explain why it is not safe for a double decker bus to carry standing passenger on the upper Decker
Answer: The reason why it's not safe for a double decker bus to carry standing passengers is because the center of gravity is raised causing increase in instability of the bus.
Explanation:
A double decker bus is a type of bus that has double decks and usually used by transportation companies for a transport business. Stability is the ability of an object to remain in its state of equilibrium after being distrubed. Stable objects are very difficult to lose balance while unstable object lose their balance easily.
For an object to be stable it must have:
--> a wide base and
--> a low center of gravity.
Therefore when an object has a wide base and a low center of gravity, it is more stable than an object that has a narrow base and a high center of gravity. Center of gravity is the average weight location of an object.
When passengers are left standing on the upper deck of the double decker bus, this INCREASES the center of gravity. This makes the center of gravity of the bus to be outside the base or edge on which it balances. The weight upon having a turning effect will cause the bus to topple and fall which can be dangerous to the lives of the passengers. Therefore it is advisable for passengers to seat while the bus is in motion to help reduce the center of gravity.
An object is placed in front of a thin converging lens of focal length 12 cm. The image of the object is upright and 2.5 times as tall as the actual object. What is the distance of the object to the lens?
7.2 cm
6.0 cm
4.2 cm
15 cm
The distance of the object to the lens will be a) 7.2 cm
What is focal length?The distance between the convex lens or a concave mirror and the focal point of a lens or mirror is called the focal length. It is the point where parallel rays of light meet or converge.
magnification (m) = h(image )/ h(object ) = image distance / object distance
given
h(image ) = 2.5 * h(object)
m = 2.5 * h(object) / h(object ) = 2.5
2.5 = image distance / object distance
2.5 = v / u
v = 2.5 u
using lens formula
given
focal length (f) = + 12cm
1/v - 1/u = 1/focal length
1/2.5 u - 1/u = 1/12
u = - 7.2 cm
correct option a)7.2 cm
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Protons have
_electric charge. Electrons have
_electric charge. Neutrons have
electric charge. Most atoms are electrically
because the number of protons
equals the number of electrons. Atoms with an electric charge become charged by gaining or losing
When this charge becomes built up on an object, we call that build up
Answer:
Protons have positive charge
Electrons have negative charge
Neutrons have no charge
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A wire 0.50 m long carrying a current of 16.0 A is at right angles to a 0.20 T magnetic field. How strong a force acts on the wire
2. A 0.02 kg ball spins in a horizontal circle of radius 0.50 m with a frequency
of 4.0 Hz. What is the tension force in the string (the centripetal force on
the ball)?
Given:
Frequency, f = 4.0 HzMass, m = 0.02 kgRadius, r = 0.50 mWe know,
→ [tex]W = 2 \pi f[/tex]
By substituting the values,
[tex]= 2 \pi \times 4.0[/tex]
[tex]= 8 \pi \ rad/s[/tex]
hence,
→ [tex]T = m \omega^2 R[/tex]
[tex]= 0.02(8 \pi)^2 (0.50)[/tex]
[tex]= 6.31654 \ N[/tex]
or,
[tex]= 6.3 \ N[/tex]
Thus the answer above is right.
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Which object orbits object E?
A
B
C
Or D
Answer:
its D I hope this helps you
He went back to the video to see what had been recorded and was shocked at what he saw.
Which statement identifies a true relationship between the various wave components?
If amplitude increases, frequency increases.
If frequency increases, wavelength decreases.
If wavelength decreases, amplitude decreases.
Answer:
it's the second one;
if the frequency increases, wavelength decreases
Explanation:
we know, v=f×lamda(wave length)
so for constant velocity Frequency f is inversely proportional to lamda
i.e.
fα 1/lamda
so as the f increases lamda decreases and vise versa
A true relationship between the various wave components is if frequency increases, wavelength decreases.
What is the frequency?The frequency is the reciprocal of the time period of the wave.
The wavelength is the distance between the two adjacent crest of the wave.
If the wavelength decreases, the number of cycles will increase in a certain time.
Hence, a true relationship between the wave components is if frequency increases, wavelength decreases.
