Answer:
29.51m
Explanation:
Suppose you walk 11.2 m straight west and then 27.3 m straight north. How far are you from your starting point
Your distance from the starting point is your displacement
Using the pythagoras theorem
d² = (-11.2)²+(27.3)²
d² = 125.44+745.29
d² = 870.73
d = √870.73
d = 29.51m
Hence your distance from the starting point is 29.51m
The kinetic energy of an object of mass 3.5 kg is 112 J. Its speed is?
64 m/s
8.0 m/s
5.67 m/s
6.8 m/s
ANSWER QUICK 30 POINTS
What force controls the movement of the planets around the sun, holds together stars grouped in galaxies, and galaxies grouped in clusters? Thoroughly explain your answer, making sure to include an example and describe how this force keeps planets in orbit. Make sure to write at least 3-5 sentences and proper conventions (spelling, grammar, punctuation, etc.) to respond. Put all answers in your own words (pls dont just take my points :( )
Answer:
Explanation:
Gravity
To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x is in meters. What is the change in potential energy ∆U when the spring is stretched 2 meters from its equilibrium position?
Answer:
64 J
Explanation:
The potential energy change of the spring ∆U = -W where W = work done by force, F.
Now W = ∫F.dx
So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)
∆U = - ∫-Fdx
= ∫F.dx
Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U = ∫₀²F.dx
= ∫₀²(40x - 6x²).dx
= ∫₀²(40xdx - 6x²dx)
= ∫₀²(40x²/2 - 6x³/3)
= ∫₀²(20x² - 2x³)
= [20x² - 2x³]₀²
= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)
= [(20(4) - 2(8)) - (0 - 0))
= [80 - 16 - 0]
= 64 J
The change in potential energy will be "64 J".
Given:
[tex]F = 40x-6x^2[/tex]The potential energy of spring:
→ [tex]P.E = \int_{0}^{x}F.dx[/tex]
[tex]= \int_{0}^{x}(40x - 6 x^2).dx[/tex]
here,
x = 2m
then,
→ [tex]P.E = 64 \ J[/tex]
Thus the above answer is correct.
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6.
A runner completed the 100.-meter dash in
10.0 seconds. Her average speed was
A) 10.0 m/s
B) 1,000 m/s
C) 0.100 m/s
D) 100 m/s
Answer:
A) 10.0 m/s
Explanation:
the formula for average speed is v=d/t
the distance is 100 meters
The time is 10.0 seconds
so v=100m/10.0s
= 10.0m/s
so the runners average speed was 10.0m/s
The average speed of the runner is equal to 10.0 m/s. Therefore, option (A) is correct.
What is speed?The speed of a body can be described as the distance covered in a definite time interval. The average speed can be defined as a scalar parameter as it possesses magnitude with no direction.
The speed can be determined from the ratio of the distance traveled by the object to the time taken to travel that distance.
The speed of any object can be calculated from the formula equation mentioned below:
Speed = total distance /time taken
Given, the distance covered by the runner = 100 m
The time taken by the runner, t = 10.0 s
The average speed of the runner can be calculated from the formula as shown below:
Speed = 100/10 = 10 m/s
Therefore, the average speed of the runner is 10.0 m/s.
Learn more about the speed, here:
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At highway speeds, a particular automobile is capable of an acceleration of about 1.6 m/s. At
this rate, how long does it take to accelerate from 36 m/s to 50 m/s? (8.8 s)
Proof please
Answer:
8.8s
Explanation:
Given parameters:
Acceleration = 1.6m/s²
Initial velocity = 36m/s
Final velocity = 50m/s
Unknown:
Time taken for the acceleration = ?
Solution:
Acceleration is the rate of change of velocity with time;
Acceleration = [tex]\frac{Final velocity - initial velocity }{time taken }[/tex]
Insert the parameters and solve for time taken;
1.6 = [tex]\frac{50 - 36}{t}[/tex]
1.6t = 14
t = 8.8s
Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo de 30.0° con la pared. ¿Cuánto trabajo (en Julios) realiza la gravedad sobre el pintor? *
Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
[tex]W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J[/tex]
Hence, the required work done is 1786.17 J.
A current of 310 is flowing through a copper wire with a resistance of 30 ohms. What is the voltage?
Answer:
930v
Explanation:
u=i*r
u=310*30=930v
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Explanation: That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
_H₂ + O₂- _ H₂O. how do you balance this equation
Answer:
first balance hydrogen.
Explanation:
2H2+O2---->2H2O
Answer:
2H^2 + 2H^2O
Explanation:
Put a 2 in front of both H's and the H's will be 4 and the O's will be 2. The chemical equation is balanced.
