Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
8. An airplane of mass 8500 kg dives has an altitude of 15,000 m. It then dives steeply to an
altitude of 11,000 m. What was the change in potential energy?
Answer:
333.5 MJ
Explanation:
ΔV = m·g·Δh
= 8500 · 9.81 · (15000-11000)
= 8500 · 9.81 · 4000
= 333 540 000
≈ 333.5 ·10⁶J = 333.5 MJ
The change in potential energy is 333.5 MJ
What is Potential Energy?
Potential energy is defined as the energy possessed by an object because of its position relative to other objects, tension within itself, its electric charge, or other factors. Any object that is raised above its rest position has energy stored in it, so it is called potential energy because it has the potential to do work when released.
It can be expressed as:
P.E.= mgh
where, m is the mass of an object measured in grams 'g'
g is the acceleration due to gravity which is [tex]9.8 m/s^2[/tex]
h is the height measured in meter 'm'
Here,
the given information is
m= 8500kg
Δh= [tex]h_2 - h_1[/tex] = 15000-11000 = 4,000m
So, P.E. = 850 x 9.8 x 4000 = 333 540 000= 333.5 ·10⁶J = 333.5 MJ
Thus, the change in potential energy is 333.5 MJ
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I will mark brainlist
If a wave’s amplitude is 2cm, then its height is equal to:
5 cm.
0 cm
4 cm.
2 cm
Answer:
4 cm
Explanation:
Amplitude is the measure from the MIDLINE to the peak of the wave....so the wave HEIGHT is twice the ampltude
2 x 2 cm = 4 cm
A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a mass of 0.63 kg, which is distributed uniformly along its length. Find the kinetic energies of the rod.
Hi there!
[tex]\large\boxed{KE =1.245 J}}[/tex]
The equation of angular kinetic energy is:
[tex]KE = \frac{1}{2}Iw^2[/tex]
Where:
I = moment of inertia (kgm²/s)
ω = angular speed (rad/sec)
A rod rotated about one of its endpoints has a standard moment of inertia of 1/3mR², so:
[tex]KE = \frac{1}{2}(\frac{1}{3}mL^2w^2)[/tex]
[tex]KE = \frac{1}{6}mL^2w^2[/tex]
Plug in the given values:
[tex]KE = \frac{1}{6}(0.63)(0.82^2)(4.2^2) = 1.245 J[/tex]
when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?
Hello!
We can use the following angular kinematic equation to solve:
α = Δω/Δt, or (ωf-ωi)/t
Plug in the given values:
α = (25 - 10)/5 = 15/5 = 3 rad/sec²
During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the hanging masses and other forces present is 0.91 J. At the initial time of observation the cart moves with speed 0.61 m/s. Determine the speed at the second point of observation. The mass of the cart is 0.80 Kg.
Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
The speed of the cart at the second point of observation is equal to 1.62 m/s.
What is the work done?Work can be demonstrated as the energy utilized when a force is exerted to make an object move through a particular displacement. Work done by this force is calculated from the product of the magnitude of the exerted force (F) and the distance (d) covered by the body
W= F.d
Where 'F' is the exerted force and 'd' is the displacement and W is work done.
Work and energy have a direct relationship with each other. Work done by an object can be expressed as a change in kinetic energy:
[tex]W =\frac{1}{2}mv_2^2- \frac{1}{2}mv_1^2[/tex]
Where m is the mass of the object, v₂ is the final velocity (m/s), and v₁ is the initial velocity (m/s).