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A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.70 ss for the boat to travel from its highest point to its lowest, a total distance of 0.660 mm . The fisherman sees that the wave crests are spaced a horizontal distance of 5.90 mm apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling ?
d. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?
Answer:
a) 1.092 m/s
b) 0.33 m
c) 0.25 m
Explanation:
To start with, from the formula of wave, we know that
v = f λ, where
v = velocity of wave
f = frequency of the wave
λ = wavelength of the wave
Again, on another hand, we know that
T = 1/f, where T = period of the wave
From the question, we are given that
t = 2.7 s
d = 0.66 m
λ = 5.9 m
Period, T = 2 * t
Period, T = 2 * 2.7
Period, T = 5.4 s
If T = 1/f, then f = 1/T, thus
Frequency, f = 1/5.4
Frequency, f = 0.185 hz
Remember, v = f λ
v = 0.185 * 5.9
v = 1.092 m/s
Amplitude, A = d/2
Amplitude, A = 0.66/2
Amplitude, A = 0.33 m
If the other distance travelled by the boat is 0.5, then Amplitude is
A = 0.5/2
A = 0.25 m
The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the building can cause windows to pop out. This happened with the original design of the John Hancock Building in Boston.
(a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m X 1.50 m. Assume the density of air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane?
(b) What force is experienced by the window pane from air if the airspeed outside is now 22.4m/s The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pressure. The higher atmospheric pressure in the still air inside the building can cause windows to pop out. This happened with the original design of the John Hancock Building in Boston.
Answer:
a. 451.58 N b. 1806.34 N
Explanation:
(a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m X 1.50 m. Assume the density of air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the window pane?
Using Bernoulli's equation
P₁ + 1/2ρv₁² + ρgh₁ = P₂ + 1/2ρv₂² + ρgh₂ where P₁ = pressure of air in building = atmospheric pressure = 1.013 × 10⁵ N/m², ρ = density of air = 1.20 kg/m³, v₁ = speed of air in building = 0 m/s(since it is still), h₁ = h₂ = h = height of building, P₂ = pressure on the outside of window pane, v₂ = speed of air outside window pane = 11.2m/s and g = acceleration due to gravity
So, since h₁ = h₂ = h
P₁ + 1/2ρv₁² + ρgh = P₂ + 1/2ρv₂² + ρgh
P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂²
Also, v₁ = 0m/s
So, P₁ + 1/2ρ(0 m/s)² = P₂ + 1/2ρv₂²
P₁ + 0 = P₂ + 1/2ρv₂²
P₁ = P₂ + 1/2ρv₂²
P₁ - P₂ = 1/2ρv₂²
So the net pressure on the window is ΔP = P₁ - P₂ = FA where F is the total force on the window pane and A is the area of the window pane = 4.00 m × 1.50 m = 6.00 m²
So, P₁ - P₂ = 1/2ρv₂²
ΔP = 1/2ρv₂²
F/A = 1/2ρv₂²
F = 1/2ρAv₂²
Substituting the values of the variables into the equation, we have
F = 1/2ρAv₂²
F = 1/2 × 1.20 kg/m³× 6.00 m² × (11.2 m/s)²
F = 1/2 × 1.20 kg/m³ × 6.00 m² × 125.44 m²/s²
F = 451.584 N
F ≅ 451.58 N
(b) What force is experienced by the window pane from air if the airspeed outside is now 22.4m/s
When v₂ = 22.4 m/s, F is
F = 1/2ρAv₂²
F = 1/2 × 1.20 kg/m³× 6.00 m² × (22.4 m/s)²
F = 1/2 × 1.20 kg/m³ × 6.00 m² × 501.76 m²/s²
F = 1806.336 N
F ≅ 1806.34 N
Which of the following statements about waves is NOT true?
Amplitude is related to the amount of energy in a wave.
Frequency is related to the number of wave cycles per unit of time.
Wavelength is the distance from the wave crest to the wave trough.
Waves transfer energy (sometimes in the form of vibrations).
Wave amplitude it the distance from the rest position to the crest or trough
This statement is not true:
"Wavelength is the distance from the wave crest to the wave trough."
The others listed are true.
A light bulb has 0.25 A of current using a voltage of 120 V. What is the power rating for this bulb?
