. The mass of an equal volume of water is 67338 Kg
4. The weight of an equal volume of water is 659912.4 Kg
5. The volume of water that would have a mass equal to the mass of air in the room is 8.588×10⁻² m³
How to determine the volume of the airWe shall determine the volume of the air in the room by finding the volume of the room.
Volume of air = volume of room
Volume of air = 4.3 × 5.4 × 2.9
Volume of air = 67.338 m³
1. How to determine the mass of the airVolume of air = 67.338 m³Density of air = 1.2754 kg/m³Mass of air = ?Density = mass / volume
Cross multiply
Mass = Density × volume
Mass of air = 1.2754 × 67.338
Mass of air = 85.88 Kg
2. How to determine the weight of the airMass of air (m) = 85.88 KgAcceleration due to gravity (g) = 9.8 m/s²Weight of air (W) = ?W = mg
W = 85.88 × 9.8
Weight of air = 841.624 N
3. How to determine the mass of equal volume of waterVolume of air = 67.338 m³Volume of water = Volume of air = 67.338 m³Density of water = 1000 kg/m³Mass of water = ?Density = mass / volume
Cross multiply
Mass = Density × volume
Mass of water = 1000 × 67.338
Mass of water = 67338 Kg
4. How to determine the weight of equal volume of waterMass of water = 67338 KgAcceleration due to gravity (g) = 9.8 m/s²Weight of water (W) = ?W = mg
W = 67338 × 9.8
Weight of air = 659912.4 Kg
5. How to determine the volume of water with equal mass of air Mass of air = 85.88 KgMass of water = Mass of air = 85.88 KgDensity of water = 1000 kg/m³Volume of water = ?Volume = mass / density
Volume of water = 85.88 / 1000
Volume of water = 8.588×10⁻² m³
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what is barometer ???
Answer:
an instrument measuring atmospheric pressure, used especially in forecasting the weather and determining altitude.
Explanation:
hope it helps ya
accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for the function in the figure?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter C for the function in the figure is 2.
What is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
f(x) = Acos(x - C)
where;
A is amplitude of the waveC is phase difference of the waveWhat is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
From the blue colored graph; at y = 1, x = -2 cm
1 = cos(2 - C)
(2 - C) = cos^(1)
(2 - C) = 0
C = 2
Thus, the value of parameter C for the function in the figure is 2.
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It's your first day on the job without your mentor as a member of mission control.
Your shift begins and your counterpart, who happens to be in their 10th year in this
position, speaks to you for about 45 minutes as they end their shift and you begin
yours. What were they talking to you about?
getting you up to date with recent mission developments
sanitation protocol that all employees must memorize
connecting you with each of the astronauts in space at the time
how to change settings for each system that's in use
Explanation:
getting me up to date with recent mission development
A thin spherical shell of mass M and radius r is allowed to roll from the edge of a hemispherical bowl of radius R = 80.0 cm. It rolls down with no slipping.
1) Find the speed of the center of mass of the spherical shell when it is at the bottom of the bowl, if r is very small.
2) Repeat part 1) if r = 10.0 cm. Moment of inertia of a thing spherical shell if 2/3Mr^2.
(a) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
(b) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
Speed of the shell at the bottom of the bowl
The speed of the shell at the bottom of the bowl is calculated by applying the principle of conservation of energy.
K.E(rot) + K.E(trans) = P.E
where;
P.E is the potential energy of the ball at the initial positionK.E(rot) is rotational kinetic energyK.E(trans) is translation kinetic energy¹/₂mv² + ¹/₂Iω² = mgh
where;
I is moment of inertia of the spherical shellh is the height of fallv is the speed at the bottomω is angular speed¹/₂mv² + ¹/₂(²/₃Mr²)(v/r)² = mgh
¹/₂v² + ¹/₂(²/₃r²)(v²/r²) = gh
¹/₂v² + ¹/₂(²/₃)(v²) = gh
¹/₂v² + ¹/₃v² = gh
⁵/₆v² = gh
v² = 6gh/5
v = √(6gh/5)
Let the vertical height from the edge of bowl to the bottom , h = R = 80 cm
v = √(6 x 9.8 x 0.8 /5)
v = 3.1 m/s
When the radius = 10 cmv = √(6gh/5)
v = √(6 x 9.8 x 0.1 /5)
v = 1.1 m/s
Thus, the speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
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Consider the f(x) = Acos(x) function shown in the figure in blue color. What is the value of amplitude A for this function?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
What is amplitude of a wave?
