Answer:
A) 2.650 km
Explanation:
The relationship between acceleration of gravity and gravitational constant is:
[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)
Where
[tex]R = 6,400 km[/tex] -- Radius of the earth.
From the question, we understand that the gravitational field of the rocket is 50% of its original value.
This means that:
[tex]g_{rocket} = 50\% * g[/tex]
[tex]g_{rocket} = 0.50 * g[/tex]
[tex]g_{rocket} = 0.5g[/tex]
For the rocket, we have:
[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]
Where r represent the distance between the rocket and the center of the earth.
Substitute 0.5g for g rocket
[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)
Divide (1) by (2)
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]
[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]
[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]
[tex]2 = \frac{r^2}{R^2}[/tex]
Take square root of both sides
[tex]\sqrt 2 = \frac{r}{R}[/tex]
Make r the subject
[tex]r = R * \sqrt 2[/tex]
Substitute [tex]R = 6,400 km[/tex]
[tex]r = 6400km * \sqrt 2[/tex]
[tex]r = 6400km * 1.414[/tex]
[tex]r = 9 049.6\ km[/tex]
The distance (d) from the earth surface is calculated as thus;
[tex]d = r - R[/tex]
[tex]d = 9049.6\ km - 6400\ km[/tex]
[tex]d = 2649.6\ km[/tex]
[tex]d = 2650\ km[/tex] --- approximated
A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.
A) What was the heat input to this engine?
B) What was its efficiency?
Answer:
a) [tex]H_{in}=39.34 kJ[/tex]
b) Efficiency=76.77%
Explanation:
a)
In order to solve this problem, we can use the following formula:
[tex]H_{in}=H_{out}+W[/tex]
the problem provides us with all the necessary information so we can directly use the formula:
[tex]H_{in}=9.14kJ+30.2kJ[/tex]
[tex]H_{in}=39.34 kJ[/tex]
b) In order to find the efficiency, we can use the following formula:
[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]
so we get:
[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]
Efficiency=76.77%
Compute the specific heat capacity at constant volume of nitrogen (N2) gas. The molar mass of N2 is 28.0 g/mol.
Answer:
724.3J/Kg.K
Explanation:
CHECK THE COMPLETE QUESTION BELOW
Compute the specific heat capacity at constant volume of nitrogen (N2) gas.and compare with specific heat of liquid water. The molar mass of N2 is 28.0 g/mol.
The specific heat capacity can be computed by using expression below
c= CV/M
Where c= specific heat capacity
M= molar mass
CV= molar hear capacity
Nitrogen is a diatomic element, the Cv can be related to gas constant with 5/2R
Where R= 8.314J/mol.k
Molar mass= 28 ×10^-3Kg/mol
If we substitute to the expression, we have
c= (5R/2)/(M)
=5R/2 × 1/M
=(5×8.314) /(2×28 ×10^-3)
=724.3J/Kg.K
Hence, the specific heat capacity at constant volume of nitrogen (N2) gas is
724.3J/Kg.K
The specific heat of liquid water is about 4182 J/(K kg) which is among substance with high specific heat, therefore specific heat of Nitrogen gas is 724.3J/Kg.K which is low compare to that of liquid water.
A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?
Answer:
Fx = 467.65N
Fy = 270N
Explanation:
Given
Force = 540N
angle of inclination = 40 degree
Horizontal component Fx = Fcos 30
Fx = 540cos30
Fx = 540(0.8660)
Fx = 467.65N
Hence the horizontal component is 467.65N
Vertical component Fy = Fsin 30
Fy = 540sin30
Fy = 540(0.5)
Fy = 270N
Hence the vertical component is 270N
A statement of the second law of thermodynamics is that:__________.
a) spontaneous reactions are always exothermic.
b) energy is conserved in a chemical reaction that has a decrease in entropy.
c) spontaneous reactions are always endothermic.
d) in a spontaneous process, the entropy of the universe increases.
Answer:
in a spontaneous process, the entropy of the universe increases.
Explanation:
Entropy is a measure of of the degree of randomness or disorderliness in a system.
The second law of thermodynamics can be stated as follows; "in any spontaneous process, the entropy of the universe increases."
The universe here refers to the system's disorder and the disorder of the surroundings. Therefore, a spontaneous process can occur, in which the entropy of the system decreases, only if the entropy increases in the surroundings.
For instance, when ice freezes, the entropy of liquid water decreases, that is, the entropy of the system decreases. However, heat is given off to the surroundings and the entropy of the surroundings increases. This is an obvious expression of this law.
What would its weight be on Jupiter?
24.9N
Answer:
1.898 × 10^27 kg
Explanation:
thats how much it ways
Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?
Answer:
The value is [tex]p = 0.7556 c[/tex]
Explanation:
From the question we are told that
The speed at which galaxy B moves away from galaxy A is [tex]v = 0.577c[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
The speed at which galaxy C moves away from galaxy B is [tex]u = 0.731 c[/tex]
Generally from the equation of relative speed we have that
[tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]
Here p is the velocity at which galaxy C recede from galaxy A so
[tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]
=> [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]
=> [tex]0.731c - 0.4218 p = p - 0.577c[/tex]
=> [tex]0.731c + 0.577c = p + 0.4218 p[/tex]
=> [tex]1.308 c = 1.731 p[/tex]
=> [tex]p = 0.7556 c[/tex]
Which of the following is true?
