(What is the weight of a 50 kg woman at the top of Jupiter's
atmosphere, where g = 24.8 N/kg? Give your answer in both
newtons and pounds.
Answer:
The correct answer is - 1240 newtons; 278.763 lbs.
Explanation:
a rugby player passes the ball 5.34 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 7.7 m/s, assuming that the smaller of the two possible angles was used
Answer:
[tex]31.035^{\circ}[/tex]
Explanation:
x = Displacement in x direction = 5.34 m
t = Time taken to travel the displacement
y = Displacement in y direction = 0
u = Initial velocity of ball = 7.7 m/s
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Displacement in x direction is given by
[tex]x=u\cos\theta t\\\Rightarrow t=\dfrac{5.34}{7.7 \cos\theta}[/tex]
Displacement in y direction is given by
[tex]y=u\sin\theta t-\dfrac{1}{2}gt^2\\\Rightarrow 0=7.7\sin\theta \dfrac{5.34}{7.7\cos\theta}-\dfrac{1}{2}\times 9.81 (\dfrac{5.34}{7.7\cos\theta})^2\\\Rightarrow 0=7.7\sin\theta-4.905\times \dfrac{5.34}{7.7\cos\theta}\\\Rightarrow 0=7.7^2\sin\theta \cos\theta-4.905\times 5.34\\\Rightarrow 0=7.7^2\dfrac{\sin2\theta}{2}-4.905\times 5.34\\\Rightarrow 0=7.7^2\sin2\theta-4.905\times5.34\times 2\\\Rightarrow \sin2\theta=\dfrac{4.905\times 5.34\times 2}{7.7^2}\\\Rightarrow 2\theta=\sin^{-1}\dfrac{4.905\times 5.34\times 2}{7.7^2}[/tex]
[tex]\Rightarrow \theta=\dfrac{62.07}{2}\\\Rightarrow \theta=31.035^{\circ}[/tex]
The angle at which the ball was thrown is [tex]31.035^{\circ}[/tex].
A ray of light strikes a flat mirror at an angle of 40° from the perpendicular. The light is reflected at
what opposite angle to the same perpendicular line?
a) 40°
b) 50°
c) 30°
d) 60°
Answer:
a
Explanation:
angle of incidence equals angle of reflection on flat surface
Answer:
40 degress
Explanation:
A turtle crawls at 4.32 m/s to cover the short 3.84 m distance to his food bowl. How long does it take?
Question 1 of 10
Which object has the most gravitational potential energy?
A. A 8 kg book at a height of 2 m
B. An 8 kg book at a height of 3 m
C. An 5 kg book at a height of 3 m
D. A5 kg book at a height of 2 m
SUBMIT
Answer:
B = An 8 kg book at a height of 3 m
Which task would be impossible without radio waves ?
Answer:
checking your e-mail on your computer using a coffee's shop WiFi. No need radio waves for this.
Explanation:
Because you are only checking it, nothing else.
g We have studied diffraction from a single slit, where light is sent through a thin opening. A similar phenomena occurs when light bends around a thin object, like a human hair. Here the width of the hair plays the role of the width of the single slit. Measurements found that when a beam of light of wavelength 632.8 nm was shone on a single strand of hair, the first dark fringe on either side of the central bright spot were 5.22 cm apart. If the screen is 1.25 meters away, how thick was this strand of hair?
Answer:
[tex]3.031\times 10^{-5}\ \text{m}[/tex]
Explanation:
[tex]y[/tex] = Distance between central maxima and first minimum
m = Order = 1
d = Thickness of hair
[tex]\lambda[/tex] = Wavelength = 632.8 nm
L = Distance between light source and screen = 1.25 m
Width of central maximum is given by
[tex]2y=5.22\times 10^{-2}\\\Rightarrow y=\dfrac{5.22\times 10^{-2}}{2}\\\Rightarrow y=0.0261\ \text{m}[/tex]
Distance between central maxima and first minimum is given by
[tex]y=L\tan\theta_{min}\\\Rightarrow \tan\theta_{min}=\dfrac{y}{L}\\\Rightarrow \tan\theta_{min}=\dfrac{0.0261}{1.25}\\\Rightarrow \theta_{min}=\tan^{-1}0.02088\\\Rightarrow \theta_{min}=1.1962^{\circ}[/tex]
Since [tex]\theta[/tex] is small [tex]\tan\theta_{min}=\sin\theta_{min}[/tex]
[tex]\sin\theta_{min}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda}{\sin\theta}\\\Rightarrow d=\dfrac{1\times 632.8\times 10^{-9}}{\sin1.1962^{\circ}}\\\Rightarrow d=3.031\times 10^{-5}\ \text{m}[/tex]
The strand of hair is [tex]3.031\times 10^{-5}\ \text{m}[/tex] thick.
An enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 190 m from where he is standing. You are 140 km from the tree. How long does it take for the sound to reach Clark
Answer:
407.61 seconds
Explanation:
Given that an enormous thunderstorm covers Dallas-Ft. Worth. Your best friend Clark is a storm chaser and heads to the center of the storm to take some readings while you stay dry at home. While Clark is at the center of the storm, he sees and hears lightning strike a tree that is 190 m from where he is standing. You are 140 km from the tree. How long does it take for the sound to reach Clark
Solution.
The distance between the Clark and you will be:
Distance = 140km - 190m
Distance = 140000 - 190
Distance = 139810 m
The radius of the earth is also considered but it will surely be insignificant as you subtracted the two distance values.
Using the formula
Speed = distance/time
Where
The speed of sound = 343 m/s
343 = 139810 / t
Make time the subject of formula
t = 139810 / 343
t = 407.61 s
Therefore, it will take 407.61 seconds for the sound to reach Clark
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 100 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg? SOLUTION (a) Let m(t) be the mass of radium-226 (in milligrams) that remains after t years. Then dm/dt = km and m(0) = 100, so this theorem gives m(t) = m(0)ekt = ekt. In order to determine the value of k, we use the fact that m(1590) = 1 2 . Thus e1590k = so e1590k = and 1590k = ln 1 2 = − ln(2) k = . Therefore m(t) = . We could use the fact that eln(2) = 2 to write the expression for m(t) in the alternative form m(t) = . (b) The mass after 1000 years is as follows. (Round your final answer to the nearest milligram.) m(1000) = ≈ mg
Answer:
See explanation
Explanation:
a) Formula for the mass of the sample that remains after t years = N= Noe^-kt
Where;
N = mass at time t years
No = mass at time t= 0
k = decay constant
t = time taken
So,
N = 100e^-kt
b) First,
t1/2 = -ln(1/2)/k
t1/2 = 0.693/k
t1/2 = half life of radium-226 =1590 years
1590 = 0.693/k
k = 0.693/1590
k = 4.36 * 10^-4
So,
N= 100e^-(4.36 * 10^-4 * 1000)
N= 65 mg
c) From
N = 100e^-kt
40 = 100e^-(4.36 * 10^-4t)
40/100 = e^-(4.36 * 10^-4t)
0.4 = e^-(4.36 * 10^-4t)
ln(0.4) = ln(e^-(4.36 * 10^-4t))
-0.9163 = -4.36 * 10^-4t
t = 0.9163/4.36 * 10^-4
t = 2102 years
What did you learn about from the NASA's perseverance Rover land on Mars project?
NASA's Mars Pathfinder mission, launched in 1996, put the first free-moving rover—called Sojourner—on the planet. ... Together, these missions have shown scientists that Mars is an active planet that is rich in the ingredients needed for life as we know it—water, organic carbon, and an energy source
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slide forward, crushing the cab, if the truck stops suddenly in an accident or even in braking. Assume, for example, that a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume that the load is not tied down to the truck, but has a coefficient of friction of 0.500 with the flatbed of the truck.
A) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck.
B) Is any piece of data unnecessary for the solution?
a) mass of the load.
b) mass of the truck.
c) velocity.
d) coefficient of static friction.
e) all are necessary.
Answer:
A)
the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B)
data that were not necessary to the solution are;
a) mass of truck and b) mass of load
Explanation:
Given that;
mass of load [tex]m_{LS}[/tex] = 10000 kg
mass of flat bed [tex]m_{FB}[/tex] = 20000 kg
initial speed of truck [tex]v_{0}[/tex] = 12 m/s
coefficient of friction between the load sits and flat bed μs = 0.5
A) the minimum stopping distance for which the load will not slide forward relative to the truck.