Given, the work done by the cart, W = 0.91 J
The mass of the cart, m = 0.80 Kg
The speed of the cart at the first point, v₁ = 0.61 m/s
[tex]0.91 =\frac{1}{2}\times 0.80 \times v_2^2- \frac{1}{2}0.80 \times (0.61)^2[/tex]
0.40 v₂² = 1.06
v₂² = 2.65
v₂ = 1.62 m/s
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You apply the same force to two objects. Object 1 has mass M and object 2 has mass 5M. The acceleration of object 2 is
Select one:
a. one-fifth as that of object 1.
b. ten times that of object 1.
c. has no relation to that of object 1.
d. the same as that of object 1.
e. five times that of object 1.
five times that of object l
)Suppose you have two insulated buckets containing the same amount of water at room temperature. You also happen to have two blocks of metal of the same mass, both at the same temperature, warmer than the water in the buckets. One block is made of aluminum and one is made of copper. You put the aluminum block into one bucket of water, and the copper block into the other. After waiting a while you measure the temperature of the water in both buckets. Which is warmer
The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.
The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.
At standard conditions;
the specific heat of aluminium = 0.215 cal/gm/° Cthe specific heat of copper = 0.0923 cal/gm/° CWe know that:
Heat loss by the metal block = Heat gained by the water bucketTherefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.
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a process generates 250 watts of useful energy and 600 watts of waste energy. WhT is the efficiency of the process?
O29percent
O2.4percent
O42 percent
O71percent
The efficiency of the process is 42%
From the question given above, the following data were obtained:
Useful energy = 250 watt
Wasteful energy = 600 watt
Efficiency =?The efficiency of the process can be obtained as illustrated below:
[tex]efficiency \: = \frac{useful \: energy}{wasteful \: energy} \times 100 \\ \\ efficiency \: = \frac{250}{600} \times 100 \\ [/tex]
Efficiency = 42%Therefore, the efficiency of the process is 42%
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Answer:
Its actually 29%
Explanation:
Took the Test
A 12.0 kg box resting on a horizontal, frictionless surface is attached to a 5.00 kg weight by a thin, light wire that passes over a frictionless pulley (Figure 1). The pulley has the shape of a uniform solid disk of mass 2.60 kg and diameter 0.500 m. A) After the system is released, find the tension in the horizontal segment of the wire. B) After the system is released, find the tension in the vertical segment of the wire. C) After the system is released, find the x-component of the acceleration of the box. D) After the system is released, find the x- and y-components of the force that the axle exerts on the pulley.
Answer:
Explanation:
Assuming x direction is horizontal any y direction is vertical
As the moment of inertia of the pulley is I = (½)mR²
The pulley mass will appear as ½ its actual mass in the system.
F = ma
5.00(9.81) = (5.00 + 2.60/2 + 12.0)a
a = 49.05 / 18.3
a = 2.6803278...
a = 2.68 m/s²
A) T = 12.0(2.68) = 32.16... = 32.2 N
B) T = 5.00(9.81 - 2.68) = 35.65 = 35.7 N
C) aₓ = 2.68 m/s²
D) Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
A)The tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
What is acceleration?Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.
The given data in the problem is;
m is the mass box resting on a horizontal =12.0 kg
the uniform solid disk of mass of 2.60 kg
d is the diameter = 0.500 m.
A)The tension in the horizontal segment of the wire will be 32.2 N
The acceleration is found as;
[tex]\rm F = ma \\\\ a= \frac{49.05}{18.3} \\\\ \rm a= 2.68 \ m/sec^2[/tex]
The tension in the horizontal segment of the wire is found as;
T=ma
T= 12.0 ×2.68
T=32.2 N
Hence the tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
T=m(g-a )
T= 5.00 × (9.8-2.68)
T=35.7 N
Hence the tension in the vertical segment of the wire will be 35.7 N
C) the x-component of the acceleration of the box. will be 2.68 m/s².
aₓ = 2.68 m/s²
D)The x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
Hence the x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
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what were your preparetion before going the different physical fitness test?
