Answer:
Nonrenewable Resources
Answer:
fossil fuels
Explanation:
Two satellites orbit earth at altitudes of 7500 km and 15000 km above earth's surface. Which satellite is faster and by what factor?
Answer:
The satellite that is 7500 km above earth's surface is 1.24 times faster than the satellite at 15000 km above earth's surface
Explanation:
Generally the gravitational force acting on each satellite is mathematically represented as
[tex]F = \frac{G * M_e * m_2 }{r^2}[/tex]
Here [tex]M_e[/tex] is the mass of the earth
[tex]r = r_e + R [/tex]
So [tex] r_e [/tex] is the radius of the satellite and R is the radius of earth with value R = 6371 km
This force is also equivalent to the centripetal force acting on each satellite which is mathematically represented as
[tex]F = \frac{mv^2}{ r^2}[/tex]
So
[tex]\frac{mv^2}{ r^2} = \frac{G * m_1 * m_2 }{r^2}[/tex]
=> [tex]v = \sqrt{ \frac{GM_e}{r} }[/tex]
So for [tex]r = 7500+ 6371 = 13871 \ km [/tex]
We have that
[tex]v_1 = \sqrt{ \frac{GM_e}{13871} }[/tex]
and [tex]r = 15000+ 6371 = 21371 \ km [/tex]
We have that
[tex]v_2 = \sqrt{ \frac{GM_e}{21371} }[/tex]
So
[tex]\frac{v_1}{v_2} = \frac{\sqrt{\frac{GM_e}{13871} } }{ \sqrt{\frac{GM_e}{21371}} }[/tex][tex]v = \sqrt{ \frac{GM_e}{r} }[/tex]
=> [tex]\frac{v_1}{v_2} = \sqrt{ \frac{21371}{13871} }[/tex]
=> [tex]v_1 = 1.24 v_2[/tex]
So satellite that is 7500 km above earth's surface is 1.24 times faster than the satellite at 15000 km above earth's surface
(1.736+9.6435)÷3.22=
The answer is 3.53400621
Reduce the expression, if possible, by cancelling common factors.
Answer: 3.53400621
Explanation: Hope this helped! :)
What is 902 in proper scientific notation?
A cheetah accelerates from rest at a rate of 9 m/s^2 for 3s. How far does it travel in this time? Assume the cheetah moves in a
straight line.
a) 27.0 m
b) 40.5 m
c) 13.5 m
d) 54.0 m
Answer:
b) 40.5 m
Explanation:
Given:
v₀ = 0 m/s
a = 9 m/s²
t = 3 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (3 s) + ½ (9 m/s²) (3 s)²
Δx = 40.5 m
4. Two people each have a mass of 55 kg. They are both in an
elevator that has a mass of 240 kg. When the elevator
begins to move, the people and the elevator have an upward
acceleration of 1.00 m/s2. What is the net force that acts on
the elevator as it accelerates upward at 1.00 m/s2?
Answer:
350 N
Explanation:
F=ma
[tex]f = force \\ m = mass \\ a = acceleration[/tex]
[tex]m = 2(55kg) + 240kg \\ a = 1.0 \frac{m}{ {s}^{2} } [/tex]
Force = 350 Newtons
The net force acting on the elevator would be 350 Newtons as it accelerates upward at 1.00 m/s2.
What is Newton's second law?
Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
The mathematical expression for Newton's second law is as follows
F = ma
As given in the problem two people each have a mass of 55 kg. They are both in an elevator that has a mass of 240 kg. When the elevator begins to move, the people and the elevator have an upward acceleration of 1.00 m/s2, then we have to find the net force acting on the elevator,
The net force acting on the elevator,
F = ma
F =(2×55 + 240)×1
= 350 Newtons
Thus, the net force acting on the elevator would be 350 Newtons as it accelerates upward at 1.00 m/s2
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What is the net Force needed to get a 16 kg box moving 4 m/s^2?
Answer:
The net force should be of a magnitude of 64 N
Explanation:
We use Newton's second Law for this:
[tex]F_{net} = m\,*\,a[/tex]
which for our case gives:
[tex]F_{net} = m\,*\,a=16\,(4)\,N= 64\,\,N[/tex]
A place kicker launches a kickoff at an angle of 25° to the horizontal and at a velocity of 20 m/s. Calculate the initial velocities in the x- and y-directions.
What are some reasons people may not prepare to be safe and comfortable when they participate in physical activity?
Answer: ummm arent you in my physical fundations class lol
Explanation:
are you or am i tripping? and sorry i dont have the answer :(
Common reasons people give for not being active include not having enough time, finding physical activity inconvenient, lacking self-motivation, etc.