A person standing on a building ledge throws a ball vertically from a launch position 55 m above the ground. It takes 2.0 s for the ball to hit the ground.
With what initial speed was the ball thrown?
Answer:
Vo = 17.69 [m/s]
Explanation:
To solve this problem we must use two equations of kinematics.
[tex]v_{f}^{2} =v_{o}^{2} +2*g*h\\v_{f}=v_{o}+g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 55 [m]
t = time = 2 [s]
Now we replace the gravity acceleration into the second equation:
[tex]v_{f}=v_{o}+9.81*2\\v_{f}=v_{o}+19.62[/tex]
And then into the first equation:
[tex](v_{o}+19.62)^{2}=v_{o}^{2}+2*9.81*55\\v_{o} ^{2}+2*v_{o}*19.62+384.94=v_{o}^{2} + 1079.1\\39.24*v_{0}=694.16\\v_{o}=17.69[m/s][/tex]
The initial speed at which the ball is thrown upward is 17.7 m/s.
According to the question the initial position of the ball, y = 55 m, and the final position of the ball is y' = 0 m. We have assumed upward direction as positive direction and downward as negative direction.
so the total displacement:
d = y'-y = 0 - 55
d = -55 m
now applying the second equation of motion:
d = ut - (1/2)gt²
where t = 2s ( given ) and g = 9.8 m/s².
-55 = 2u - 0.5×9.8×4
-55 = 2u - 19.6
u = -17.7 m/s
the negative sign indicated that the initial velocity is opposite to the direction of displacement.
This means the initial velocity is upward as it should be.
Therefore, the initial speed of the ball is 17.7m/s
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Heredity is best described as th-
Explanation:
The most established layers are on the base, and the most youthful layers are on the top. Since dregs here and there incorporate once-living creatures, sedimentary stone regularly contains a great deal of fossils. Fossils are once living beings that have been transformed into rock, fit as a fiddle or type of the creature can in any case be seen.
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in the range of 30-300 GHz. What are the free-space wavelengths that correspond to this frequency range
Answer:
It corresponds to 1mm-10 mm range.
Explanation:
Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:[tex]v = \lambda * f (1)[/tex]
Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:[tex]\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)[/tex]
[tex]\lambda_{high} = \frac{c}{f_{low}} = \frac{3e8m/s}{30e9Hz} = 10 mm (3)[/tex]
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diameter 5.0 m. The room is spinning horizontally about an axis through its center at a rate of 60 rev/min when the floor drops so that it no longer provides any support for the riders. What is the minimum coefficient of static friction between the wall and the rider required so that the rider does not slide down the wall?
Answer:
μsmín = 0.1
Explanation:
There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.This friction force has a maximum value, that can be written as follows:[tex]F_{frmax} = \mu_{s} *F_{n} (1)[/tex]
where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.This force has the following general expression:[tex]F_{c} = m* \omega^{2} * r (2)[/tex]
where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:
[tex]F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)[/tex]
When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:[tex]m* g = m* \mu_{smin} * \omega^{2} * r (4)[/tex]
(The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)Cancelling the masses on both sides of (4), we get:[tex]g = \mu_{smin} * \omega^{2} * r (5)[/tex]
Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:[tex]60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)[/tex]
Replacing by the givens in (5), we can solve for μsmín, as follows:[tex]\mu_{smin} = \frac{g}{\omega^{2} *r} = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)[/tex]
True or false. Only large celestial bodies have gravity
A rifle is aimed horizontally at the center of a large target 60 m away. The initial speed of the bullet is 240 m/s. What is the distance from the center of the target to the point where the bullet strikes the target
Answer:
The distance from the center of the target to the point where the bullet strikes the target is 0.31 m.
Explanation:
Given;
the position of center of the target, x₀ = 60 m
initial velocity of the bullet, u = 240 m/s
The time the bullet strikes the center of the large target is calculated as;
[tex]t = \frac{x_o}{u} =\frac{60}{240} = 0.25 \ s[/tex]
The distance from the center of the target to the point where the bullet strikes the target is calculated as;
x = vt + ¹/₂gt²
where;
v is the final velocity of the bullet when it strikes the target = 0
x = 0 + ¹/₂(9.8)(0.25²)
x = 0.31 m
Therefore, the distance from the center of the target to the point where the bullet strikes the target is 0.31 m.
How big is Earth’s gravitational field
Answer:
Earth's gravity goes back as how old it is, so 4.5 billion light years.
Explanation:
Identify the heat transfer processes that determine the temperature of an asphalt pavement on a summer day. Write an energy balance for the surface of the pavement.