We can't tell what RATING is printed on the package of bulbs, but right now, the bulb is dissipating IxE= 30 watts of heat and light.
The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c, space-dock personnel measure the rocket ship to be 43.3 m long. Part A The rocket ship Adventure travels to a star many light-years away, then turns around and returns at the same speed. When it returns to the space dock, who would have aged less: the space-dock personnel or ship's captain?
Complete question:
Part A:) The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c , space-dock personnel measure the rocket ship to be 43.3 m long. The rocket ship Adventure travels to a star many light-years away, then turns around and returns at the same speed. When it returns to the space dock, who would have aged less: the space-dock personnel or ship's captain?
Part B: What is the momentum of a proton traveling at 0.62 c ?
Answer
a)Who would have aged less=The Captain would have aged less
b) [tex]p=3.96*10^{-19}kgm/s[/tex]
Explanation:
From the question we are told that
Length measured by captain [tex]l_c=50m[/tex]
Speed of rocket past tje space dock [tex]V=0.5c[/tex]
Length measured by space-dock personnel [tex]l_c=43.3m[/tex]
a)
Generally time moves slower when moving at speed of light, due to time dilation or variation.
Who would have aged less=The Captain would have aged less
b)
Generally the equation for Relativistic Momentum is mathematically given as
[tex]p=\frac{m*v}{1 - v^2/c^2}[/tex]
[tex]p=\frac{1.67*10^2-27*0.62*3.0*10^8)}{\sqrt{ 1 -0.6^2}}[/tex]
[tex]p=3.96*10^{-19}kgm/s[/tex]
Albert and Emmy are looking to purchase some new spaceships. They pick out two identical single-seat space racers and take them out to space for a test drive. After flying around for a while, they decide to do some experiments. As Albert flies past Emmy at a very high but constant velocity, Emmy measures the length of Albert's ship and compares it to the length of her own ship.
Emmy observes that Albert's ship is __________ her own ship.
At the same time, Albert observes Emmy's ship and compares it with his ship.
Albert observes that Emmy's ship is ___________ his own ship.
Answer:
shorter than,
shorter than
1) option B. Emmy observes that Albert's ship is shorter than her own ship.
2) option B. Albert observes that Emmy's ship is shorter than her own ship
Explanation:
Applying the principle of length contraction where the length of a moving object is seen as being contracted (shorter than its real length),the length of the object in the rest frame does not appear to change.
In this case, the space racer in his own ship measures a reduced length of the other ship flying past it at a high but constant speed. Moreover, the space racer in a ship measures the real length of his own ship and not a contracted length.
Therefore, Emmy will observe that Albert's ship is shorter than her own ship while Albert will also observe that Emmy's ship is shorter than his own ship. It is the principle of length contraction that holds
A vector starts at the point (0.0) and ends at (3,-3). What is the angle of the displacement?
4.24
18
45 degrees
90 degrees
Answer:
135degrees
Explanation:
Given that a vector starts at the point (0.0) and ends at (3,-3)
y component value = -3
x component value = 3
angle of the displacement theta = arctan(y/x)
theta = arctan(-3/3)
theta = arctan (-1)
theta = -45degrees
Since tan is negative in the second quadrant
theta = 180 - 45
theta = 135degrees
Hence the angle of the displacement is 135degrees
Tawny notices that Jim has been forgetting to check two forms of
identification for new patients. How could she best communicate this
effectively to Jim in a way that avoids conflict?
A. "I forget this a lot, too, so I'm not blaming you or anything, but you
need to check two forms of identification."
B. "Did you forget your training? We always have to check two forms
of identification for each patient."
C. "We have to check two forms of identification, so stop forgetting
to do it."
D. "I think the proper procedure is to check two forms of
identification for each patient."
Answer: D. "I think the proper procedure is to check two forms of
identification for each patient."
Explanation: took the quiz
Define specific heat capacity and latent heat of fusion.
Answer:
A substance's basic heat of fusion is the amount of heat needed to transform a unit mass of the substance from liquid to gaseous at its boiling point. B)... A substance's latent heat is known as the amount of energy emitted or consumed when the substance changes states.
Explanation:
Select the correct arrows. Identify the arrows that show the correct direction of heat transfer.