What is physical activity?Any consensual bodily mobilization generated by skeletal muscles that necessitate caloric expenditure is characterized as physical activity.
Physical activity includes all activities of any frequency, at any moment of day or night. It incorporates both strength training and incidental activity into the daily routine.
Physical activity can improve your brain health, help you manage your weight, lower your risk of disease, strengthen your bones and muscles, and improve your ability to do everyday tasks.
Adults who sit less and engage in moderate-to-vigorous physical activity reap health benefits.
Internal barriers were classified into three types namely, a lack of energy, a lack of motivation, and a lack of self-efficacy.
External barriers were also classified into three categories namely a lack of resources, a lack of social support, and a lack of time.
Thus, these are some reasons that may not prepare to be safe and comfortable when they participate in physical activity
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You must exert a force of 5N on a book to slide it across a table. If you do 2.5 J of work in the process, how far has the book moved?
Answer:
0.5 m
Explanation:
The following data were obtained from the question:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) =?
Work done (W) can be defined as the product of force (F) and distance (s) moved in the direction of the force. From the above definition, work done (W) can be represented mathematically as:
Work done (W) = Force (F) × Distance (s)
W = F × s
With the above formula, we can obtain the distance moved by the book as shown below:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) =?
W = F × s
2.5 = 5 × s
Divide both side by 5
s = 2.5/5
s = 0.5 m
Therefore, the book moved a distance of 0.5 m.
A car has an initial velocity of 50 m/s and a constant
acceleration of 5 m/s2. What is the car's velocity after 3
seconds?
A train rounding a curve at a steady speed, balanced or unbalanced force?
Answer:
Unbalanced
Explanation:
It would be speeding and under the influence of an imbalanced force if it were rounding a curve. The train must be subject to an unbalanced force in order for it to continue moving at a constant speed.
What curves a steady speed, balanced or unbalanced force?Newton's first law is only this one. Unless operated upon by an imbalanced force, an item at rest remains at rest and an object in motion maintains constant speed in the same direction.
The car tends to maintain its travel at a steady speed and direction because the two forces balance each other out and cancel each other out.
The forces are out of balance when an object's motion changes. Equal in size and directed in the opposite direction, balanced forces are. When forces are evenly distributed, motion remains unchanged.
Therefore, One of your scenarios from the previous section involved pushing or pulling an object with the equal amount of force in opposite directions.
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An automobile is driven on a straight road, and the distance traveled by the automobile after time t=0 is given by a quadratic function a where a(t) is measured in feet and t is measured in seconds for 0 <= t <= 12. Of the following, which gives the best estimate of the velocity of the automobile, in feet per second, at time t = 8 seconds?
a. s(8)
b. s(8)/8
c. s(12)- s(2)/ 12-2
d. s(9)- s(7)/9-7
Answer:
[tex]Velocity = \frac{s(8)}{8}[/tex]
Explanation:
Given
[tex]0 \leq t \leq 12[/tex]
Required
Determine the velocity when t = 8
This type of velocity is referred to as an instantaneous velocity.
In this case, it is calculated using
[tex]Velocity = \frac{Distance\ at\ 8 second}{t = 8}[/tex]
Given that s(t) models the distance;
s(8) = distance at 8 seconds
So;
[tex]Velocity = \frac{s(8)}{8}[/tex]
Option B answers the question
iv. When you use a hand pump to inflate the tires of your bicycle, the pump gets warm after a while. Why? What happens to the temperature of the air in the pump as you compress it? Why does this happen? When you raise the pump handle to draw outside air into the pump, what happens to the temperature of the air taken in? Again, why does this happen?reason
Answer:
The temperature rises because for a given volume of gas, a rise pressure of the gas in pressure results in a proportionate rise in the temperature of the gas
Similarly when the handle is raised to draw air causes a fall in pressure that results in proportionate fall in temperature, for a given volume of gas
Explanation:
From Gay-Lussac's law, states that the pressure of a given mass of gas is directly proportional to its Kelvin temperature, provided that the volume is held constant
Mathematically, the law states that Pressure ∝ Temperature, at constant Volume
Therefore;
P₁/T₁ = P₂/T₂
Similarly, by kinetic theory of gases, we have;
The
[tex]P = \dfrac{n \cdot MW \cdot v_{rms}^2}{3 \cdot V}[/tex]
[tex]v_{rms} = \sqrt{\dfrac{3 \cdot R \cdot T}{MW} }[/tex]
Therefore, as in order for the hand pump to inflate the bicycle tires, the air in the pump has to be compressed to force it into the tire, thereby increasing the pressure, of the air in a given volume of the pump which results in the raising of the temperature of the air in the pump, which raises the temperature of the wall of the pump.
