Answer:
applied force
Explanation:
I think that's right
What form of braking is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake
trail braking
controlled braking
threshold braking
coasting
Answer:
Controlled braking
Explanation:
CONTROLLED BRAKING occur in a situation where a person or an individual driving a vehicle releases the brake and slowly apply smooth as well as firmly pressure on the brake without the wheels been locked which is why CONTROLLED BRAKING are often used for emergency stops by drivers reason been that it helps to reduce speed when driving as fast as possible while the driver maintain the steering control of the vehicle.
Therefore the form of braking which is used to bring a vehicle to a smooth stop by applying smooth,steady pressure to the brake is called CONTROLLED BRAKING.
What's the difference between uk and us
Answer:
one has a k and one has a s
Explanation:
Answer:
the uk refers to the indians and the us refers to the state, (i think srry if its wrong)
Unlike potential energy, kinetic energy cannot —
Answer choices:
a. be heard as sound waves.
b. be stored in atomic bonds.
c. travel in light waves.
d. travel through electrons moving through a wire.
Answer:Which type of wave vibrates parallel to the direction the energy travels, like a slinky ... ex: Rainbow is composed of different frequencies of visible light ... ex: sledding down a hill - potential energy TRANSFORMS into kinetic energy ... C mechanical. D heat. B Chemical energy is energy stored in the bonds of a chemical ...
Explanation:
A 100-kg block being released from rest from a height of 1.0 m. It then takes it 1.40 s to reach the floor. What is the mass m of the other block
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass of the other block is [tex]m_1 = 81.14 \ kg[/tex]
Explanation:
From the question we are told that
Mass of the first block is [tex]m_1 = 100 \ kg[/tex]
The height is [tex]s = 1.0 \ m[/tex]
The time it takes it is [tex]t = 1.40 \ s[/tex]
Generally from kinematic equation
[tex]s = ut + \frac{1}{2} at^2[/tex]
Here u is the initial velocity which zero given that it was at rest initially
So
[tex]s = 0 * t + \frac{1}{2} at^2[/tex]
=> [tex]s = \frac{1}{2} at^2[/tex]
=> [tex]1 = \frac{1}{2}* a * (1.40 )^2[/tex]
=> [tex]a = 1.0204 \ m/s^2[/tex]
Generally from the diagram the resultant force due to the weight of the first object and the tension on the string is mathematically represented as
[tex]mg - T = ma[/tex]
=> [tex]T = m g - ma[/tex]
=> [tex]T = m(g - a)[/tex]
=> [tex]T = 877.96 \ N[/tex]
Generally from the diagram the resultant force due to the weight of the second object and the tension on the string is mathematically represented as
[tex]T - m_1g = m_1 a[/tex]
=> [tex]877.96 = m_1 (a + g)[/tex]
=> [tex]877.96 = m_1 (1.0204 + 9.8 )[/tex]
=> [tex]m_1 = 81.14 \ kg[/tex]
What kind of energy do plants convert light energy from the sun
Answer: chemical energy
Explanation:
Blank parasites can cause serious Illnesses like West Nile virus and Lyme diseases
Exto
Endo
Gastro
Epi
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 5.20m away from the slits.
1. Which laser has its first maximum closer to the central maximum?
2. What is the distance Δymax--max between the first maxima (on the same side of the central maximum) of the two patterns?
3. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?
Answer:
1) aser 1 has the maximum closest to the center
2) Δy = 0.0866 m , 3) Δy = 0.693 m
Explanation:
The interference phenomenon is described by the expression
d sin θ = m λ for constructive interference
d sin θ = (m + ½) λ for destructive interference
We can use trigonometry to find the angle
tan θ = y / L
in trigonometry experiments the angles are small
tam θ = [tex]\frac{sin \theta}{cos \theta} = sin \theta[/tex]
sin θ = y / L
we substitute
d y / L = m λ (1)
1) Let's find the first maximum that corresponds to m = 1 for each laser
laser 1 λ = d / 20
d y₁ / L = 1 d / 20
y₁ = L / 20
y₁ = 5.20 / 20
y₁ = 0.26 m
Laser 2 λ= d / 15
d y₂ / L = 1 d / 15
y₂ = d / 15
y₂ = 5.20 / 15
y₂ = 0.346 m
Therefore laser 1 has the maximum closest to the center
2) the difference between these maxima
Δy = y₂ - y₁
Δy = 0.3466 - 0.26
Δy = 0.0866 m
3) we look for the second maximum m = 2 of laser 1, we substitute in equation 1
y₃ = 2 5.20 / 20
y₃ = 0.52 m
now let's find the third minimum m = 3 of laser 2
d y₄ / L = (m + ½) λ
d y₄ / 5.20 = (3 + ½) d / 15
y₄ = 3.5 5.20 / 15
y₄ = 1.213 m
Δy = y₄ -y₃
Δy = 1.213 - 0.52
Δy = 0.693 m
A 1,100 kg car comes uniformly to a stop. If the vehicle is accelerating at -1.2 m/s2 , which force is closest to the net force acting on the vehicle?
