Answer:
67 kg.
Explanation:
From the question,
Total momentum = (mass of the pilot+ mass of the helicopter)×velocity.
I = (m+M)v............................... Equation 1
Where I = Total momentum of the pilot and the helicopter, m = mass of the pilot, M = mass of the helicopter, v = velocity.
make m the subject of the equation
m = (I/v)-M.............................. Equation 2
Given: I = 7200 kg.m/s, v = 60 m/s, M = 53 kg.
Substitute these values into equation 2
m = (7200/60)-53
m = 120-53
m = 67 kg.
Hence the mass of the pilot is 67 kg
What is the magnitude of the force that is exerted on a 20 kg mass to give it an acceleration of 10.0
m/s2?
Answer:Mass of the body = 20 kg.
Final Velocity = 5.8 m/s.
Initial velocity = 0
Time = 3 seconds.
Using the Formula,
Acceleration = (v - u)/ t
= (5.8 - 0)/ 3
= 1.6 m /s².
Now, Using the Formula,
Force = mass × acceleration
= 20 × 1.6
=
Explanation: I REALLY HOPE THIS HELPS I'M KINDA NEW AT THIS :] :]
The magnitude of requires force, that is exerted on a 20 kg mass to give it an acceleration of 10.0 m/s^2 is 200 Newton.
What is force?The definition of force in physics is: The push or pull on a mass-containing item changes its velocity.
An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force.
Given parameters:
Mass of the object: m = 20 kg.
Acceleration of the object: a = 10.0 m/s^2.
Hence, according to Newton's 2nd law of motion:
the magnitude of requires force = mass ×acceleration
= 20 × 10 Newton
= 200 Newton.
Hence, the magnitude of requires force is 200 Newton.
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Please answer :>
40 POINTS
Answer:
rotates
Explanation:
I'm so bored
yrfgggghhgghhyuj
19. In a turtle race, a 20 kg turtle moves with a velocity of 0.1 m/s. What is the kinetic energy of the turtle?
A. 20 J
B. 0.2 J
C. 0.1 J
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
ASAP
If a cart is released with some initial
velocity, it will slowly come to a stop.
What happens to the energy of the cart's
motion?
If the cart is pulled along at a constant velocity, no acceleration, the energy of the carts motion is constant but the force pulling it adds energy continuously. Where does the extra energy go?
Answer:
A) Energy is dissipated into heat and sound energy due to Friction
B) The energy goes into heat and sound energy due to friction again, otherwise the cart would accelerate due to an unbalanced force. Therefore, we know there's friction, and the friction causes energy loss.
___________is
objects total kinetic and potential energy.
Answer:
Bowling ball
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.
(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 spring stretches 5 cm when a 300-N mass is suspended from it. Calculate the spring constant in N / m .
Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of movement of Object 1 or of Object 2?
Answer:
I think its object 1
Explanation:
Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.
when a constant force is applied to an object, the acceleration of the object varies inversely with its mass. When a certain constant force acts upon an object with a mass 12 kg, the acceleration of the object is 6 m/s. If the same force acts on another object whose mass is 9kg, what is the objects acceleration
Answer:
8 m/s²
Explanation:
From the question,
Since the same force act on both object,
F = ma = m'a'.............................. Equation 1
Where F = force action on the obeject, m = mass of the first object, a = acceleration of the first object, m' = mass of the second object, a' = acceleration of the second object.
make a' the subject of the equation
a' = ma/m'................... Equation 2
Given: m = 12 kg, a = 6 m/s², m' = 9 kg.
Substitute these values into equation 2
a' = 12(6)/9
a' = 8 m/s².
Hence the acceleration of the second object is 8 m/s²
Which of the following requires constant agonist-antagonist muscle contraction
Answer: Dynamic balance
Explanation: Dynamic balance movements are movements in which constant agonist-antagonist muscle contractions occur in order to maintain a certain position or posture. ISSA pg 121
Physics, calculating net force. Please work it out for me
Answer:
F = 2985.125 N
Explanation:
Given that,
The radius of curvature of the roller coaster, r = 8 m
Speed of Micheal, v = 17 m/s
Mass of body, m = 65 kg We need to find the net force acting on Micheal. Net force act the bottom of the circle is given by :
[tex]F=\dfrac{mv^2}{r}+mg\\\\F=m(\dfrac{v^2}{r}+g)\\\\=65(\dfrac{17^2}{8}+9.8)\\\\=2985.125\ N[/tex]
So, the net force is 2985.125 N.
In which scenario is the greatest amount of work done on a wagon?
A force of 55 N moves it 8 m.