Answer:
Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened
g A 150-kg object and a 450-kg object are separated by 4.90 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 39.0-kg object placed midway between them. 130000000000 Incorrect: Your answer is incorrect. Draw a picture of the three objects, including the forces exerted on the third. N (b) At what position (other than an infinitely remote one) can the 39.0-kg object be placed so as to experience a net force of zero from the other two objects
Answer:
BBbbb
Explanation:
B is the correct answer:))))
2. Conner flips a coin up in the air (to determine if he or his sister needs to do the dishes) at an upward velocity of 4.00 m/s. He fails to catch it on the way down and it falls down an additional 1.6 m to the floor. Calculate the maximum height and maximum magnitude (could be positive or negative) velocity of the coin.
Answer:
5.6
Explanation:
Not so sure
Figure 1 shows that the two vectors in different direction respectively. Determine the resultant for the two vectors
Answer:
15-45/2 = -7.5 downwards
Explanation:
cos(60) x 45 = z
z = 45/2
sin (30) x 70 = y
y = 15
15-45/ 2 = -7.5 downwards
i’m confused on this worksheet can someone help
Answer:
well for C the horse is at rest at 10 to 15
D is 0 to 10
E is 15 to 25
Explanation:
If you search up examples it will help with this worksheet.
If I am work then I appologize
3. pinapatupad ng inyong lungsod ang pagpapanatili ng mga tao sa bahay sa araw ng Linggo
dahil sa nararanasang pandemya. Alin ang dapat gawin ng mga mamamayan?
a. Maagang pumunta sa dagat at doon magpalipas ng maghapon.
b. Magpunta sa simbahan at doon manatili sa buong araw.
a Pumunta sa kamag anak sa kabilang lungsod.
d. Manatili sa loob ng bahay sa buong araw.
Answer:
d because u now
Explanation:
sana makatulong
which statement is true about heat and light energy
Answer:
statements?
Explanation:
Light bulbs produce useful energy in the form of light and waste energy in the form of heat. What kind of light bulb would you use
to incubate eggs?
A. low-efficiency light bulbs, because they produce more light
B. low-efficiency light bulbs, because they produce more heat
C. high-efficiency light bulbs, because they produce more heat
D. high-efficiency light bulbs, because they produce more light
Answer:
C) high efficiency light bulbs, because they produce more heat
Light bulbs produce useful energy in the form of light and waste energy in the form of heat, the kind of light bulb would you use to incubate eggs are low-efficiency light bulbs, because they produce more heat, therefore the correct answer is option B.
What is thermal energy?It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy.
The type of light bulb you would use to incubate eggs is a low-efficiency light bulb since they produce more heat; consequently, option B is the right response. Light bulbs produce useful energy in the form of light and waste energy in the form of heat.
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calculate 18% of 2758 correct to 4 significant figure
Answer:
......the answer is 496.4
A object travels at constant negative acceleration. What does the graph of the object's velocity as a fun
(1 point)
O a curved line increasing from left to right
O a straight line with negative slope
O a straight line with positive slope
O a curved line decreasing from left to right
SOMEONE HELP! If I don't get this done i will fail, I have been trying for hours.
p.s. both questions have no answer choices.
1. If a baseball has a mass of 0.075kg and is traveling with an acceleration of 25 m/s2. What is the force that the baseball will hit the bat?
````````````````````````````````````````
2. Given the force diagram of the basketball, what can you tell about the basketball motion? (Diagram below)
Thanks in advance!!
Answer:
1. The baseball will hit the bat at a force of 1.875 Newtons (N)
2. The basketball is being pushed to roll to the right of its original position with an applied rightward force. There is a force of friction to the left, a normal force of the floor acting upwards on the ball, and a gravity force of the Earth acting downwards on the ball.
Explanation:
1. To calculate force, you just have to multiply mass (m) and acceleration (a). F = m * a --> F = 0.075kg * 25 m/s^2 = 1.875 N.
2. Use your knowledge of different types of forces and free-body diagrams to come to this conclusion.
Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball is thrown and travels 30 inches before bouncing, It bounces and travels 50% of the distance traveled. It continues to do this until coming to rest. What is the total distance traveled by ball? Also it's cold and dark and you have no phone nor anyone to ask. You have no ball
60 inches
because if 30=50% then 60=100% so its 60 inches
The sum of the series allows to find the result for the total distance that the ball bounces is:
total distance = 59.52 in
A series is a set of things or numbers related by a specific operation.