Answer choices:
A. -9600N
B. -1300N
C. -900N
D. -94N
Answer:
Explanation:
D
The net force acting on the car is -1320 N. Option B is the closest force that is acting on the car.
How do you calculate the force?Given that the mass m of the car is 1100 kg and its acceleration a is -1.2 m/s2. The force on the car is calculated by Newton's second law of motion.
Newton's second law states that the force on an object is equal to the products of its mass and acceleration.
[tex]F = ma[/tex]
Substituting the values, we get the force on the car.
[tex]F = 1100 \times -1.2[/tex]
[tex]F = -1320 \;\rm N[/tex]
Hence we can conclude that the net force acting on the car is -1320 N. Option B is the closest force that is acting on the car.
To know more about Newton's second law of motion, follow the link given below.
https://brainly.com/question/8898885.
Can someone help me pls??
Answer:
newton's first law (sorry its really late i know you prob don't need it anymore
Explanation:
If the top of the oil is 25.0 cm above the bottom of the tube, what is the height of the top of the water above the bottom of the tube?
This question is incomplete, the complete question is;
In the Manometer tube in the figure below, the oil in the right-hand arm is olive oil of density 916 kg/m³
If the top of the oil is 25.0 cm above the bottom of the tube, what is the height of the top of the water above the bottom of the tube?
Answer:
the height of the top of the water above the bottom of the tube is 22.9 cm
Explanation:
from the diagram
P_oil = 916 kg/m³
h_oil = 25 cm = 0.25 m
p_water = 1000 kg/m³
Now, in equilibrium;
pressure at p due to water = pressure at p due to oil
⇒h_water × p_water × g = h_oil × p_oil × g
h_water = h_oil × p_oil × g / p_water × g
h_water = (0.25 × 916) / 1000
h_water = 0.229 m ≈ 22.9 cm
Therefore the height of the top of the water above the bottom of the tube is 22.9 cm
How is it possible for man made things to move?
Answer:
Explanation:
Sure!
Cars do it all the time. Do you mean without some sort of guidance system, or some sort of computerized brain? That's getting a little tougher. It's possible I think, if you let some sort of timer turn it on.
Callisto is a moon of Jupiter (mass = 1.90 x 1027 kg), which orbits the planet with a period of 16.9 days. What is the radius of its orbit?
The radius of its orbit = 8.27 x 10¹³ m
Further explanationGiven
mass Jupiter=1.9 x 10²⁷ kg
T = 16.9 days=1.46 x 10⁶ s
Required
the radius =r
Solution
To hold the moon in its orbit, the gravitational force between two objects (jupiter and moon) will be equal to the centripetal force
[tex]\tt G\dfrac{M.m}{r^2}=m.\dfrac{v^2}{r}\rightarrow v=\dfrac{2.\pi.r}{T}\\\\M=\dfrac{r^3.4\pi^2}{T^2.G}\rightarrow r^3=\dfrac{GMT^2}{4\pi^2}[/tex]
G = 6.67 x 10⁻¹¹ N/m²kg²
Input the value :
[tex]\tt r^3=\dfrac{6.67\times 10^{-11}\times 1.9\times 10^{27}\times (1.46\times 10^6)^2}{4\pi^2}\\\\r^3=6.85\times 10^{27}\rightarrow r=8.27\times 10^{13}[/tex]
I need help
Pls for me
1 point
7. A student had an attendance at school of 80% in week A. In week B their
attendance increased to 95%. Calculate the percentage increase in
attendance.
