Answer:
Explanation:
Time period of rotation
T = 2πR/ V where R is radius of orbit and V is orbital velocity
Orbital velocity V = √ ( GM/R ) , m is mass of the earth .
T = 2πR √R / GM
T² = 4π²R³ / GM
Putting the values
( 126 x 60 )² = 4 x 3.14² x R³ / 6.67 x 10⁻¹¹ x 5.97 x 10²⁴
57.15 x 10⁶ = 39.44 x R³ / 39.82 x 10¹³
R³ = 577 X 10¹⁸
R = 8.325 x 10⁶ m
= 8325 km
Radius of earth = 6400 km
height of satellite = 8325- 6400 = 1925 km .
An object A with mass 200 kg and an another object B with mass 1000 kg are moving with same speed. The ratio of kinetic energy of object A to B is
Answer:
Ratio of kinetic energy of object A to B = 1:5
Explanation:
Given:
Mass of object A = 200 kg
Mass of object B = 1,000 kg
Find:
Ratio of kinetic energy of object A to B
Computation:
Kinetic energy = (1/2)(m)(v²)
Kinetic energy of object A = (1/2)(200)(v²)
Kinetic energy of object A = (100)(v²)
Kinetic energy of object B = (1/2)(1,000)(v²)
Kinetic energy of object B = (500)(v²)
Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B
Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)
Ratio of kinetic energy of object A to B = 100 / 500
Ratio of kinetic energy of object A to B = 1/5
Ratio of kinetic energy of object A to B = 1:5
The photograph shows part of the Great Plains of North America. How do
plains form?
A. Magma rises at a divergent plate boundary.
B. Tectonic movements make folds and faults in the crust.
C. Continental plates slip past each other at a transform boundary.
D. Water, wind, or glaciers either remove or deposit crust.
SUBMIT
2 PREVIOUS
Answer:
Some plains form as ice and water erodes, or wears away, the dirt and rock on higher land. Water and ice carry the bits of dirt, rock, and other material, called sediment, down hillsides to be deposited elsewhere. As layer upon layer of this sediment is laid down, plains form. Volcanic activity can also form plains.
g A velocity selector consists of crossed electric and magnetic fields. The electric field has a magnitude of 480 N/C and is in the negative z direction. What should the magnetic field (magnitude and direction) be to select a proton moving in the negative x direction with a velocity of 3.50 cross times 10 to the power of 5 m/s to go un-deflected
Answer:
B = 1.37 mT
Explanation:
Given that,
The magnitude of the electric field, E = 480 N/C
The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]
We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,
[tex]qE=qvB[/tex]
Where
B is the magnetic field
[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]
So, the magnetic field is equal to 1.37 mT.
It takes 20 seconds to fill a two-liter bottle with water from your kitchen faucet. What is the mass flow rate from the faucet if water has a density of 1000 fraction numerator k g over denominator m cubed end fraction?
a. 0.1kg/sec.
b. 0.01kg/sec.
c. 1g/sec.
d. 1kg/sec.
Answer:
0.1 kg/s.
Explanation:
The density of water, d = 1000 kg/m³
Volume, V = 2 L
Time, t = 20 s
We need to find the mass flow rate from the faucet. We know that the density of an object is given by :
[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]
So, the mass flow rate is equal to 0.1 kg/s.
. Estimate the buoyant force that air exerts on you. (To do this, you can estimate your volume by knowing your weight and by assuming that your weight density is a bit less than that of water.)
Answer:
[tex]0.886[/tex] N buoyant force is exerted by air
Explanation:
My weight is [tex]75[/tex] Kg
Weight = mass * gravity
As we know
Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body
Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3
Volume is equal to mass / density
[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]
[tex]= 0.0751[/tex] * g
Buoyant Force
= Volume * g * density
[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3
[tex]= 0.886[/tex] N
A 101 kg basketball player crouches down 0.380 m while waiting to jump. After exerting a force on the floor through this 0.380 m, his feet leave the floor and his center of gravity rises 0.920 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity (in m/s) when he leaves the floor. m/s (b) What average force (in N) did he exert on the floor
Answer:
[tex]4.25\ \text{m/s}[/tex]
[tex]3391.22\ \text{N}[/tex]
Explanation:
y = Height of compression = 0.38 m
m = Mass of basketball player = 101 kg
h = Height of center of gravity after jump = 0.92 m
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Energy balance of the system is given by
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]
The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]
[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]
The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. Bifocal lenses are used to correct his vision. A diverging lens is used in the top part of the lens to allow the person to clearly see distant objects.