A force of 60 N moves it 6 m.
A force of 50 N moves it 5 m.
A force of 40 N moves it 10 m.
Answer:
The first scenario!
Explanation:
W=F*d
a) 55*8= 440J
b) 60*6= 360J
c) 50*5= 250J
d) 40*10= 400J
The scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.
What is Work done?Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;
W = F × d
Where F is force applied and d is distance travelled.
From the question;
A force of 55 N moves it 8 m
W = 55N×8m = 440N.m
A force of 60 N moves it 6 m.
W = 60N×6m = 360N.m
A force of 50 N moves it 5 m.
W = 50N×5m = 250N
A force of 40 N moves it 10 m.
W = 40N×10m = 400N.m
Therefore, the scenario with the greatest amount of work done on a wagon is "A force of 55 N moves it 8 m" and the Work done is 440N.m.
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How does the current in a series circuit compare
Plsss I want answer???
Answer:
s=136.89/2g meter
s=6.98 meter (correct to 3 sig.fig. taking g=9.81ms^-2)
Explanation:
u= + 11.7 ms^-1
a= - g ms^-2
At highest point: v=0ms^-1
v^2=u^2+2as
0=11.7^2+2(-g)s
s=136.89/2g meters
During which sequence of phases does the moon's visibility increase
During the "waxing" phases, the two weeks immediately following the New Moon.
New Moon ==> waxing crescent ==> First Quarter ==> waxing gibbous ==> Full Moon
The phase of the moon decides the moon's visibility. The sequence of phases does the moon's visibility increases is; the waxing crescent, first quarter, waxing gibbous, full Moon.
What is the moon phase?The moon changes shape every day. This is due to the fact that the celestial body has no light of its own and can only reflect sunlight.
Only the side of the moon facing the sun can reflect this light and seem bright. The opposite side appears black. this is a full moon.
We can only see the black section when it lies between the sun and the earth when a new moon occurs. We witness intermediate phases like a half-moon and crescent in between these two extremes.
Following are the sequence of phases does the moon's visibility increase is;
1. Waxing crescent
2. First Quarter
3. Waxing gibbous
4. Full Moon
The phase of the moon decides the moon's visibility. Hence the visibility changes with the change in the phase of the moon.
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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.
A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.
Answer:
The acceleration is 2.448 meters per square second and is vertically upward.
Explanation:
The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:
[tex]\Sigma F = F - m\cdot g = m\cdot a[/tex] (1)
Where:
[tex]F[/tex] - Buoyant force, measured in newtons.
[tex]m[/tex] - Mass of the plastic ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]a[/tex] - Net acceleration, measured in meters per square second.
If we know that [tex]F = 5\,N[/tex], [tex]m = 0.408\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the net acceleration of the plastic ball is:
[tex]a = \frac{F}{m} - g[/tex]
[tex]a= 2.448\,\frac{m}{s^{2}}[/tex]
The acceleration is 2.448 meters per square second and is vertically upward.
How did Einstein’s and Newton’s theories differ in terms of explaining the cause of gravity?
thank you
Answer:
Newton's theory identified mass as the factor that causes gravity. On the other hand, Einstein's theory identified the curvature of space-time as the factor that causes gravity.
Answer:
Hey mate...
Explanation:
This is ur answer....
In the 17th century Newton concluded that objects fall because they are pulled by Earth's gravity. Einstein's interpretation was that these objects do not fall. According to Einstein, these objects and Earth just freely move in a curved spacetime and this curvature is induced by mass and energy of these objects.
Hope it helps you,
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A spring has a spring constant of 65.5 N/m and it is
stretched with a force of 15.3 N. How far will it stretch?
14. Which of the following is not an example of work being done?
A. pushing a basketball away from your body
B. holding a coffee mug
C. carrying boxes across a warehouse floor
Answer:
B. holding a coffee mug
Explanation:
Something must move a distance for work to be done.
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.
He-Ne Laser device emits photons of wave length 632.8 nm by rate 4.5 x 1020 photon/s, so
the power of the laser beam =
a. 3.14 W
b. 141.3 W
c. 314.1 w
d. 431.4 W
Answer: Option b.