16
5%
19%
h
15%
Answer:
i need a girlfreind
Explanation:
An air tank of volume 1.5 m3 is initially at 800 kPa and 208C. At t 5 0, it begins exhausting through a converging nozzle to sea-level conditions. The throat area is 0.75 cm2 . Estimate (a) the initial mass fl ow in kg/s, (b) the time required to blow down to 500 kPa, and (c) the time at which the nozzle ceases being choked.
Answer:
(a) m = 0.141 kg/s
(b) t = 47.343 s
(c) t = 143.745 s
Explanation:
Given that:
The volume of air in the tank V = 1.5 m³
The initial temperature in the tank is supposed to be 20° C and not 208 C;
So [tex]T_o = 20^0 C = ( 20 +273) K = 293K[/tex]
The initial pressure in the tank [tex]P_o= 800 \ kPa[/tex]
The throat area [tex]A_t[/tex] = 0.75 cm²
To find the initial mass flow in kg/s.
Lets first recall that:
Provided that [tex]\dfrac{P_{amb}}{P_{tank}}< 0.528[/tex], then the flow is choked.
Then;
[tex]\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}[/tex]
[tex]\dfrac{P_{amb}}{P_{tank}}= 0.1266[/tex]
From what we see above, it is obvious that the ratio is lesser than 0.528, therefore, the flow is choked.
Now, for a choked nozzle, the initial mass flow rate is determined by using the formula:
[tex]m = \rho \times A \times V[/tex]
where;
[tex]\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}[/tex]
[tex]\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}[/tex]
[tex]\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}[/tex]
[tex]\rho =9.51350( 0.8333 ) ^{2.5}[/tex]
[tex]\rho =6.03 \ kg/m^3[/tex]
[tex]T = T_o \bigg ( \dfrac{2}{k+1}\bigg)[/tex]
where;
[tex]T_o = 293 \ K[/tex]
[tex]T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)[/tex]
[tex]T = 293 \bigg ( \dfrac{2}{2.4}\bigg)[/tex]
[tex]T = 293 ( 0.8333)[/tex]
T = 244.16 K
During a critical condition when Mach No. is equal to one;
[tex]V = a = \sqrt{kRT}[/tex]
[tex]V = \sqrt{1.4 \times 287 \times 244.16}[/tex]
[tex]V = \sqrt{98103.488}[/tex]
V = 313.214 m/s
Thus, the initial mass flow rate [tex]m = \rho \times A \times V[/tex]
m = 6.03 × 0.75 × 10⁻⁴ × 313.214
m = 0.141 kg/s
(b)
The mass balance formula for the control volume surrounding the tank can be expressed as:
[tex]\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)[/tex]
[tex]= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m[/tex]
When the air mass flow rate is:
[tex]m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex]
Thus; replacing [tex]m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex] in the previous equation; we have:
[tex]\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex]
[tex]\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt[/tex]
Taking the differential of both sides from 0 → t
[tex]In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t[/tex]
[tex]In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t[/tex]
[tex]\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )[/tex]
So, when the pressure P = 500 kPa, the time required is:
[tex]\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )[/tex]
t = 47.343 s
(c)
Let us recall that:
The choking on the nozzle occurred when [tex]\dfrac{P_{amb}}{P_{tank}} = 0.528[/tex]
[tex]\dfrac{P_{amb}}{0.528} = P_{tank}[/tex]
[tex]\dfrac{101.35}{0.528} = P_{tank}[/tex]
[tex]P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa[/tex]
From [tex]\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )[/tex]; the time required for [tex]P_{tank} \simeq 192 \ kPa[/tex] is:
[tex]\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )[/tex]
By solving:
t = 143.745 s
describe how wet and dry barometers work
Answer: Wet barometer - The tool works by measuring atmospheric pressure to predict incoming weather. Since the glass is only filled halfway with water, the other half is exposed to the atmosphere. When the outdoor atmospheric pressure rises, the pressure in the glass decreases, and causes the water to move down the spout.
Dry barometer - A Torricellian barometer (sometimes called a mercury barometer) is an inverted (upside-down) glass tube standing in a bath of mercury. Air pressure pushes down on the surface of the mercury, making some rise up the tube. The greater the air pressure, the higher the mercury rises.
I hope this helps!