The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
Amplitude of the red colored waveFrom the graph, the amplitude of the red colored wave is 1 unit.
Amplitude of the blue colored waveFrom the graph, the amplitude of the red colored wave is 2.1 unit.
Thus, the amplitude of the red colored wave is 1 unit and the amplitude of the red colored wave is 2.1 unit.
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A converging lens with a focal length of 4.0 cm is to the left of a second identical lens. When a feather is placed 12 cm to the left of the first lens, the final image is the same size and orientation as the feather itself. What is the separation between the lenses?
The distance of separation between the lenses is determined as 12 cm.
Distance of image formed by the first lens
The distance of the image formed by the first lens is calculated as follows;
1/d₁ = 1/f - 1/di₀
1/di₁ = 1/4 - 1/12
1/di₁ = 2/12
1/di₁ = 1/6
di₁ = 6 cm
Find the final position of the imageUse magnification formula as shown below;
M = (di₁di₂)/(d₀₁d₀₂)
1 = (di₁di₂)/(d₀₁d₀₂)
di₂ = (d₀₁d₀₂)/(di₁)
di₂ = (12 x d₀₂)/(6)
di₂ = 2d₀₂
1/f = 1/ d₀₂ + 1/di₂
1/f = 1/ d₀₂ + 1/2d₀₂
1/f = 1.5/d₀₂
d₀₂ = 1.5f
d₀₂ = 1.5(4 cm) = 6 cm
Distance between the two converging lensD = di₁ + d₀₂
D = 6 cm + 6 cm
D = 12 cm
Thus, the distance of separation between the lenses is determined as 12 cm.
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An object with an initial velocity of
0.12rad accelerates at 0.11rad over a
distance of 0.25 radians. What is the
final angular velocity of the object?
rad
S
Answer:
100rad because it angular velocity
An ultracentrifuge is spinning at a speed of 80,000 rpm. The rotor that spins with
the sample can be roughly approximated as a uniform cylinder of 10 cm radius
and 8 kg mass, spinning about its symmetry axis). In order to stop the rotor in
under 30 s from when the motor is turned off, find the minimum braking torque
that must be applied.
O-19.2 Nm
-17.2 Nm
O -15.2 Nm
O-11.2 Nm
O None of the above
D. The minimum braking torque that must be applied is -11.2 Nm.
Moment of inertia of the uniform cylinderThe moment of inertia of the uniform cylinder is calculated as follows;
I = ¹/₂MR²
where;
M is mass of the cylinderR is radius of the cylinderI = (0.5)(8)(0.1²)
I = 0.04 kgm²
Minimum braking torqueτ = -Iα
where;
α is angular accelerationα = ω/t
α = (80,000 x 2π/rev x 1 min/60s) / (30 s)
α = (80,000 x 2π)/(60 x 30)
α = 279.25 rad/s²
τ = - ( 0.04 kgm²) x (279.25)
τ = -11.2 Nm
Thus, the minimum braking torque that must be applied is -11.2 Nm.
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The Sun subtends an angle of about 0.5∘ to us on Earth, 150 million km away.
Estimate the radius of the Sun.
Express your answer to two significant figures and include the appropriate units.
The radius of the Sun is approximately 654497.9 kilometers.
A subtended angle in geometry is an angle created by a common point (in this case, the Earth) that crosses two points of a nearby circular arc (the Sun in this case). Below is a visual illustration of the subtended angle.
The law of cosine can be used to calculate the sun's radius, which is 654497.950 kilometers in kilometers.
The Sun's radius is nearly 109 times bigger than the Earth's. The Earth's radius is 6378 kilometers. Or to put it another way, both dimensions are 1: 109 ratios.
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Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back down to its initial temperature. If the circuit element can withstand a temperature change of 5.1 ∘C without being damaged, what is the maximum rate at which energy can be added to the circuit element?
The maximum rate at which energy can be added to the circuit element mathematically given as
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
What is the maximum rate at which energy can be added to the circuit element?Generally, the equation for P is mathematically given as
[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]
Therefore
[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]
[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]
[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]
Max temp Change
[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]
[tex]\text { time }=3 \times 60[/tex]
t=180s
In conclusion, Max Energy Rate
[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]
[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]
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Friction typically __________ objects. A. Speeds up B. Slows down C. Doesn't affect D. Destroys
Answer:
B
Explanation:
friction opposes motion
Friction typically Slows down objects
Option "B"How does friction affect speed?Friction and Speed
While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.
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