Answer:
Explanation:I Did the test and it said this was the right answer so it probally is
Answer:
That's the question that I wanted to ask!!
Explanation:
A sodium surface is illuminated with light of wavelenght 300nm. The work function of the metal is 2.4eV.
1. Determine the K.E
2. The cutt-off wavelenght of sodium
Answer:
1) 1.67eV
2) 505nm
Explanation:
The maximum kinetic energy of photoelectrons ,
KEmax=λhc−W=(0.3×10−6)(1.6×10−19)(6.62×10−34)(3×108)eV−2.46eV=1.67eV
If λ0 is the cut-off wavelength, W=λ0hc
or λ0=Whc=2.46×1.6×10−19(6.62×10−34)(3×108)=5.05×10−7m=505×10−9=505nm
We know that the work function is the minimum photon energy for taking place of photoelectric effect.
A car has a mass of 1.00x10 to the 3rd power kilograms, it has an acceleration of 4.5 meters/seconds, what is the net force on the car?
Explanation:
Net force on the car= mass of the car × acceleration
F=1×10^3×4.5
=4.5×10^3 N
A person hears sounds with a frequency in the range of about 16 Hz to 20000 Hz. What is the wavelength of sound in air and water corresponding to these frequencies?
Please let me know the answer and how you got the answer (what formula you used)
Answer:
v=f×λ
Explanation:
v=f×λ
v is the speed of sound
f is the frequency of sound
λ is the wavelength
The speed of sound in air and water is different, but the frequency can be calculated from the given range above.
HAVE FUN LEARNING!!!( sorry im just to lazy to get the number so...)
A flag is waved 3.2 m above the surface of a flat pool of water. A swimmer in the pool sees the flag at what distance above the water? Let the indices of refraction nwater = 1.33 and nair = 1.00.
3.2 cm
2.4 cm
5.1 cm
4.3 cm
Answer: 4.3cm
Explanation:
just took the test
Type of bacteria that attacks the throat- causing fever sore throat rash
Answer:
Streptococcus Pyogenes
Explanation:
Strep throat!
A musical note has a frequency of 512 Hz. If the wavelength of the note is 0.685 m, what is the speed of the sound of that note?
F. O 0.00120 m/s
G.O 351 m/s
H
345 m/s
J.
841 m/s
Answer:351
Explanation:trust
A musical note has a frequency and the wavelength of the note. The speed of the sound of that note will be 351m/s. The correct option is G.
What is frequency?The frequency is the number of cycle per second.
The wavelength is related to frequency and speed as
f = v/λ
Given is the frequency f = 512Hz and wavelength λ= 0.685m, then the speed of note will be
v =512 Hz x 0.685m
v =350.72 m/s
Speed is approximately 35 m/s.
Thus, the correct option is G.
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The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many
grams were in the original sample?
Answer:
400 g
Explanation:
The computation of the number of grams in the original sample is shown below:
Given that
half-life = 5.26 years
total time of decay = 15.8 years
final amount = 50.0 g
Now based on the above information
number of half-lives past is
= 15.8 ÷ 5.26
= 3 half-lives
Now
3 half-lives = 1 ÷ 8 remains = 50.0 g
So, the number of grams would be
= 50.0 g × 8
= 400 g
A pendulum consists of a 2.0-kg block hanging on a 1.5-m length string. A 10-g bullet moving with a horizontal velocity of 900 m/s strikes, passes through, and emerges from the block (initially at rest) with a horizontal velocity of 300 m/s. To what maximum height above its initial position will the block swing
Answer:
The maximum height above its initial position is:
[tex]h_{max}=1.53\: m[/tex]
Explanation:
Using momentum conservation:
[tex]m_{b}v_{ib}=m_{B}v_{fB}+m_{b}v_{fb}[/tex] (1)
Where:
m(b) is the mass of the bulletm(B) is the mass of the blockv(ib) is the initial velocity of the bulletv(fb) is the final velocity of the bulletv(fB) is the final velocity of the blockLet's find v(fb) using equation (1)
[tex]m_{b}(v_{ib}-v_{fb})=m_{B}v_{fB}[/tex]
[tex]v_{fB}=\frac{m_{b}(v_{ib}-v_{fb})}{m_{B}}[/tex]
[tex]v_{fB}=\frac{0.1(900-300)}{2}[/tex]
[tex]v_{fB}=30\: m/s[/tex]
We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.