The temperature of the air in the pump also falls as the pressure in the pump is reduced by raising the pump handle, to reduce the air pressure inside the pump and and allow air to be taken into the pump.
An external reward is known as extrinsic motivation.
ОА.
True
OB. False
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
2.13 m/s^2.
Find the remaining answer in the explanation
Explanation:
Given that a sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line.
1.) The maximum horizontal acceleration will be when she attained her maximum speed. That is,
Speed = 41 km/h
Convert km/h to m/s
(41×1000) / 3600
Speed = 11.39 m/s
Using speed formula to calculate time
Speed = distance/time
11.39 = 61/t
t = 61/11.39
t = 5.36s
Where the distance = 61 m
Maximum acceleration = velocity /time
Maximum acceleration = 11.39/5.36
Maximum acceleration = 2.13 m/s^2
The maximum acceleration value occurs when the sprinter is starting from rest and attaining the maximum speed.
What acceleration would a 18 kg box have if we pushed it with a force of 250 N across a surface that
has a coefficient of friction of 0.8?
N
Answer:
[tex]a=0.08\ m/s^2[/tex]
Explanation:
Given that,
Mass of a box, m = 18 kg
It is pushed with a force of 250 N
The coefficient of friction is 0.8
We need to find the acceleration of the box.
Normal force acting on the box,
N = mg
M = 18 × 10
N = 180 N
The force acting on the force in terms of coefficient of friction is given by :
[tex]F=\mu mg\\\\\text{or}\\\\ ma=\mu mg\\\\a=\dfrac{\mu }{g}\\\\a=\dfrac{0.8}{10}\\\\a=0.08\ m/s^2[/tex]
So, the acceleration of the box is [tex]0.08\ m/s^2[/tex]
5 blocks east and 2 blocks north and 6 blocks and then 8 blocks south. what's the total distance traveled?
Answer:
The answer is 21
Explanation:
5 + 2 + 6 + 8 = 21
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
What is the time interval of marble?
Answer:
t = 6.09 seconds
Explanation:
Given that,
Speed, v = 44.1 cm/s
Distance, d = 269 cm
We need to find the time interval of the marble. Speed is distance per unit time.
[tex]v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s[/tex]
Hence, the time interval of the marble is 6.09 seconds.
Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temperature), and measured terminal speed of 2.64X10^-5 m/s and weight-neutralizing voltage of 35.0 V. How many electrons have been attached to, or detached from the initially neutral plastic sphere
Following are the calculation to the given question:
Given:
Please find the given question.
To find:
value=?
Solution:
With terminal voltage, no force equals zero.
So,
[tex]\to mg = 6\times \pi \times \eta \times r\times \vartheta \\\\[/tex]
As [tex]35\ V[/tex] is weight neutralizing voltage
[tex]\to mg= q\times v \times r\\\\[/tex]
So,
[tex]\to q\times v \times r= 6 \times \pi \times \eta \times r \times \vartheta \\\\\to q\times v = 6 \times \pi \times \eta \times \vartheta \\\\\to q =\frac{6 \times \pi \times \eta \times \vartheta}{v} \\\\[/tex]
[tex]=\frac{6 \times 3.14 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 18.84 \times \eta \times 2.64 \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\=\frac{ 49.7376 \times \eta \times 10^{-5}\ \frac{m}{s}}{35} \\\\ =3.2 \times 10^{-10}\ coulombs[/tex]
Therefore
[tex]\to q= n \times e\\\\[/tex]
Hence n comes to be [tex]2\times 10^9\[/tex] electrons.
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An electronic line judge camera captures the impact of a 57.0-g tennis ball traveling at 32.2 m/s with the side line of a tennis court. The ball rebounds with a speed of 21.6 m/s and is seen to be in contact with the ground for 3.94 ms. What is the magnitude of the average acceleration of the ball during the time it is in contact with the ground
Answer:
average acceleration is 1.365 × [tex]10^{4}[/tex] m/s²
Explanation:
given data
initila speed u = -32.2 m/s
final speed v = 21.6 m/s
time taken t = 0.00394 s
solution
we get here average acceleration that will be express as
v = u + at ..........................1
put here value and we get
21.6 = -32.2 + a × 0.00394
solve it we get
a = 1.365 × [tex]10^{4}[/tex] m/s²
so average acceleration is 1.365 × [tex]10^{4}[/tex] m/s²
what is surface tension
Answer:
Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects, usually denser than water, to float and slide on a water surface.
What happens if we increase the value of the resistor in forward bias connection?
What is the speed of an object that travels 60 meters in 4 seconds?
Answer:
15 m/s
Explanation:
s=d/t
s=60/4
s=15 m/s
Answer:
The answer would be 15 m/s
Explanation:
s=d/t
s=60/4
s= 15 m/s
Have a beautiful day beautiful
You welcome.