1. What power lens (in diopters) should be used in the top half of the lens to allow her to clearly see distant objects?
2. What power lens (in diopters) should be used in the bottom half of the lens to allow him to clearly see objects 25 cm away?
Answer:
1) P₁ = -2 D, 2) P₂ = 6 D
Explanation:
for this exercise in geometric optics let's use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
1) to see a distant object it must be at infinity (p = ∞)
[tex]\frac{1}{f_1} = \frac{1}{q}[/tex]
q = f₁
2) for an object located at p = 25 cm
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]
We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm
we substitute in the equations
1) f₁ = -50 cm
2)
[tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]
[tex]\frac{1}{f_2}[/tex] = 0.06
f₂ = 16.67 cm
the expression for the power of the lenses is
P = [tex]\frac{1}{f}[/tex]
where the focal length is in meters
1) P₁ = 1/0.50
P₁ = -2 D
2) P₂ = 1 /0.16667
P₂ = 6 D
why the speed of light decreases as it passes from air into another substance?
Answer:
If light enters any substance with a higher refractive index (such as from air into glass) it slows down. The light bends towards the normal line. If light travels enters into a substance with a lower refractive index (such as from water into air) it speeds up. The light bends away from the normal line.
A 25.0kg girl pushes a 50.0kg boy so that he accelerates at 4.00m/s2. What is the force of the boy on the girl? A. 200N B. 100N C. 12.5 D. 400N
Answer:
a
Explanation:
so the answer is 200N
and I hope it is correct
A phase change is when a substance changes from one state of mind to nother because of the adding or removal of thermal energy
True
False
Answer:
true it all changes
Explanation:
________________________
In many places on Earth, humans are responsible for the removal of grasses, shrubs, trees, and other plants with roots that hold soil in place. This activity is best described by which of the following? *
A) deforestation
B) urbanization
C) air pollution
D) rise in sea level
Instead of changing the focal length of the lens, the eyes of amphibians work in a different manner: a set of muscles changes the shape of the eye which increases the distance between the front of the eye and the retina. The world's largest frog, the Goliath frog of west Africa, has an eye with a maximum size similar to a human's: 2.5cm. However, unlike a human, where the focal length is also 2.5cm, the focal length of the Goliath frog's eye is 2.146 cm. What is the maximum distance this frog can see
Answer:
The answer is "15.56 cm".
Explanation:
[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]
Calculating object of length is x so:
[tex]u= -x[/tex]
Using formula:
[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]
can someone help
pls !
Tameika makes a table about sensory organs
Eye
skin
brain
tongue
Which organ should be removed from the table?
A. eye
B. skin
C. brain
D. tongue
Answer:
I think its d
Explanation:
I'm not sure I'm sorry if I'm wrong
In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The minimum horizontal velocity required is "2.6 m/s".
First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height = 3 m
vi = initial vertical speed = 0 m/s
t = time interval = ?
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]
t = 0.78 s
Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:
[tex]s = vt\\\\v = \frac{s}{t}[/tex]
where,
s = horizontal distance = 2 m
t =0.78 s
v = minimum horizontal velocity = ?
Therefore,
[tex]v = \frac{2\ m}{0.78\ s}[/tex]
v = 2.6 m/s
Learn more about equations of motion here:
brainly.com/question/20594939?referrer=searchResults
The attached picture shows the equations of motion in the horizontal and vertical directions.
A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2
. Calculate
the length of this wire required to make a 10 Ω coil.
Answer:
the length of the wire is 134.62 m.
Explanation:
Given;
resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm
cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²
resistance of the wire, R = 10Ω
The length of the wire is calculated as follows;
[tex]R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m[/tex]
Therefore, the length of the wire is 134.62 m.
