Answer: no because you have left the number
Explanation:
A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 kg of exhaust at a speed of 230 m/s relative to the space probe. What is the final velocity of the probe?
Answer:
179.47m/s
Explanation:
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity
Substitute
7750(179)+72(230) = (7750+72)v
1,387,250+16560 = 7822v
1,403,810 = 7822v
v = 1,403,810/7822
v= 179.47m/s
Hence the final velocity of the probe is 179.47m/s
Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?
1.When they are halfway to the bottom
2.When they reach the bottom of the fall
Answer:
2.When they reach the bottom of the fall
Explanation:
The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).
Thus, the correct option is "2" When they reach the bottom of the fall
(c) The plates are moved farther apart with each plate maintaining the same net charge. In a coherent paragraph-length response, apply concepts of work and energy to explain how the electric potential difference between the plates changes, if at all, when the plates are moved farther apart.
Answer:
Energy is applied on the charge to do work.
Explanation:
Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..
A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s. The frictional force between the block and the floor is 10 N. The net work done on the block in 1 minutes is most nearly:
Answer:
5× 3 over 10 = 1.5
Explanation:
F = 1m/s × kg over n
The net work done on the block in 1 minutes is most nearly 1800 Joule.
What is work?
A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction. The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item.
Work has no direction and only magnitude. Work is a scalar quantity as a result.
Given that: A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s.
As the block is moving with uniform velocity; this force is equal to the frictional force in magnitude.
So, The net work done on the block in 1 minutes is =
force× displacement
= force × velocity × time
= 10 × 3 × 60 Joule
= 1800 Joule.
Learn more about work here:
https://brainly.com/question/18094932
#SPJ2
A jet pilot puts an aircraft with a constant speed into a vertical loop. (a) Which is greater, the normal force exerted on the seat by the pilot at the bottom of the loop or that at the top of the loop
Answer:
A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.
Explanation:
Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).
When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:
?F(c) = ma(c) = n - mg
n = mg + ma(c)
= m[g + a(c)]
Since centripetal acceleration equals v² / r, the equalization enhances:
n = m[g + (v² / r)]
PLEASE HELP!!! Which process in living things evolved as oxygen levels increased in Earth’s atmosphere?
glycolysis
photosynthesis
fermentation
aerobic respiration
Answer: Areobic Resporation
Explanation; -Aerobic respiration is the process in living things that evolved as oxygen levels increased in earth's atmosphere.
1 What is the resistance of the inductive coil
takes 5A current across 240V, 50Hz supply
at 0.8 power factor?
A 48
B 4250
C 38.40
D 26.60
Answer:
38.40
Explanation:
This is the answer for this question
A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.
Answer:
The value is [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
Explanation:
From the question we are told that
The force is [tex]F = 22.7 \ N[/tex]
The value of room temperature is [tex]T = 298 \ K[/tex]
Generally the rate at which its entropy changes as it stretches is mathematically represented as
[tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]
=> [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]
=> [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
why did the following change occur? the snow that covered the ground all winter began to melt, temperatures rose, and flowers started to bloom
A.) Days became shorter.
B.) The Earth tilted away from the sun
C.) The sun's rats became for concentrated.
D.) Earth completed one revolution
Answer:C
Explanation:
Answer:
B
Explanation:
Because the earth shifts axes points so it would make it cllser hotter etc
Two spherical balls are placed so that their centers are 3.61 m apart. The
force between them is 1.65 x 10-7 N. If the mass of the smaller ball is 81 kg,
what is the mass in kilograms of the other ball?
Answer:
The mass of the other ball is 397.775 kg.
Explanation:
Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.
The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.
The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.
In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:
[tex]F=G\frac{M1*M2}{d^{2} }[/tex]
where:
F = It is the module of the force exerted between both bodies, and its direction is found on the axis that joins both bodies. G = It is the constant of Universal Gravitation, whose value is 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]In this case:
F= 1.65*10⁻⁷ NG= 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]M1= 81 kgM2= ?d= 3.61 mReplacing:
[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]
Solving for M2:
[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]
M2= 397.775 kg
The mass of the other ball is 397.775 kg.
The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 310 mW/m2. The maximum value of the magnetic field in the wave is closest to:
Answer:
5.096*10^-8
Explanation:
Given that
The average value of the electromagnetic wave is 310 mW/m²
To find the maximum value of the magnetic field the wave is closest to, we say
Emax = √Erms
Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]
Emax = √233.7648
Emax = 15.289
Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer
15.289 / (3*10^8) = 5.096*10^-8 T
Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8
______ is the total distance traveled divided by the total time of travel.
(5 Points)
speed
total speed
average speed
displacement
which law of thermodynamics would be violated if heat were to spontaneously flow between two abject which are in thermal equilibrium?
Answer:
The law zero of thermodynamics.
Explanation:
The law zero of thermodynamics, which tells us that heat flows from a body at a higher temperature to another body with lower temperature, when the heat transfer is zero, it is said that the two bodies are in thermal equilibrium, their temperatures are equal
a 60N at an angle of 30°from
horizontal
Explanation:
Force, F = 60 N
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find horizontal and vertical component of the force.