Answer:
Heat transfer processes: Radiation - Natural Convection
Energy balance:
[tex]\dot Q_{rad} -\dot Q_{conv} = \frac{dU_{sys}}{dt}[/tex]
[tex]\epsilon\cdot \sigma\cdot A_{s}\cdot (T_{s}^{4}-T_{a}^{4})-h\cdot A_{s} \cdot (T_{a}-T_{\infty}) = m\cdot c\cdot \frac{dT_{a}}{dt}[/tex]
Explanation:
The asphalt is heated due to the radiation from the sun and cooled by natural convection of the wind. By the First Law of Thermodynamics, we have the following model that represents the system:
[tex]\dot Q_{rad} -\dot Q_{conv} = \frac{dU_{sys}}{dt}[/tex] (1)
Where:
[tex]\dot Q_{rad}[/tex] - Heat transfer by radiation, measured in watts.
[tex]\dot Q_{conv}[/tex] - Heat transfer by natural convection, measured in watts.
[tex]\frac{dU_{sys}}{dt}[/tex] - Rate of change of thermal energy of the asphalt, measured in watts.
By applying the definitions for conduction, radiation and thermal energy, we expand (1) below:
[tex]\epsilon\cdot \sigma\cdot A_{s}\cdot (T_{s}^{4}-T_{a}^{4})-h\cdot A_{s} \cdot (T_{a}-T_{\infty}) = m\cdot c\cdot \frac{dT_{a}}{dt}[/tex] (2)
Where:
[tex]\epsilon[/tex] - Emissivity of the asphalt.
[tex]\sigma[/tex] - Stefan-Boltzmann constant, measured in watts per square meter-quartic Kelvin.
[tex]A_{s}[/tex] - Surface area of the asphalt, measured in square meter.
[tex]T_{s}[/tex] - Temperature of the sun, measured in Kelvin.
[tex]T_{a}[/tex] - Temperature of the asphalt, measured in Kelvin.
[tex]T_{\infty}[/tex] - Temperature of the wind, measured in Kelvin.
[tex]h[/tex] - Natural convection constant, measured in watts per square meter-Kelvin.
[tex]m[/tex] - Mass of the asphalt, measured in kilograms.
[tex]c[/tex] - Specific heat of the asphalt, measured in joules per kilogram-Kelvin.
I need help with these four problems.. anyone??
A football game starts with a kickoff. If the ball is kicked with a velocity of 72 m/s 35 o to
the horizontal. How far will the football travel? What is the speed an angle of impact?
Explain please
Answer:
Explanation: very far
Two objects, each of mass 2 kg, are moving in the same straight line but in opposite directions. They are moving with velocities 10 m/s and 5 m/s respectively before the collision during which they stick together. The velocity of the combined object after collision is,
Answer:
2.5m/s
Explanation:
Using the law of conservation of energy;
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute the given values in the formula
2(10)-2(5) = (2+2)v
20-10 = 4v
10 = 4v
v = 10/4
v = 2.5m/s
Hence the velocity of the combined object after collision is 2.5m/s
arrange the values according to magnitude
1.8 X 10^5
4.3 X 10^-2
8.4 X 10^-6
9.0 X 10^-6
59000
Answer:
[tex]59000[/tex]
[tex]1.8\times 10^5[/tex]
[tex]4.3\times 10^{-2}[/tex]
[tex]9.0\times 10^{-6}[/tex]
[tex]8.4\times 10^{-6}[/tex]
Explanation:
The given numbers are
[tex]1.8\times 10^5=18000[/tex]
[tex]4.3\times 10^{-2}=0.043[/tex]
[tex]8.4\times 10^{-6}=0.0000084 [/tex]
[tex]9.0\times 10^{-6}=0.000009[/tex]
[tex]59000[/tex]
The arrangement of the numbers from largest to smallest is
[tex]59000[/tex]
[tex]1.8\times 10^5[/tex]
[tex]4.3\times 10^{-2}[/tex]
[tex]9.0\times 10^{-6}[/tex]
[tex]8.4\times 10^{-6}[/tex].
Three charges, each separated by 100 m from adjacent charges, are located along a horizontal line: a -3.00 C charge on the left, a 2.00 C charge in the middle, and a 1.00 C charge on the right. What is the resultant force on the 1.00 C charge due to the other two
Answer:
F= 11.25*10⁵ N to the right.