Explanation:
We know:
Wavelength = 632.8 nm
Fluence = 4.5*10^20 photon/s
The energy of a single photon of wavelength λ is:
E = (h*c)/λ
where:
h = 6.6*10^(-34) J*s
c = 3*10^8 m/s
And we should rewrite the wave length in meters, so:
λ = 632.8 nm = 632.8*10^(-9) m
replacing these in the energy equation, we get:
E = (6.6*10^(-34)J*s)*(3*10^8 m/s)/(632.8*10^(-9) m) = 3.13*10^(-19) J
So each one of the 4.5x10^20 photon that flow each second have this energy, then the power is:
P = (3.13*10^(-19) J)*(4.5*10^20 /s) = 140.85 J/s
and 1 W = 140.85 J/S
Then the power is:
P = 140.85 W
Then the correct answer is the option b, where the units are a little bit different than mine because I used really simplified values for c and h.
is electricity matter
Answer:
Yes it is
Explanation:
Electricity is the positive and negative matter that's found in protons and electrons.
Answer:
yes
Explanation:
because electricity is a positive and negative proton
A basketball player jumps straight up for a ball. To do this, he lowers his body 0.250 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.960 m above the floor. (a) Calculate his velocity (in m/s) when he leaves the floor. 4.33 Correct: Your answer is correct. m/s (b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.
Answer:
a. 3.73 m/s b. 27.8 m/s²
Explanation:
(a) Calculate his velocity (in m/s) when he leaves the floor.
Using the conservation of energy principles,
Potential energy gained by basketball player = kinetic energy loss of basket ball player
So, ΔU + ΔK = 0
ΔU = -ΔK
mg(h' - h) = -1/2m(v'² - v²)
g(h' - h) = -1/2(v'² - v²) where g = acceleration due to gravity = 9.8 m/s², h' = 0.960 m, h = 0.250 m, v' =0 m/s (since the basketball player momentarily stops at h' = 0.960 m) and v = velocity with which the basketball player leaves the floor
Substituting the values of the variables into the equation, we have
9.8 m/s²(0.960 m - 0.250 m) = -1/2((0 m/s)² - v²)
9.8 m/s²(0.710 m) = -1/2(-v²)
6.958 m²/s² = v²/2
v² = 2 × 6.958 m²/s²
v² = 13.916 m²/s²
v = √(13.916 m²/s²)
v = 3.73 m/s
(b) Calculate his acceleration (in m/s2) while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.250 m.
Using v² = u² + 2as where u = initial speed of basketball player before lengthening = 0 m/s, v = final speed of basketball player after lengthening = 3.73 m/s, a = acceleration during lengthening and s = distance moved during lengthening = 0.250 m
So, making, a subject of the formula, we have
a = (v² - u²)/2s
Substituting the values of the variables into the equation, we have
a = ((3.73 m/s)² - (0 m/s)²)/(2 × 0.250 m)
a = (13.913 m²/s² - 0 m²/s²)/(0.50 m)
a = 13.913 m²/s²/(0.50 m)
a = 27.83 m/s²
a ≅ 27.8 m/s²
Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?
Answer:
2.2 ma
Explanation:
Given :
Length of the rope = L
Mass of the rope = m
Mass of the object pulled by the rope = M
M = 1.5 m
So, L [tex]$\rightarrow$[/tex] m
For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]
∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.3 m
And for remaining 0.7 L = [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.7 m
By Newtons law of motion,
F - T = ( 0.3 m) a .........(1)
T = ( M + 0.7 m) a
T = ( 1.5 m + 0.7 m) a
T = ( 2.2 m ) a ..............(2)
So from equation (1) and (2), we have
Tension on the rope
T = 2.2 ma
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].
Two automobiles, each of mass 1000 kg, are moving at the same speed, 20 m/s, when they collide and stick together. In what direction and at what speed does the wreckage move (a) if one car was driving north and one south (b) if one car was driving north and one east?
A. The wreckage after collision is moving at the speed 18 m/s to the south.
B. The wreckage after collision is moving at the speed 9.0 m/s to the north.
C. The wreckage after collision is moving at the speed 9.0 m/s to the south.
D. The wreckage after collision is moving at the speed 18 m/s to the north.
E. The wreckage after collision is motionless.
Answer:
The reckage after collision is motionless (E)
Explanation:
The first law of thermodynamics states that energy is neither created nor destroyed but is converted from one form to another.
The kind of collision described in the question above is known as a perfectly inelastic collision, and in this type of collision, the maximum kinetic energy is lost because the objects moving in opposite directions have a resultant momentum that is equal, but in opposite directions hence they cancel each other out.
The calculation is as follows:
m₁v₁ + m₂v₂
where:
m₁ = m₂ = 1000kg
v₁ = 20 m/s
v₂ = -20 m/s ( in the opposite vector direction)
∴ resultant momentum = (1000 × 20) + (1000 × -20)
= 20000 - 20000 = 0
∴ The reckage after collision is motionless
Answer:
The wreckage after collision is moving at the speed 18 m/s to the south.
Explanation:
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.