Horizontal component,
[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]
Vertical component,
[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]
So, the horizontal and vertical component are 51.96 N and 30 N respectively.
Which list ranks the three types of radiation from lowest penetrating power to
highest penetrating power?
A. Gamma < alpha < beta
B. Beta < alpha < gamma
C. Alpha < beta < gamma
D. Gamma < beta < alpha
Answer:gamma>beta>alpha
Explanation: Just took the test
g A proton and an alpha particle are released from rest at different locations from the negative plate of a charged parallel plate capacitor. The plates are 15 mm apart, and they are charged to a potential difference of 680 V. The alpha particle is placed at the point where the potential is 600 V. Where do you need to place the proton so that both particles reach the negative plate with the same speed
Answer:
d₂ = 6.18 mm
Explanation:
We can work on this interesting exercise using the concepts of energy conservation
alpha (1) particle, with have a charge q = 2e
Starting point
Em₀ = U = q V₁
final point
[tex]Em_{f}[/tex] = ½ m v²
Em₀ = Em_{f}
2eV₁ = ½ m₁ v²
v² =[tex]\frac{4 \ e \ V_{1}}{m_{1} }[/tex]
proton (2) particle, with have a charge q = e
Starting point
Em₀ = qV₂
Final point
Em_{f} = ½ m₂ v²
Em₀ = Em_{f}
eV₂ = ½ m₂ v²
v² = [tex]\frac{2 \ e \ V_{2} }{m_{2} }[/tex]
in the exercise, it should be noted that the two particles have the same velocity when reaching the plate, therefore let us solve the velocity in each equation and equal
\frac{4 \ e \ V_{1}}{m_{1} }= \frac{2 \ e \ V_{2} }{m_{2} }
[tex]V_{2} = \frac{2 m_{2} }{m_{1} } \ V_{1}[/tex]
the alpha particle is composed of two protons and two neutrons, therefore in first approximation
m₁ = 4 m₂
subtitute
V₂ =[tex]\frac{1}{2}[/tex] V₁
let's calculate
V₂ = [tex]\frac{1}{2}[/tex] 600
V₂ = 300 V
To find the distance we use the relationship between the electric field and the potential difference
V = -d E
The electric field between the plates is constant, so
E = - V / d = - V₂ / d₂
d₂ = [tex]\frac{V_{2} }{V} \ d[/tex]
let's calculate
d₂ = [tex]\frac{300}{680} \ 15[/tex]
d₂ = 6.18 mm
A 2.00 kg copper pot is placed on a glass shelf and then one side of the shelf falls so that it becomes angled at 35.0° to the horizontal. The coefficient of static friction between copper and glass is 0.680 and the coefficient of kinetic friction is 0.530. Will the pot slide, and if so, what is its acceleration?
Answer:
Explanation: about 2 Mph when it falls and sorry if I’m wrong
The intensity of sound is measured on the decibel scale, dB. The equation dB=10 log I represents the decibel level, where I is the ratio of the sound to the human hearing threshold. A noise is 150,000 times greater than the human hearing threshold. Which shows a valid step in the process of finding the decibel level of the noise?
a. 150,000 = 10 log I
b. 15,000 = log I
c. dB = 10 log 150,000
d. 10dB = log 150,000
e. 10/dB= log 150,000
Answer:
The correct option is c
Explanation:
From the question we are told that
The ratio of the noise to human hearing threshold is [tex]I = 150 000[/tex]
Generally from the equation given we have that
dB = 10 log I
So
dB = 10 log 150000
The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct
Given the equation for calculating the intensity of sound which is measured in decibel expressed as:
dB=10 log I
where;
I is the ratio of the sound to the human hearing threshold
Given that noise is 150,000 times greater than the human hearing threshold.
Substitute I = 150,000 into the expression above;
dB = 10 log 150,000
Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000
Learn more here: https://brainly.com/question/14468501
A harmonic oscillator starts with an amplitude of 20.0 cm. After 10.0 s, the amplitude decreases to 15.0 cm. If the linear damping coefficient is 2.00 Ns/m, how much mass is oscillating
Answer: the amount of mass is oscillating is 34.8 kg
Explanation:
Given that;
amplitude A = 20.0 cm
time t = 10 s
amplitude decreases x = 15.0 cm
damping coefficient b = 2.00 N.s/m
amount of mass is oscillating = ?
