Answer:
momentum=mass x velocity= 500 x 3.5 = 1750kgm/s
You've always wondered about the acceleration of the elevators in the 101 story high Empire State Building. One day, while visiting New York, you take your bathroom scale into the elevator and stand on it. The scale reads 177 lb as the door closes. As the elevator moves upward the scale reading varies between a minimum value of 138 lb and a maximum value of 227 lb as the elevator travels 101 floors.
a. What is the magnitude of the acceleration as the elevator starts upward?
b. What is the magnitude of the acceleration as the elevator brakes to a stop?
Answer:
a) 2.8 m/s2
b) 2.2 m/s2
Explanation:
The lecture on the bathroom scale is equal to the normal force that the scale exerts on you, and is directed upward.There is another force acting on you, which is gravity.This force (which we call weight) is the product of the mass times the acceleration due to gravity, g, and is directed downward, opposite to the normal force.When the elevator is at rest, no acceleration takes place, so according Newton's 2nd Law, no net force must be exerted on you.In this condition, the normal force Fn₀ must be equal to the weight:[tex]F_{no} = m * g = 177 lb (1)[/tex]Since the normal force takes any value needed to satisfy Newton's 2nd law, the two extreme lectures can be expressed as follows, in terms of the two forces acting on you while the elevator is moving upward:[tex]F_{n1} = m * (g +a_{1}) = 227 lb (2)[/tex]
[tex]F_{n2} = m * (g +a_{2}) = 138 lb (3)[/tex]
Replacing m by 177lb/g (a given), and rearranging, we can solve (1) for a₁, as follows:[tex]a_{1} = (\frac{227lb}{177lb} *g) - g = 0.28*g = 0.28*9.8m/s2 = 2.8 m/s2 (4)[/tex]
As it can be seen, the normal force takes a larger value in order to be compliant with the upward acceleration that opposes to gravity.In the same way, we can find the magnitude of the downward acceleration when the elevator brakes to a stop, from (3):[tex]a_{2} = (\frac{138lb}{177lb} *g) - g = -0.22*g =- 0.22*9.8m/s2 = -2.2 m/s2 (5)[/tex]
In this case, the normal force takes a lower value than at rest, due to the acceleration has the same direction as gravity.*15 points*
How much louder is the sound of heavy street traffic (60dB) than the sound of normal conversation(40dB)?
A. 20 times louder
B. 100 times louder
C. 240 times louder
D. 2 times louder
The sound of heavy street traffic is 100 times louder than the sound of normal conversation.
What is street traffic?Street traffic is the movement of vehicles and pedestrians in a designated public space, such as a street or alley. Traffic is generally regulated by signs, signals, and lane markings. It is important to obey these rules, as they are intended to keep pedestrians and drivers safe. Street traffic can involve many types of vehicles, including cars, buses, motorcycles, bicycles, and trucks.
The sound of heavy street traffic is 60dB, while the sound of normal conversation is 40dB. This means the sound of the heavy street traffic is 20dB louder than the sound of a normal conversation. To calculate the difference in loudness, we need to take the difference in decibels and raise it to the power 10. This means that the sound of heavy street traffic is 100 times louder than the sound of normal conversation.
To learn more about street traffic
https://brainly.com/question/9010462
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A car wheel turns through 12pi in 10.7s calculate the angular speed of the wheel
Answer:
The answer should be D..
Explanation:
Rolled cookies are cut into fancy shapes.
True
False
Answer:
false
Explanation:
the answer is false
Answer:
true
Explanation:
cause why not, try new things init
Plants are known as producers because they can make their own food. Which of the following provides a list of basic needs a plant must have for photosynthesis to happen?
A.
water and oxygen in the air
B.
soil, energy from the sun, and oxygen in the air
C.
energy from the sun, carbon dioxide in the air, water
D.
soil, carbon dioxide in the air, energy from the sun
HELP PLEASE
Answer:
C .energy from the sun, carbon dioxide in the air, water
hope my answer helps
be sure to follow me
stayy safe
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Answer:
6
Explanation:
just add
Can someone help me with this one please I need help I will mark brainless
Answer:
Your answer is A) 3.33ft/min E
Explanation:
i know this because i know to find the velocity you have to divide distance and time. So for this question your distance=10 and your time=3.
A U-tube with a cross-sectional area of 1.00 cm2 is open to the atmosphere at both ends. Water is poured into the tube until the water rises part-way along the straight sides, and then 5.00 cm3 of oil is poured into one end. As a result, the top surface of the oil ends up 0.550 cm higher than the water surface on the other side of the U. What is the density of the oil
Answer:
0.89 g/cm^3 = 890 kg/m^3
Explanation:
Cross sectional area of U-tube ( A ) = 1.00 cm^2
volume of oil ( V ) = 5.00 cm^3
change between top surface = 0.550 cm
height of oil = 5 cm ( volume / area )
height of water = 5 - 0.550 = 4.45 cm
pressure at the oil-water junction = Pressure on the second side of the U-tube at same level
Po * g * Hoil = Pw * g * Hwater
Po * 5 = 1 * 4.45
∴ Density of oil ( Po ) = 4.45 / 5 g/cm^3 = 0.89 g/cm^3
Which is a tool that use in medicine to view internal organs and pregnancy
Which device increases or decreases voltage as electricity is transmitted from power plants to homes and businesses?
generator
power line
transformer
turbine
Answer:Transformer
Explanation:All the other devices generate their own electricity apart from power line and Transformer, power line just carries electricity. A transformer can be used to step up or step down voltages from a source.