What is the order of the events for the water cycle on a typical warm day?
А
rain, snow, sleet
B
precipitation, evaporation, rain
с
evaporation, condensation, precipitation
D
condensation, evaporation, precipitation
A 1.1-kg object is suspended from a vertical spring whose spring constant is 120 N/m. (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 m and released from rest. Find the speed with which the object passes through its original position on the way up.
Answer:
e = 0.0898m
v = 2.07m/s
Explanation:
a) According to Hooke's law
F = ke
e is the extension
k is the spring constant
Since F = mg
mg = ke
e = mg/k
Substitute the given value
e = 1.1(9.8)/120
e = 10.78/120
e = 0.0898m
Hence it is stretched by 0.0898m from its unstrained length
2) Total Energy = PE+KE+Elastic potential
Total Energy = mgh +1/2mv²+1/2ke²
Substitute the given value
5.0= 1.1(9.8)(0.2)+1/2(1.1)v²+1/2(120)(0.0898)²
Solve for v
5.0 = 2.156+0.55v²+0.48338
5.0-2.156-0.48338= 0.55v²
2.36 =0.55v²
v² = 2.36/0.55
v² = 4.29
v ,= √4.29
v = 2.07m/s
Hence the required velocity is 9.28m/s
What voltage would be measured across the 15 ohm resistor?
A)
2.5 volts
B)
5.0 volts
C)
7.5 volts
D)
10 volts
Answer:
7.5 volts
Explanation:
I did it on USA Testprep
If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror
Answer:
U=30cm
Explanation:
All you have to do is to put
Mirror formula , 1/f=1/u + 1/v
You should be careful in sign convention .
Virtual image is negative
we take focal length of convex lens negative even if its not given and so on...
If a reflected ray is 55 degrees from the normal line, they what is the angle of the
incident ray from normal?
Answer:
xplanation:
Angle of reflection is measured between the incident ray and the angle which it makes with the normal at the point where incident ray strikes the mirror surface.
Further on reflection, it makes the same angle i.e. angle of reflection is equal to angle of reflection.
Hence, as angle of incidence is 55∘ angle of reflection too is 55∘ and the angle between the incident ray and the reflected ray is 55∘+55∘=110∘
Which waves can travel through space?
a. Electromagnetic waves only
b. Mechanical waves only
c. Electromagnetic and mechanical waves
d. Longitudinal and electromagnetic waves
Answer:
electromagnetic waves only
Explanation:
I just took the test, Hope it helps!
Answer:
A: Electromagnetic waves only
Explanation:
Protons, neutrons, electrons, and a nucleus are
A wave has a frequency of 2 Hz. Find its period
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]
Also, at any instant t
[tex]\Rightarrow l^2=x^2+y^2[/tex]
differentiate w.r.t.
[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]
How much mechanical work is required to catch a 14.715N ball traveling at a velocity of 37.5m/s?
To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.
What is work?Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.
In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that
Work is the product of the component of the force acting in the displacement's direction and its magnitude.
Weight of the ball = 14.715 N.
Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.
Velocity of the ball = 37.5 m/s
Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.
Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.
Learn more about work here:
https://brainly.com/question/18094932
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At what speed do a bicycle and its rider, with a combined mass of 90 kg
k
g
, have the same momentum as a 1500 kg
k
g
car traveling at 4.0 m/s
m
/
s
?
Answer:
2
Explanation:
49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n times as great as that measured at corner B. What is the ratio L1/L2 of the lengths of the sides of the rectangle when n
Answer:
[tex]\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}[/tex]
Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
[tex]a_A = n \ a_B[/tex]
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A = [tex]\sqrt{L_1^2 + L_2^2}[/tex]
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
Please help me someone !
Answer:
The object is moving at constant speed.
Explanation:
The spaces between the dots are equal.
What is the Lorentz force law used for? A. To find the voltage of a battery B. To find the magnitude of a magnetic force C. To find the velocity of an electromagnetic wave. D. To find the direction of a magnetic field
Answer:
B
Explanation:
The Lorentz force is the sum of the electric force and magnetic force.
F = qE + qvB
qE represents electric force and qvB represents magnetic force