Explanation:
Assuming that the three charges can be treated like point charges, they must obey Coulomb's Law.Due to the linearity of this Law, we can use superposition in order to find the resultant force on the 1.00 C charge due to the other two.First, we find the force that the -3.00 C charge (located 200 m to the left) exerts on the 1.00 C, as follows:[tex]F_{13} = \frac{K*q_{1}*q_{2}}{r_{13} ^{2} } = \frac{9e9*(-3.00C)(1.00C)}{(200m)^{2}} = -6.75e5 N (1)[/tex]
Then, in the same way, we can find the force that the 2.00 C exerts on the 1.00 C charge, located 100 m away to the left:[tex]F_{23} = \frac{K*q_{3}*q_{2}}{r_{23} ^{2} } = \frac{9e9*(2.00C)(1.00C)}{(100m)^{2}} = 18e5 N (2)[/tex]
Since both vectors are on the same line, their sum is directly the algebraic sum, as follows:F₃ = F₁₃ + F₂₃ = -6.75*10⁵ N + 18.00*10⁵ N = 11.25*10⁵ N to the right, assuming this direction as positive.Helppp
In Static Electricity, The charges do not ____
Answer:
basically I will tell you the definition
Explanation:
so when charges are unbalanced statistic energy is formed positive attract negative and negative attracts positive like repell while unlike attract .
The micrometer (1 μm) is often called the micron. (a) How many microns make up 3.0 km? (b) How many centimeters equal 3.0 μm? (c) How many microns are in 3.0 yd?
Answer:
3 x 10⁻⁹km
3 x 10⁻⁴cm
2.73 x 10⁶μm
Explanation:
A micron is a subunit of measurement usually for length dimensions.
1μm = 1 x 10⁻⁶m
a. How many microns make up 3km;
Now convert to meter first;
1000m = 1km
So, 3km will be made up of 3000m
So;
1 x 10⁻⁶m = 1μm
3000m = [tex]\frac{3000}{1 x 10^{-6} }[/tex] = [tex]\frac{3 x 10^{3} }{ 1 x 10^{-6} }[/tex] = 3 x 10⁻⁹km
b. How many centimeters equal 3.0 μm?
Since;
1μm = 1 x 10⁻⁶m
3μm = 3 x 1 x 10⁻⁶ = 3 x 10⁻⁶m
So;
100cm = 1m;
1m = 100cm
3 x 10⁻⁶m = 3 x 10⁻⁶ x 10² = 3 x 10⁻⁴cm
c. How many microns are in 3.0 yd?
1yd = 0.91m
3yd = 3 x 0.91 = 2.73m
So;
1 x 10⁻⁶m = 1μm
2.73m will give [tex]\frac{2.73}{1 x 10^{-6} }[/tex] = 2.73 x 10⁶μm
Why was Cassini launched toward Venus if it is going to explore Saturn?
The gravity of Venus will boost the speed of the space probe
The orbit of Venus is between Earth and Saturn
Venus is needed to reflect the radio signals to the probe
Answer: A. The gravity of Venus will boost the speed of the space probe
Explanation: Hope this helped in time (:
A student draws an energy diagram to explain why the temperature changes between a when a hot flask is placed inside of a cold beaker. What is wrong with the energy diagram below?
Answer:
there is not a picture below but it would be because the beaker was put in cold water then the temp decreased
Explanation:
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the: A. propagation speed of the radiation. B. wavelength of t
Answer:
D.number of photons emitted.
Explanation:
These are the options for the question
A.propagation speed of the radiation.
B.wavelength of the radiation.
C.polarization of photons emitted.
D.number of photons emitted.
.
Electromagnetic energy of any radiation is proportional to the photons present. And we know that intensity is ratio of energy and unit time.
Hence, The intensity of the radiation emitted by the oxygen sensor is directly proportional to the number of photons emitted
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the number of photons emitted. The correct answer is D.
The intensity of radiation refers to the amount of energy carried by the radiation per unit of time and unit of area. In the context of the oxygen sensor, the intensity of the radiation emitted by the sensor is directly proportional to the number of photons emitted.
The more photons emitted, the higher the intensity of the radiation. The intensity is not directly related to the propagation speed, wavelength, or polarization of the photons emitted, but rather the quantity or number of photons being emitted.
Therefore, The intensity of the radiation emitted by the oxygen sensor is directly proportional to the number of photons emitted. The correct answer is D.
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The complete question is:
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the
A. propagation speed of the radiation.
B. wavelength of the radiation.
C. polarization of photons emitted.
D. number of photons emitted.
A 1.5 x 10 3kilogram car is moving at 10 meters per second east. A braking force acts on the car for 5.0 seconds, reducing its velocity to 2.0 meters per second east. ??
Answer:
What should we find state that first.
4. What is the mass of the block
of iron illustrated below?
Ty in advance
What happens to a species if the death rate is higher then the birthrate?
1.The species will start to migrate.
2.The population size decreases.
3.The population size increases.
4.The species becomes extinct.
Answer:
The population size decreases.
Explanation:
If more of a species are dying than being born, the population size will decrease.