Answer:
transformer!! :)
Explanation:
How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?
50 J,
12.5 J,
25 J,
49 J
Answer:
The correct option is (b).
Explanation:
We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.
We know that, the work done is equal to the kinetic energy of an object i.e.
[tex]W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J[/tex]
So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.
A 105-kg rubber mat is pushed on a dry concrete floor. What is the coefficient of static friction, if the force required to start the mat sliding is 1030 N?
Answer:
[tex]\mu=1[/tex]
Explanation:
Given that,
Mass of a rubber mat, m = 105 kg
We need to find the coefficient of static friction, if the force required to start the mat sliding is 1030 N.
The applied force is given by :
[tex]F=\mu mg\\\\\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{1030}{105\times 9.8}\\\\\mu=1[/tex]
So, the required coefficient of static friction is equal to 1.
On a spaceship designed to support a multiyear voyage to the outer planets of the solar system, plants will be grown to provide oxygen and to recycle carbon dioxide. Because the spaceship will be so far from the sun, an artificial light source will be needed to support photosynthesis. Which wavelengths of light should be used to maximize plant growth?
Answer:
A mixture of blue & red light.
Explanation:
During photosynthesis, the oxygen delivered emanates from water particles and if a weighty isotope of oxygen atom was noticed in delivered sub-atomic oxygen, the water atoms were marked with the hefty isotope.
In order to maximize the growth rate of the plant, the required wavelength of light to be used is a mixture of blue & red light. This is on the grounds that as the absorption optima of plant's photoreceptors are at wavelength frequency of red and blue light, subsequently the combination of red and blue light would be ideal for plant growth and development.
The productivity of red (650–665 nm) LEDs on plant development is straightforward on the grounds that these wavelength frequencies entirely fit with the retention pinnacle of chlorophylls and phytochrome, while the enhanced blue light presented the possibility that development under regular light could be mirrored utilizing blue and red LEDs with negligible use of energy.
Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has mass 2mm and is hanging at rest with its rope vertical. Sphere BB has mass mm and is held so that its rope makes an angle with the vertical that puts BB a vertical height HH above AA. Sphere BB is released from rest and swings down, collides with sphere AA, and sticks to it.
In terms of H,H, what is the maximum height above the original position of A reached by the combined spheres after their collision?
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m [tex]v_{b}^2[/tex]
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b = [tex]\sqrt{2gh}[/tex]
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓ [tex]\sqrt{2gh}[/tex]
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’= [tex]\frac{v^2}{2g}[/tex]
h’ = [tex]\frac{1}{3^2} \ \frac{ 2gh}{2g}[/tex]
h’ = 1/9 h
If a force is applied, but the object does
not move, what can we say about the
amount of work that is done?
A. Some work is done.
B. No work is done.
C. The work equals the force.
D. A large amount of work is done.
Answer:
doesn't move, no work is done; if a force is applied and the object moves a distance d in a direction other than the direction of the force, less work is done than if the object moves a distance d in the direction of the applied.
Answer:
B. No work is done.
Explanation:
Work is only done when the object moves in the direction of the force.
consider an electromagnetic wave of frequency f=3x10^6 Hz. does this radiation belong to the visible range to the ultraviolet or to the infrared of radio waves ?why?
Answer:
eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Explanation:
44e
PLEASE HELP ASAP!!!
An object has a kinetic energy of 50 J and a velocity of 5 m/s, what is the mass of the
object?
A. 4 kg
B. 55 kg
С. 10 kg
D. 1 kg
Answer:
0.5mv^2=50, v=5, 25/2×m=50, m=50×2/25, So, the answer is 4
An experimental procedure calls for 35.9 g sodium chloride and 15.2 g silver nitrate. What tool should be used to measure these substances?
Answer:
Electronic Balance
Explanation:
1. Budding is a way that some very tiny organisms can reproduce.
True
False
____________________________________
Male-female reproduction needs two parent cells to begin.
True
False
Answer:1.. true 2 false
Explanation:
How many joules of gravitational potential energy does a person have after they went up stairs that are 10 meters tall? Assume the person has a mass of 60 kilograms. And assume this takes place on earth.
Answer:
P = 5880 J
Explanation:
Given that,
The mass of a person, m = 60 kg
The height of the stairs, h = 10 m
We need to find the gravitational potential energy of the person. The formula is as follows :
[tex]P=mgh[/tex]
Substitute all the values,
[tex]P=60\times 9.8\times 10\\\\P=5880\ J[/tex]
So, the required gravitational potential energy is equal to 5880 J.