The wet barometer can be used to measure the pressure of incoming weather where as dry barometer was used to measure the air pressure for sealed bellows.
What is wet and dry barometers?Barometer
The instrument barometer can be used to measure the pressure in the atmosphere or the environment. Thus, this pressure can be also known to be Barometric pressure.
Wet barometer
A water barometer or a wet barometer is a device that are used measure barometric pressure. This device can be used with the force of gravity, combined with pressure in the air, to create a vacuum, to have a value of the pressure that can be described as with the barometric pressure. The procedure of the wet barometer can be, the glass is only filled halfway with water, the other half is exposed to the atmosphere. When the outdoor atmospheric pressure rises, the pressure in the glass decreases, and causes the water to move down the spout.
Dry barometer
A Dry barometer or a Torricellian barometer or a mercury barometer which can be the inverted glass tube standing in a bath of mercury not the water. Air pressure will pushes down on the surface of the mercury, and makes some rise up the tube. If we give more pressure on the barometer, the mercury will gets increased. Only by this way the value of pressure in the dry barometer can be determined.
Learn more about Barometer,
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Which has a greater buoyant force on it, a 31.0-cm3cm3 piece of wood floating with part of its volume above water or a 31.0-cm3cm3 piece of submerged iron
Answer:
The submerge iron has a greater buoyant force on it
Explanation:
Buoyant force is regarded as the force which is exerted on an object when the object is partly or wholly immersed in a fluid. It can be regarded as that force that is responsible for floating of an object. Buoyant force is one as a result of the differences in pressure that is acting on opposite sides of an object when immersed in a static fluid.
According to Archimedes' principle which states that when a body is fully or partially immersed in fluid, the upward buoyant force that is exerted on that body is proportional to the weight of the fluid that is been displaced by the object
FB= -ρgv
Where FB= buoyant force
ρ= density of fluid
V= volume of fluid
g= acceleration due to gravity
NOTE: from this principle, the magnitude of the buoyant force is proportional to weight of the displaced fluid
✓Volume of both wood and iron=31.0 cm^3
✓the piece of the iron is submerged totally
✓the piece of wood is partially submerged
ANSWER:
The buoyant force that is exerted on the body depends on both the volume of the fluid displaced and the density, but we know that both piece of iron and the wood posses the same volume of 31.0 cm^3 and we also know that the piece of iron displaces more water compare to the wood , simply because the iron is totally submerged, Hence the piece of iron has a greater buoyant force on it.
Suppose an astronaut floating in space throws a rock. What will happen to the rock?
Answer:
C
Explanation:
Two spheres of equal mass, A and B. are projected off the edge of a 2.0 m bench. Sphere A has a horizontal velocity of 5.0 m/s and sphere B has a horizontal velocity of 2.5 m/s.
If both spheres leave the edge of the table at the same instant, sphere A will land where in relation to B?
A) Both Spheres will land at the same distance from the table.
B) Sphere A will go farther than Sphere B.
C) There isn't enough information to answer this question.
D) Sphere B will go farther than Sphere A.
Answer:
the answer is B the second one
For an atom's electrons, how many energy sub levels are present in the principal energy level n=4?
A. 4
B. 9
C. 10
D. 16
E. 32
Answer:
the principal energy level is d because electrons are 2 times more
Which two options describe physical properties of matter?
Answer:
A and E.
Explanation:
Physical properties have to do with things that are not done chemically.
A has to do with light that you can see.
B has to do with Ph (If not A this is your next answer)
C mentions a Patina which is a chemical reaction known as oxidation
D has to do with chemical reactions
E is always correct. One of the fundamental laws about matter is that it must always have mass.
I am 99% confident in A and E as your answer, but if it is wrong go with B and E.
Hope this helps!
What is the velocity of a wave that traveled 36 meters south in 6 seconds?
O 6 meters per second
O 30 meters per second
O 42 meters per second
O 216 meters per second
Answer:
A
Explanation:
A 1 kg object can be accelerated at 10 m/s^2. If you apply this same force to a 4kg object, what will its acceleration be?
Answer:
[tex]a2 = 2.5~m/s^2[/tex]
Explanation:
Mechanical Force
According to the second Newton's law, the net force F exerted by an external agent on an object of mass m is:
F = m.a
Where a is the acceleration of the object.