They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.
Let's build a table to observe the sequence.
drop height rebound
1 30 15
2 15 7.5
3 7.5 3.75
If we call the first term y₀
The first bounce can be found.
y₁ = [tex]\frac{y_o}{2}[/tex]
The second bounces.
[tex]y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}[/tex]
The third bounce.
[tex]y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}[/tex]
By observing this table we can construct a series of the form
Total distance = [tex]y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )[/tex]
The sum of the serie has a result of
sum = 127/64 = 1,984
Let's calculate
distance total = 30 1,984
Distance total = 59.52 cm
In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:
total distance = 59.52 in
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A patient comes to an outpatient laboratory for a physician-ordered fasting test. The patient indicates that he forgot that the test was to be fasting and ate a candy bar 2 hours ago. The patient insists that you should draw the test because he cannot come back at another time. What should you do?
The patient eating a candy bar instead of fasting for the test should be told
that the test results will be wrong and he may receive a wrong diagnosis.
The medical practitioners told him to fast when coming for a reason and he
forgetting and eating something means the objective for the test may have
been defeated.
This is why it's best to explain to him about the consequences of his actions
which is likely getting a wrong result and diagnosis.
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Calculate the torque produced by a 50.0 N perpendicular force at the end of a 0.300 m long wrench.
Answer:
Torque = 50N x 0.3m = 15Nm
Explanation:
Torque = Force x length of lever arm. To obtain the torque simply multiply the two given values.
Why does the sky appear red and yellow during sunsets.
1. The reds and yellows are caused by the Sun's light having to pass through dust particles near closer to the horizon than when the Sun is higher in the sky.
2. At sunset the prismatic effect of the atmosphere at low angles produces longer red and yellow wavelengths.
3. At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.
4. The red and yellows are appear at sunset because the light must pass through more atmosphere which slows down the wavelength frequency turning it to slower red and yellow frequencies.
Answer:
The answer should be 3-At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.
Answer:
3. At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.
9
are things that you can achieve quickly.
O A.
Dreams
OB.
Long-term goals
O c.
Short-term goals
OD.
Plans
Answer:
d
Explanation:
A dreams
.......wkkwkwkwkwkwnwnsksk
Answer:
C. short term goal
Explanation:
took a midterm quiz on canvas, Hope this helps <3
QUESTION 5 When an instrument is sounded together with a turning fork of frequency 260Hz , 2 beats are heard. When the same instrument is sounded with a fork of frequency 256H2, , 6 beats are heard. Find the frequency of the instrument .
Answer:
Explanation:
4126h2.
knives, axe ,blades and nails are examples of
knives, axe ,blades and nails are examples of potential energy
or sharp objects
A stone is allowed to fall from the top of the tower 100m high at the same moment another stone is projected vertically upward with a velocity of 25m/s. Where and when will the two cross each other
Answer:
both the stone will meet at a distance of 80 m from the top of tower.
Explanation:
let "t" = time after which both stones meet
"S" = distance travelled by the stone dropped from the top of tower
(100-S) = distance travelled by the projected stone.
◆ i) For stone dropped from the top of tower
-S = 0 + 1/2 (-10) t²
or, S = 5t²
◆ ii) For stone projected upward
(100 - S) = 25t + 1/2 (-10) t²
= 25t - 5t²
Adding i) and ii) , We get
100 = 25t
or t = 4 s
Therefore, Two stones will meet after 4 s.
◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get
S = 5 × 16
= 80 m.
Thus , both the stone will meet at a distance of 80 m from the top of tower.
(Hope this helps can I pls have brainlist (crown)☺️)
Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?
Answer:
215mph
Explanation:
multiply 112milws by 2 and multiple that by seven, then divide it by two
The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m
Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N