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity as measured in W/m2 changes by a multiplicative factor. The number of decibels increases by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
The question is incomplete. Here is the complete question.
To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is
[tex]\beta=10log(\frac{I}{I_{0}} )dB[/tex],
where [tex]I_{0}[/tex] is a reference intensity. for sound waves, [tex]I_{0}[/tex] is taken to be [tex]10^{-12} W/m^{2}[/tex]. Note that log refers to the logarithm to the base 10.
Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. [tex]I=10I_{0}[/tex]? Express the sound intensity numerically to the nearest integer.
Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. [tex]I=100I_{0}[/tex]? Express the sound intensity numerically to the nearest integer.
Part C: Calculate the change in decibels ([tex]\Delta \beta_{2},\Delta \beta_{4}[/tex] and [tex]\Delta \beta_{8}[/tex]) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.
Answer and Explanation: Using the formula for sound intensity level:
A) [tex]I=10I_{0}[/tex]
[tex]\beta=10log(\frac{10I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(10 )[/tex]
β = 10
The sound Intensity level with intensity 10x is 10dB.
B) [tex]I=100I_{0}[/tex]
[tex]\beta=10log(\frac{100I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(100)[/tex]
β = 20
With intensity 100x, level is 20dB.
C) To calculate the change, take the f to be the factor of increase:
For [tex]\Delta \beta_{2}[/tex]:
[tex]I=2I_{0}[/tex]
[tex]\beta=10log(\frac{2I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(2)[/tex]
β = 3
For [tex]\Delta \beta_{4}[/tex]:
[tex]I=4I_{0}[/tex]
[tex]\beta=10log(\frac{4I_{0}}{I_{0}} )[/tex]
[tex]\beta=10log(4)[/tex]
β = 6
For [tex]\Delta \beta_{8}[/tex]:
[tex]I=8I_{0}[/tex]
[tex]\beta=10log(\frac{8I_{0}}{I_{0}} )[/tex]
β = 9
Change is
[tex]\Delta \beta_{2},\Delta \beta_{4}[/tex], [tex]\Delta \beta_{8}[/tex] = 3,6,9 dB
A ball with mass kg is thrown upward with initial velocity m/s from the roof of a building m high. Neglect air resistance. Use m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. s
Answer:
Explanation:
The question is incomplete. Here is the complete question.
A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 17 m high. Neglect air resistance. Use g=9.8 m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.
a) Using the equation of motion formula;
v² = u²+2gH where;
u is the initial velocity
v is the final velocity
theta is the angle of launch
g is the acceleration due to gravity.
H is the maximum height reached by the ball
Since the ball is thrown upwards, the acceleration due to gravity will be negative. The equation then becomes;
v² = u²-2gH
Given
v = 0m/s
u = 22m/s
g = 9.8m.s²
0² = 22²-2(9.8)H
-22² = -19.6H
H = -22²/-19.6
H = 24.69m
If the biuliding is 17m high, the maximum height above the ground that the ball reaches will be;
Hmax = 24.69+17
Hmax = 41.69m
b) The time it takes to hit the ground can be expressed using the formula
v = u-gt
0 = 22-9.8t
-22 = -9.8t
t = -22/9.8
t = 2.45secs
What is the correct answer choice to the question above?
Answer: A. a straight line inclined to the time axis
Explanation:
Willie swift drives a hockey puck into the goal at a speed of 34.8 m/s. From a distance of 60.0 m, how long will it take to reach the net?
Explanation:
distance = rate*time
time = distance divided by rate
now just substitute
time = distance (60 m) divided by rate (34.8 m/sec)
a cart travels 3.00 meters east and then 4.00 meters north. the magnitude
Answer: 5 m
Explanation:
X^2= 9+16
X = 5
According to pythagoras law
Which inference about Arthur is supported by the text?
A wire 6.60 m long with diameter of 2.05 mm has a resistance of 0.0310 Ω.
Find the resistivity of the material of the wire.
rho= Ω*m
Answer:
1.551×10^-8 Ωm
Explanation:
Resistivity of a material is expressed as shown;.
Resistivity = RA/l
R is the resistance of the material
A is the cross sectional area
l is the length of the wire.
Given;
R = 0.0310 Ω
A = πd²/4
A = π(2.05×10^-3)²/4
A = 0.000013204255/4
A = 0.00000330106375
A = 3.30×10^-6m
l = 6.60m
Substituting this values into the formula for calculating resistivity.
rho = 0.0310× 3.30×10^-6/6.60
rho = 1.023×10^-7/6.60
rho = 1.551×10^-8 Ωm
Hence the resistivity of the material is 1.551×10^-8 Ωm