Now, using the expression
Fs,max = μs [tex]F_{N}[/tex] -------------let this be equation 1
where [tex]F_{N}[/tex] = normal force = mg
so
Fs,max = μs mg
ma[tex]_{max}[/tex] = μs mg
divide through by mass
a[tex]_{max}[/tex] = μs g ---------- let this be equation 2
in equation 2, we substitute in our values
a[tex]_{max}[/tex] = 0.5 × 9.8 m/s²
a[tex]_{max}[/tex] = 4.9 m/s²
now, from the third equation of motion
v² = u² + 2as
[tex]v_{f}[/tex]² = [tex]v_{0}[/tex]² + 2aΔx
where [tex]v_{f}[/tex] is final velocity ( 0 m/s )
a is acceleration( - 4.9 m/s² )
so we substitute
(0)² = (12 m/s)² + 2(- 4.9 m/s² )Δx
0 = 144 m²/s² - 9.8 m/s²Δx
9.8 m/s²Δx = 144 m²/s²
Δx = 144 m²/s² / 9.8 m/s²
Δx = 14 m
Therefore, the minimum stopping distance for which the load will not slide forward relative to the truck is 14 m
B) data that were not necessary to the solution are;
a) mass of truck and b) mass of load
A 10 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true of the motion of the combined objects immediately after the collision?
Answer:
"Cannot be determined," if that's an answer choice. It depends on the velocities of both objects, since momentum=mass*velocity.
Explanation:
Using the American Engineering system of units (AES), a) Calculate the weight of a 170.5 lbm person on the surface of the earth, where the local acceleration due to gravity is 32.174 ft/s2 . Report your answer in pound-force, lbf. b) What would be the weight of a 170.5 lbm astronaut on the moon, where the local acceleration due to gravity is 5.32 ft/s2 . Report your answer in pound-force, lbf.
Answer:
a) the weight of the person is 170.5 lbf
b) weight of the astronaut on the moon is 28.2 lbf
Explanation:
Given the data in the question;
a)
we know that;
weight on the surface of the earth = m[tex]g_{earth[/tex]
given that m = 170.5 lbm and g = 32.174 ft/s²
we substitute
weight on the surface of the earth = 170.5 lbm × 32.174 ft/s²
= 5485.667 lbm-ft/s²
1 lbf = 32.174 lbm-ft/s²
so
weight on the surface of the earth = (5485.667 / 32.174) lbf
weight on the surface of the earth = 170.5 lbf
Therefore, the weight of the person is 170.5 lbf
b)
given that;
weight on the surface of the earth = m[tex]g_{moon[/tex]
m = 170.5 lbm and g = 5.32 ft/s²
weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²
= 907.06 lbm-ft/s²
1 lbf = 32.174 lbm.ft/s²
weight on the surface of the earth = ( 907.06 / 32.174 ) lbf
weight on the surface of the earth = 28.2 lbf
Therefore, weight of the astronaut on the moon is 28.2 lbf
Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to
A. Paul Costa
B. Hans Eysenck
C. Gordon Allport
D. Robert McCrae
The answer is b I just took the test
Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to Hans Eysenck.
What is Hans Eysenck's theory?Eysenck's theory of personality is based on three logical attributes namely:
a) Introversion vs. extroversion – Extroversion leads to sociable life , while introversion causes need of being alone and limited interactions
b) Neuroticism vs. stability – Neuroticism leads to anxiousness and an overactive sympathetic nervous system while stability leads to emotional stability.
c) Nsychoticism vs. socialization. - psychoticism leads to independent thinking, and hostility. While socialization leads to co-operative and conventional behaviour.
Therefore Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to Hans Eysenck.
To know more about Hans Eysenck's theory follow
https://brainly.com/question/11912668
#SPJ2
[O.05]Which of these Earth spheres interact with oceans during beach erosion?
geosphere
hydrosphere
cryosphere
atmosphere
Answer:
hydrosphere of these Earth spheres interact with oceans during beach erosion.
Answer:
Geosphere
Explanation:
The geosphere is the solid parts of the earth including rocks, minerals, and sand on the beach. The shoreline is where the geosphere meets the ocean. This is how they interact during erosion.
It is not the hydrosphere because the ocean is a part of the hydrosphere.
A 20g bullet moving at 200m/s hits a bag of sand and comes to rest in 0.011s, calculate the momentum of the bullet just before hitting the bag
Answer:
momentum = mass * velocity ... kg•m/s
acceleration = v / t = 200 m/s / 11 ms ... m/s^2
force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N
Explanation:
What happens when a moving object is moving a particular direction experiences a net force opposite the direction?