we know that; amplitude can be expressed as;
x = Ae^-(∝t)
we substitute
15 = 20e^-∝(10)
∝ = 0.02877 s⁻¹
Hence mass m will be;
m = b/2∝
we substitute
m = (2 N.s/m) / ( 2 × 0.02877 s⁻¹)
m = 34.8 kg
Therefore the amount of mass is oscillating is 34.8 kg
A solid sphere made of plastic density of 1350 kg/m3 has a radius of 35.0 cm. It is suspended by a massless cord. 75% of its volume is in water and 25% of it is in oil. The density of water is 1000 kg/m3 and the density of the oil is 850 kg/m3. What is the total buoyant foce on the sphere
Answer:
B_total = 1694 N , T = - 937.7 N
Explanation:
This is an exercise of the Archimedes principle that establishes that the thrust of a liquid is equal to the weight of the dislodged volume
B = ρ g V_body
For this case we will assume that the weight of the body is in equilibrium with the thrust of the liquids and the tension of the rope
B₁ + B₂ - W + T = 0
suppose liquid 1 is water and liquid 2 is oil
ρ₁ g V₁ + ρ₂ g V₂ = W
for body weight let's use the definition of density
ρ = m / V
m = ρ V
we substitute
T = ρ g V - (ρ₁ g V₁ + ρ₂ g V₂)
In the problem we are told that the volume in the water is 75% and the volume in the oil is 25% of the body's volume.
T = ρ g V - g (ρ₁ 0.75 V + ρ₂ 0.25 V)
let's calculate the volumes
Body
V = 4/3 π r³
liquid 1 (water)
V₁ = 0.75 V = ¾ V
liquid 2 (oil)
V₂ = 0.25 V = ¼ V
we substitute
T = ρ g [tex]\frac{4}{3}[/tex] π r³ - g (ρ₁ [tex]\frac{3}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³ + ρ₂ [tex]\frac{1}{4}[/tex] [tex]\frac{4}{3}[/tex] π r³)
T = [tex]\frac{4}{3}[/tex] ρ g r³ - g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)
The term of the floating force is
B_total = g π r³ (ρ₁ + [tex]\frac{1}{3}[/tex] ρ₂)
let's calculate its value
B_total = 9.8 π 0.35³ (1000 + ⅓ 850)
B_total = 1694 N
therefore the tension in the string is
T = [tex]\frac{4}{3}[/tex] 1350 9.8 0.35³ - 1694
T = 756.3 - 1694
T = - 937.7 N
The negative sign indicates that the system tends to come out of the liquid, for which a downward force must be applied to keep it in position.
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Answer:
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Can we use a hydrometer to
measure the density of milk?
Answer:
yes
Explanation:
i hope this helps not sure im right
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]
[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]
The second distance, r₂, can be determined from sound intensity formula given as;
[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]
Therefore, the second distance of the sound from the source is 431.78 m.
A ball is thrown horizontally at 30 m/s from a height of 45m. How long is it in the air?
(i) How fast is it moving horizontally when it hits the ground?
Answer:
1.25s
1) 90.9m
Explanation:
From the question, we are given the following;
Speed v = 30m/s
Maximum height H = 45m
Required;
Time
Using the equation of motion S = ut + 1/2gt²
Substitute;
45= 30t + 1/2(9.8)t²
45 = 30t + 4.9t²
4.9t² + 30t - 45 = 0
Factorize;
t = -30 ±√30²-4(-45)(4.9)/2(4.9)
t = -30 ±√900+882/9.8
t = -30 ±42.21/9.8
t = -30 + 42.21/9.8
t = 12.21/9.8
t = 1.25secs
Hence the ball spent 1.25secs in air.
i) To get the horizontal distance, we will use the formula;
R = U√2H/g
R = 30√2(45)/9.8
R = 30√90/9.8
R = 30√9.18
R = 30(3.03)
R = 90.9m
Hence the horizontal distance is 90.9m
5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m/s??
Answer:
470.59 kgExplanation:
The the mass of the car can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{400}{0.85} \\ = 470.58823...[/tex]
We have the final answer as
470.59 kgHope this helps you
If vector A= aj^ and vector B = bj^, then vector A×B is equal to
Answer:
0 because j×j =0
Explanation:
so a×b is 0
4. A box of old textbooks is on the middle shelf in the bookroom 1.3 m from the floor. If the janitor relocates the box to the shelf that is 2.6 m from the floor, how much work does he do on the box. The box has a mass of 10.0 kg.
Answer:
127.53 J
Explanation:
Given that
Height of the floor, h = 1.3 m
Acceleration due to gravity, g = 9.81 m/s²
Mass of the box, m = 10 kg
To find the work done on the box by the janitor, we use the formula for potential energy. Remember that potential energy is actually the energy stored in an object as a result of its position.
P.E = mgh, substituting each parameters for its value, we then have
P.E = 10 * 9.81 * 1.3
P.E = 127.53 J
Therefore, we can conclude that the amount of work done on the box by the janitor is 127.53 J
What are earth crossing asteroids and how do we monitor them
Answer:
Earth-crossing asteroid, asteroid whose path around the Sun crosses Earth's orbit. Two groups of such asteroids—Aten and Apollo asteroids—are distinguished by the size of their orbits and how closely they approach the Sun.
Explanation:
Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT
Answer:
(A)
Explanation:
Answer:
275 x 10"N
Explanation:
when is a bandage best suited to be used?