[tex]\frac{1}{2}m_{B}v_{fB}=m_{B}gh_{max}[/tex]
[tex]h_{max}=\frac{1}{2g}v_{fB}[/tex]
[tex]h_{max}=\frac{1}{2(9.81)}30[/tex]
[tex]h_{max}=1.53\: m[/tex]
I hope it helps you!
The maximum height above its initial position is: [tex]h_{max}[/tex]=1.53m
What is simple harmonic motion?Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
Using momentum conservation:
[tex]m_bv_{1b}=m_Bv_{fB}+m_bv_{fb}[/tex]
Where:
m(b) is the mass of the bullet
m(B) is the mass of the block
v(ib) is the initial velocity of the bullet
v(fb) is the final velocity of the bullet
v(fB) is the final velocity of the block
Now for finding the value of final velocity VfB
[tex]m_b(v_{1b}-v_{fb})=m_Bv_{fB}[/tex]
[tex]v_{fB}=\dfrac{m_b(v_{1b}-v_{fb})}{m_B}[/tex]
[tex]v_{fB}=\dfrac{0.1(900-{300})}{2}[/tex]
[tex]v_{fB}=30\ \frac{m}{s}[/tex]
We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.
[tex]\dfrac{1}{2}m_Bv_{fB}=m_Bv_{fB}[/tex]
[tex]h_{max}=\dfrac{v_{fB}}{2g}[/tex]
[tex]h_{max}=\dfrac{30}{2\times 9.81}[/tex]
[tex]h_{max}=1.53\ m[/tex]
Thus the maximum height above its initial position is: [tex]h_{max}[/tex]=1.53m
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Different tissues working together to perform a particular job are called:
A: Organ systems
B: Organelles
C: Organs
Answer:
C. Organs
Explanation:
Organs are groups of tissues that work together to perform a particular job.
hope this helps and is right; p.s. i really need brainliest :)
A 125-g coin is placed 8.0 cm from the axis of rotation of a horizontally rotating turntable as shown. The coefficient of static friction between the coin and the turntable is μs . The turntable makes exactly 1.0 revolution per second and the coin remains in place without slipping. [a] (8 pts) Draw a free-body force diagram for the coin. [b] (10pts)What is the frictional force acting on the coin? [c] (12pts)When the rotational speed is increased to 1.2 revolutions per second, the coin starts to slip. Calculate the coefficient of static friction μs between the coin and the turntable?
Answer:
zhvshshisvdiscdoscd if vdidg
Ali is sitting on the chair, identify the forces acting on him.
A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of mass 1.51 kg and length 1.79 meters spinning clockwise with an angular velocity of 5.12 rad/s is dropped on the spinning disk and stuck to it (the centers of the disk and the rod coincide). The combined system continues to spin with a common final angular velocity. Calculate the magnitude of the loss in rotational kinetic energy due to the collision
Answer:
The loss in rotational kinetic energy due to the collision is 36.585 J.
Explanation:
Given;
mass of the disk, m₁ = 1.64 kg
radius of the disk, r = 0. 61 m
angular velocity of the disk, ω₁ = 17.6 rad/s
mass of the rod, m₂ = 1.51 kg
length of the rod, L = 1.79 m
angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)
let the counter-clockwise be the positive direction
let the clock-wise be the negative direction
The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;
m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)
where;
ωf is the common final angular velocity
1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)
21.1328 = ωf(3.15)
ωf = 21.1328 / 3.15
ωf = 6.709 rad/s
The moment of inertia of the disk is calculated as follows;
[tex]I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2[/tex]
The moment of inertia of the rod about its center is calculated as follows;
[tex]I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2[/tex]
The initial rotational kinetic energy of the disk and rod;
[tex]K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J[/tex]
The final rotational kinetic energy of the disk-rod system is calculated as follows;
[tex]K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J[/tex]
The loss in rotational kinetic energy due to the collision is calculated as follows;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J[/tex]
Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.