Answer:
stop
Explanation:
object stop
Answer:
If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement. Two or more opposite forces are balanced forces if their effects cancel each other and they do not cause a change in an object's motion.
How does the amplitude of the wave change as you get farther from the speaker?
Answer:
The amplitude decreases
Explanation:
From the inverse square law, we know that;
I ∝ 1/d²
Where;
I is intensity
d is distance
Also, we know that;
A² ∝ I
Where A is amplitude
Thus, we can say that;
A ∝ 1/d²
Thus, amplitude is inversely proportional to the distance.
So the larger the distance, the lesser the amplitude and the lesser the distance, the higher the amplitude.
Thus, as you move farther from the speaker the distance increases and therefore the amplitude decreases
Why might video games increase creativity while the use of cell phones, the internet, or computers do not?
Explanation:
Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.
Which one of the following statements concerning the buoyant force on an object submerged in a liquid is true?
A) The buoyant force depends on the mass of the object.
B) The buoyant force depends on the weight of the objeet.
C) The buoyant force is independent of the density of the liquid.
D) The buoyant force depends on the volume of the liquid displaced.
E) The buoyant force will increase with depth if the liquid is incompressible.
Answer:
D) The buoyant force depends on the volume of the liquid displaced.
Explanation:
Buoyancy can be defined as an upward force which is created by the water displaced by an object.
According to Archimede's principle, it is directly proportional to the amount (weight) of water that is being displaced by an object.
Thus, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.
The buoyancy of an object is given by the formula;
[tex] Fb = pgV [/tex]
[tex] But, \; V = Ah [/tex]
[tex] Hence, \; Fb = pgAh [/tex]
Where;
Fb = buoyant force of a liquid acting on an object.g = acceleration due to gravity. p = density of the liquid.v = volume of the liquid displaced. h = height of liquid (water) displaced by an object. A = surface area of the floating object.The unit of measurement for buoyancy is Newton (N).
Additionally, the density of a fluid is directly proportional to the buoyant force acting on it i.e as the density of a liquid decreases, buoyancy decreases and vice-versa.
Hence, the statement concerning the buoyant force on an object submerged in a liquid which is true is that the buoyant force depends on the volume of the liquid displaced.
If an object is completely submerged, the volume of the object is equal to the volume of liquid displaced.
what is the mass of a ball rolling at 4.5m/s if it's momentum is 3.5 kg•m/s?
Answer:
The ball has a mass of 7/9 kg
Explanation:
Momentum is just mass times velocity, so to find its mass, we can simply divide the given momentum by the given velocity.
[tex]3.5 kg\frac{m}{s} \div 4.5\frac{m}{s} = \frac{7}{9}kg[/tex]
What is the name of the theory describing how the lithosphere is broken into segments, or plates, which "float" on the asthenosphere, and associates the interactions between these plates with earthquakes and volcanic activity and the formation of mid-ocean ridges, trenches, mountains, and chains of volcanic islands?
A. Conservation of Matter
B. Big Bang Theory
C. Plate Tectonics
D. Theory of Evolution
Answer:
C. Plate Tectonics
Explanation:
The theory of plate tectonics is when the lithosphere is separated into plates. These plates move over or float over the asthenosphere. The movement of these plates cause earthquakes and can interact with the volcanic activity.
A ray of light is incident on a plane mirror at an angle of 30 degree state the angle of reflection?
Answer:
30°
Explanation:
The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.
∠i = ∠r
so as ∠i = 30°
∠r will also be 30°
cosmic Microwave Background radiation is one piece of supporting evidence for the Big Band Theory. Are these microwaves at the longer or shorter end of the spectrum?
are they long or short?
Answer:
it is long.
Explanation:
I know the answer but I don't know how to explain it
An open pipe is 1.42 m long
What fundamental frequency
does it produce?
(Speed of sound = 343 m/s)
(Unit = Hz)
Answer:
the fundamental frequency produced by the open pipe is 120.78 Hz
Explanation:
Given;
length of the open pipe, L = 1.42 m
speed of sound in air, v = 343 m/s
The length of the open pipe for the fundamental frequency is equivalent to half of wavelength;
[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]
The fundamental frequency produced by the open pipe is calculated as;
[tex]f_o = \frac{v}{\lambda} \\\\f_o = \frac{v}{2L} \\\\f_o = \frac{343}{2 \times 1.42} \\\\f_o = 120.78 \ Hz[/tex]
Therefore, the fundamental frequency produced by the open pipe is 120.78 Hz
Explain what happens to the movement, energy and arrangement of particles during the process
of distillation.