Assume we apply some given force F to an object of m1=1 Kg that produces an acceleration [tex]a1=10 m/s^2[/tex], then:
F = m1.a1
The same force F is now applied to a second object m2=4 Kg that produces an acceleration a2, then:
F = m2.a2
Dividing both equations:
[tex]\displaystyle 1=\frac{m1.a1}{m2.a2}[/tex]
Solving for a2:
[tex]\displaystyle a2=\frac{m1.a1}{m2}[/tex]
Substituting values:
[tex]\displaystyle a2=\frac{1*10}{4}[/tex]
[tex]a2 = 2.5~m/s^2[/tex]
Responding to an alarm, a 765-N firefighter slides down a pole to the ground floor, 3.5 m below. The firefighter starts at rest and lands with a speed of 3.8 m/s . Find the average force exerted on the firefighter by the pole.
Answer:
Explanation:
Let acceleration of fall be a .
v² = u² + 2as
v = 3.8 m /s
u = 0
s = 3.5 m
3.8² = 0 + 2 x a x 3.5
a = 2.06 m /s²
Since this acceleration is less than g , an upward force is acting on the firefighter in the form of friction . Let this force be F . Let mass of the firefighter be m .
m = 765 / 9.8
= 78.06 kg
mg - F = ma
765 - F = 78.06 x 2.06
765 - F = 160.8
F = 604.2 N .
A block is being pulled a spring scale. The block has a 10g mass. Which
change will create a larger force reading on the spring scale?
A. Sliding the block 60 cm
B. Attaching a larger spring scale
C. Using a 20 g mass block
D. Using a circular piece of wood
Clear
Answer:c
Explanation:
F = ma
If you increase the mass but hold the acceleration constant you will increase the force and thus the spring scale will read higher
Suppose astronomers discover a type-M star with a very large luminosity. What type of star is it likely to be
Answer: A red supergiant
Explanation:
Red supergiants are the stars that have a supergiant luminosity which has a class of either K or M spectral type. In terms of volume, they are regarded as the largest stars on Earth even though they are not the most luminous.
Red supergiants are formed when a star collapses after the hydrogen fuel that the star has in its core runs out and
then fusion begins when the outer shells of hydrogen gets hot.
Light of wavelength 510 nm is used to illuminate normally two glass plates 24.3 cm in length that touch at one end and are separated at the other by a wire of radius 0.023 mm. How many bright fringes appear along the total length of the plates.
Answer:
[tex]m = 180 \ bright \ fringes[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 510 \ nm = 510 *10^{-9} \ m[/tex]
The length of the of the plate is [tex]l = 24.3 \ cm = 0.243 \ m[/tex]
The radius of the wire is [tex]r= 0.023 \ mm = 2.3 *10^{-5} \ m[/tex]
Generally the diameter of the wire which is the distance between the glass plates is mathematically represented as
[tex]d = 2 * r[/tex]
=> [tex]d = 2 * 2.3 *10^{-5}[/tex]
=> [tex]d = 4.6 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is mathematically represented as
[tex]2 *d = m \lambda[/tex]
=> [tex]m = \frac{2d}{\lambda}[/tex]
=> [tex]m = \frac{2 * (4.6* 10^{-5})}{ 510 *10^{-9} }[/tex]
=> [tex]m = 180 \ bright \ fringes[/tex]
Abus has velocity 20m/stowards east and another bus has velocity 15m/s
in west direction. If they start to move from a point simultaneously, what
distances do they cover in 2 minute? What will be their separation?
Answer:
20x120=2400m equivalent to 2km
15x120 =1800 m = 1.8km
2km +1.8 km = 3.8 km
The weight of an object is 118N. If the object is moved to a
distance that is twice its original distance from the planet's
center, what will the new weight be?
Answer:
half the weight depending on planet gravitational force. weight wouldn't change though if gravitational force is strong
Explanation:
if the evaporator oulet temperature on an r-410A system is 50f and the evaporator superheat is 10f, what is the evaporating pressure of the refrigerant in the system
Answer: 143 psig
Explanation:
Given data:
System heat = 50f
Evaporator superheat = 10f
System = r410a.
Solution:
Evaporating pressure of the refrigerant in the system
= system heat - evaporator superheat.
= 50f - 10f
= 40f.
Using the r-410a temperature chart
40f = 143 psig.