15. Which tectonic boundary would have many normal faults associated
with it?
A. Convergent
B. Divergent
C. Transform
what makes 60% of your arm and what makes the other 40%
Electromagnetic waves used in cell phones are called
a. radio waves
b.microwaves
c. Gamma rays
d. Ultraviolet rays
Answer:
A.Radio waves
Explanation:
A temperature scale has an upper fixed point of 260mm and the lower fixed point of 50mm.what will be the reading on this scale when the thermometer read 125°F
Answer:
the reading on the scale is 158.5 mm.
Explanation:
Given;
upper fixed point of the temperature scale, x₁ = 260 mm
lower fixed point of the temperature scale, x₂ = 50 mm
upper temperature scale, T₁ = 212 °F
lower temperature scale, T₂ = 32 °F
thermometer reading, t = 125 °F, let the reading on the scale = x
Interpolate as follows to determine the value of "x"
[tex]\frac{X_1 - X}{X_1-X_2} = \frac{T_1-t}{T_1-T_2} \\\\\frac{260-X}{260-50} = \frac{212-125}{212-32} \\\\\frac{260 -X}{210} =\frac{87}{180} \\\\180(260-X) = 87(210)\\\\46800 -180 X = 18270 \\\\180X = 46800-18270\\\\180X = 28530\\\\X = \frac{28530}{180} \\\\X = 158.5 \ mm[/tex]
Therefore, the reading on the scale is 158.5 mm.
Constructive interference occurs when the crest of 2 waves meet. This can occur when 2 troughs meet as well. When this occurs, water waves look
A- the same (no change)
B- larger
C- invisible
D- smaller
Answer:
B; LargerExplanation:
I took the K-12 quiz, and that was the correct answer. Hope this helped!You push
on box A with a horizontal
100-N force (Fig. Q4.31). Box A
weighs 150 N, and box B
weighs 50 N. Is the force that
box A exerts on box B equal to 100 N, greater than 100 N, or less than 100 N? Explain.
called.
4) The object that changes its position relative to a fixed point with time is called ...
3) Theratiobetweentotaldisplacementandtotaltimeisсalled..
5) The motion that repeats itself in equal intervals of time is called
6) If the body moves at uniform velocity its acceleration is equal to
7) Acceleration of an object's motion when it changes its velocity with unequal
amounts in equal time is called.
8) When an object started motion from rest, its initial velocity is equal to
9) When a body moves with uniform velocity its initial velocity and its final velocity
are ....
10If an object is thrown upwards its horizontal range reached by it is
Answer:
4) Object in motion
5) Velocity
6) 0 m/s²
7) non - uniform acceleration
8) 0 m/s
9) Equal
10) distance covered by the object along the horizontal plane before it reaches the same vertical position from which it started.
Explanation:
4) Object in motion
5) Ratio between displacement and time is velocity. Displacement is in metres(m) and time is in seconds(s). Thus, displacement/time is in m/s which is velocity.
6) Acceleration = 0 m/s² since it loves with uniform velocity and therefore doesn't accelerate.
7) Since the object changes its velocity with unequal amounts in equal time, the acceleration is called non - uniform acceleration since the acceleration will differ at different points.
8) Initial velocity of an object that starts motion from rest is zero.
9) They are equal since velocity is uniform.
10) The horizontal range will be distance covered by the object along the horizontal plane before it reaches the same vertical position from which it started.
In 1994, Leroy Burrell of the United States set what was then a new world record for the men’s 100 m run. He ran the 1.00 102 m distance in 9.5 s. Assuming that he ran with a constant speed equal to his average speed, and his kinetic energy was 3.40 103 J, what was Burrell’s mass?
Answer:
61.33 Kg
Explanation:
From the question given above, the following data were obtained:
Distance = 1×10² m
Time = 9.5 s
Kinetic energy (KE) = 3.40×10³ J
Mass (m) =?
Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:
Distance = 1×10² m
Time = 9.5 s
Velocity =?
Velocity = Distance / time
Velocity = 1×10² / 9.5
Velocity = 10.53 m/s
Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:
Kinetic energy (KE) = 3.40×10³ J
Velocity (v) = 10.53 m/s
Mass (m) =?
KE = ½mv²
3.40×10³ = ½ × m × 10.53²
3.40×10³ = ½ × m × 110.8809
3.40×10³ = m × 55.44045
Divide both side by 55.44045
m = 3.40×10³ / 55.44045
m = 61.33 Kg
Thus, the mass of Leroy Burrell is 61.33 Kg
The tundra is located _______.
a.
north of the equator
b.
south of the equator
c.
along the equator
d.
both north and south of the equator
Please select the best answer from the choices provided
A
B
C
D
Answer:
d: both north and south
Explanation:
If distance between two charges increased 5 times then force between them
decreased 5 times
Explanation: if the force increases 5 times between them would decrease 5 times