Answer:
Distillation separates a liquid from a solution. For example, water can be separated from salty water by simple distillation. This method works because the water evaporates from the solution, but is then cooled and condensed into a separate container. The salt does not evaporate and so it stays behind.
To determine the coefficient of static friction between two materials, an engineer places a small sample of one material on a horizontal disk whose surface is made of the other material and then rotates the disk from rest with a constant angular acceleration of 0.4 rad/s2. If she determines that the small sample slips on the disk after 9.903 s, what is the coefficient of friction
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the coefficient of friction is 0.32
Explanation:
Given the data in the question;
we make use of kinematic equation of motion;
ω = ω₀ + ∝t
we substitute
ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )
ω = 3.9612 rad/s
The centripetal force acting on the sample is;
Fc = mrω²
from the image; r = 200 mm = 0.2 m
so we substitute
Fc = m(0.2 m ) ( 3.9612 rad/s )²
Fc = (3.13822 m/s²)m
we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;
f = Fc
μN = Fc
μmg = (3.13822 m/s²)m
μ = (3.13822 m/s²)m / mg
μ = (3.13822 m/s²) / g
acceleration due to gravity g = 9.8 m/s²
so
μ = (3.13822 m/s²) / 9.8 m/s²
μ = 0.32
Therefore, the coefficient of friction is 0.32
Can someone plz explain :
Two cars have the same mass, the first car is moving towards the east and the second is
stationary. If the two cars merge together after a collision and head towards the east their speed
after the collision is equal to.......
A)1/4vi b)1/2vi c)vi d) 2vi
Answer: Option b, the final velocity is half of the initial velocity.
Explanation:
Here we will use the conservation of the total momentum of a system.
This means that the total momentum at the beginning must be the same as the final momentum.
Where momentum is:
P = M*v
Initially, we have two cars, both with the same mass M, and only one of them has a velocity v.
Then the initial momentum is:
P = M*v + M*0 = M*v
After the collision, the two cars move together. Then the total mass that is moving is equal to the sum of the masses of the cars, this is 2*M
and we can suppose that the two cars move at a final velocity v'
Then the final momentum is:
P' = (2*M)*v'
Now we use the conservation of momentum, then:
P = P'
M*v = (2*M)*v'
Now we need to solve this for v'
(M*v)/(2*M) = v'
v/2 = v'
This means that the final velocity is half of the initial velocity.
Then the correct option is option b.
A 7.80 g bullet has a speed of 620 m/s when it hits a target, causing the target to move 6.30 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force (in N) that stops the bullet. (Enter the magnitude.) N (b) Assuming the force is constant, determine how much time elapses (in s) between the moment the bullet strikes the target and the moment it stops moving. s
Answer:
[tex]23796.19\ \text{N}[/tex]
[tex]0.0002032\ \text{s}[/tex]
Explanation:
F = Force
s = Displacement = 6.3 cm
m = Mass of bullet = 7.8 g
v = Velocity of bullet = 620 m/s
t = Time taken
Work done is given by
[tex]W=Fs[/tex]
Kinetic energy is given by
[tex]K=\dfrac{1}{2}mv^2[/tex]
Using work energy considerations we get
[tex]Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}[/tex]
The average force that stops the bullet is [tex]23796.19\ \text{N}[/tex].
Force is given by
[tex]F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}[/tex]
The time taken to stop the bullet is [tex]0.0002032\ \text{s}[/tex].
As a sports car travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the sports car by the air and the road. If the power developed by the engine is 0.824 hp, calculate the total friction force acting on the sports car (in N) when it is moving at a speed of 30 m/s. One horsepower equals 746 W.
Answer:
20.49 N
Explanation:
From the question given above, the following data were obtained:
Power (P) = 0.824 hp
Velocity (v) = 30 m/s
1 h = 746 W
Force (F) =?
Next, we shall convert 0.824 hp to Watts (W). This can be obtained as follow:
1 h = 746 W
Therefore,
0.824 hp = 0.824 hp × 746 W / 1 h
0.824 hp = 614.704 W
Finally, we shall determine the force as follow:
Power (P) = 614.704 W
Velocity (v) = 30 m/s
Force (F) =?
P = F × v
614.704 = F × 30
Divide both side by 30
F = 614.704 / 30
F = 20.49 N