A
The Atlantic, Pacific, Indian, Arctic, and Southern Oceans are completely separate
from each other.
B
The ocean covers about half of the Earth's surface.
с
Scientists have studied most of the ocean, but a tiny bit remains unexplored.
D
Scientists know more about the moon than they do the ocean.
Answer:
options B,C,D are true
Explanation:
A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?
A 300N
B 202N
C 398N
D 98N
Answer:98
Explanation:hope this helps!
The fact that our preconceived ideas contribute to our ability to process new information best illustrates the importance of: the serial position effect. O repression iconic memory . semantic encoding . retroactive interference .
Answer:
It’s a
Explanation:
Don’t actually put that i needed the points mb
Which objects cannot be observed in detail without a microscope?
Answer:
partecls
Explanation:
because they are to small to see with plain eyes
What is the acceleration of the the object during the first 4 seconds?
Answer:
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
Explanation:
What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
As a bicycle is ridden west in a straight line with decreasing speed,the acceleration of the bicycle must be
Answer:
Decreasing
Hope this helps! :)
What is the distance between a 900 kg compact car and a 1600 kg pickup truck if the gravitational force between them is about 0.0001 N?
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance is 0.96m
Explanation:
Given
m1= 900kg
m2= 1600kg
Force F= 0.0001nN
G=6.67430*10^-11 Nm^2/kg^2
Required:
The distance r
Step two:
the formula for the force is given as
F = Gm1m2/r2
make r subject of the formula
[tex]r= \sqrt{\frac{Gm1m2}{F} }[/tex]
[tex]r= \sqrt{\frac{6.67430*10^-11*900*1600}{0.0001} }\\\\r= 0.00009610992/0.0001`}\\\\r= 0.96m[/tex]
Answer:
The distance between the compact car and pickup truck is 0.96048 m
Explanation:
The gravitational force is directly proportional to the product of the masses of the interacting object, it is also inversely proportional to the square of the distance between them. This is shown in equation 1;
[tex]F =G \frac{m_{1} X m_{2} }{d^{2} }[/tex]............ 1
Where F is the gravitational force = 0.0001 N
G is the gravitational constant = 6.673 x [tex]10^{-11} Nm^{2} kg^{-2}[/tex]
[tex]m_{1}[/tex] is the mass of the compact car = 900kg
[tex]m_{2}[/tex] is the mass of the pickup truck = 1600kg
d is the distance and its unknown ?
Let us make d the subject formula in equation 1
[tex]d = \sqrt{G\frac{m_{1} m_{2} }{F } }[/tex] .... 2
Substituting into equation 2 we have
[tex]d = \sqrt{\frac{6.673x10^{-11} x 900 x 1600}{0.0001N} }[/tex]
d = 0.96048m
Therefore the distance between the compact car and pickup truck is 0.96048 m
A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m from where it was launched.
-Find the hang time for the projectile.
-Find the initial speed of a projectile.
-What are the x and y components of the projectile’s velocity the moment before it strikes the ground?
-At what speed will the projectile strike the ground?
Answer:
a)t = 1,43 s
b) V = 10,49 m/s
c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s
d) Vf = 17,5 m/s
Explanation:
According to the problem statement
V₀ = V₀ₓ and V₀y = 0
And at the end of the movement t = ? the distance y = 10 m
Therefore as
h = V₀y - (1/2)*g*t²
Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²
10 = 4,9*t²
t² = 10/4,9 ⇒ t² = 2,04 s
t = 1,43 s
a) 1,43 s is the time of movement
b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )
Just before touching the ground, the horizontal distance is
hd = 15 = Vₓ * t
Then 15 /1,43 = Vₓ = V₀ₓ
Vₓ = 10,49 m/s
Then initial speed is V = 10,49 m/s since V₀y = 0
Vf² = Vₓ² + Vy²
Vyf = V₀y - g*t
Vyf = 0 - 9,8 *1,43
Vyf = - 14,01 m/s
And finally the speed when the projectile strike the ground is:
Vf² = Vₓ² + Vy²
Vf = √ (10,49)² + (14,01)²
Vf = 17,50 m/s
Which is the largest gas that occurs in our atmosphere?
Helium
Nitrogen
Other Gases
Oxygen
Answer:
OXYGEN
Explanation:brainlyist me
Answer:
Nitrogen
Explanation:
Oxygen is second
What is the initial vertical velocity of the ball?
A.
0 m/s
B.
9.81 m/s
C.
20.0 m/s
D.
60.0 m/s
A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m. The force of friction between the refrigerator and the floor is 230 N. How much work has been performed by the mover on the refrigerator?
Given :
A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.
The force of friction between the refrigerator and the floor is 230 N.
To Find :
How much work has been performed by the mover on the refrigerator.
Solution :
Since, refrigerator is moving with constant velocity.
So, force applied by the mover is also 230 N ( equal to force of friction ).
Now, work done in order to move the refrigerator is :
[tex]W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m[/tex]
Hence, this is the required solution.
convert 100 Newton into dyne
Answer:10000000
Explanation:
An element or compound used to enhance a semiconductor is called a(n) ____.
The element named boron can be used to enhance the properties of semiconductors.
What is a semiconductor?A semiconductor is a material that has electronic properties and has the value that falls in between a conductor. It can be a metallic copper or an insulator.
The rise in temperatures leads to a fall in resistivity. The element named boron can improve the electrical properties of the semiconductor as they form the impurities.
Find out more information about the element.
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