1.
"An estimate of the future value of some variable" is the definition of
A. procurement
B. logistics.
C. a forecast.
D. capacity
find the expression for pressure exerted by fluid with proper description
Answer: Let us calculate the pressure exerted on the bottom by the weight of the fluid. That pressure is the weight of the fluid mg divided by the area A supporting it (the area of the bottom of the container): P=mgA P = m g A .
Explanation:
The refractive index of water is 1.33
and that of diamond is 2.42. Draw a labelled diagram to show how a light ray bends when it travels from water
into diamond.
Explanation:
light travel slower in daimond
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 35 m in front of you. You reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 10 \mathrm { m } / \mathrm { s } ^ { 2 }10m/s 2 . a. How much distance is between you and the deer when you come to a stop
Answer:
Explanation:
Discount the time here; it's not important. It doesn't tell you how long it takes the car to stop, it only refers to reaction time, which means nothing in the scheme of things.
The useful info is as follows:
initial velocity = 20 m/s
final velocity = 0 m/s
a = -10 m/s/s
and we are looking for the displacement. Use the following equation:
[tex]v^2=v_0^2+2a[/tex]Δx
where v is the final velocity, v₀ is the initial velocity, a is the deceleration (since it's negative), and Δx is displacement. Filling in:
[tex]0^2=(20)^2+2(-10)[/tex]Δx and
0 = 400 - 20Δx and
-400 = -20Δx so
Δ = 20 meters
2. Speed limits on curves help to reduce the effect of ........force on a vehicle
A. enertia
b. centrifugal
c. gravity
Answer:
enertia
Explanation:
enertia
The correct answer is (B) Centrifugal Force
The centrifugal force on a vehicle on the curves is outwards or can say away from the curve and pulls the out of the track.
the centrifugal force is given by [tex]F=\frac{mv^{2} }{r}[/tex], which is directly proportional to speed v.
Speed limits on the curves help in reducing the centrifugal force by reducing the speed.
Learn more about centrifugal force:
https://brainly.com/question/25546106
determjne the density of liquid whose relative density is 1.25 given that the density is 1000kgm-3
Answer:
divide the density of solution by density of water
EXPLANATION:
LIKE:
1.25÷1000kgm-3
Hideki had normal vision for most of his life, but now that he is in his 60s, he has started to have difficulty focusing on near objects. He went to an optometrist, who explained that his vision problem was the result of the lenses in his eyes losing elasticity due to aging. Which condition does Hideki have
Answer:
He is suffering from hypermetropia.
Explanation:
There are some defects of vision.
Longsightedness of hypermetropia : It is the defect of vision in which the person is not able to see the nearby objects clearly but can see the far off objects clearly. This is due to the elongation of size of eye ball. It is cured by using convex lens of suitable focal length.
Nearsightedness or Myopia : It is the defect of vision in which a person is not able to see the far off objects clearly but can see the nearby objects clearly. It is due to the contraction in the size of eye ball. It is cured by using concave lens of suitable focal length.
So, Hideki is suffering from hypermetropia. So, he should use the convex lens of suitable focal length.
b) A force is represented in magnitude and direction as (6N, 250degrees. Find both the vertical and horizontal components of the force.
Answer:
Explanation:
To find the horizontal component, the x component specifically, use the formula:
[tex]V_x=Fcos\theta[/tex] and for the vertical component, the y component, use the formula:
[tex]V_y=Fsin\theta[/tex]
where F is the magnitude of the force and theta is the angle in degrees.
For the x-component:
[tex]V_x=6cos250[/tex] so
[tex]V_x=-2.1[/tex] and depending upon whether this is a displacement vector or a velocity vector, the label would be meters/feet or m/s, respectively.
For the y-component:
[tex]V_y=6sin250[/tex] so
[tex]V_y=-5.6[/tex]
A 5.41 kg ball is attached to the top of a vertical pole with a 2.37 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.75 m/s in a horizontal circle with the string remaining taut. Calculate the angle θ, between 0∘ and 90∘, that the string makes with the pole. Use g=9.81 m/s2.
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ = [tex]\frac{m \ v^2}{L \ sin \theta}[/tex]
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L [tex]\frac{sin^2 \theta}{cos \theta}[/tex] = v² / g
(1 -cos²)/ cos θ = [tex]\frac{v^2 }{g \ L}[/tex]
1 - cos² θ = [tex]\frac{4.75^2}{9.81 \ 2.37}[/tex] cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x= [tex]\frac{-0.97 \pm \sqrt{0.97^2 - 4 1} }{2}[/tex]
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ = [tex]\frac{m \ v^2}{ r}[/tex]
T cos θ = m g
resolved
tan θ = [tex]\frac{v^2}{ r g}[/tex]
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
What is the purpose of the lab the importance of the topic and the question you are trying to answer?
perpose of lab is to store apparatus and do some experiment
Helps someone to know the exert lengh of something
The atomic bomb dropped on Hiroshima converted about 7.00x10-4kg of mass to energy. How much energy did that bomb produce?
A)2.10x10^5J
B)7.78x10^-21J
C)6.30x10^13J
D)2.10x10^61J
Answer:
[tex] \sf \: given \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bf \: mass \: \: \: m \: = 7.00 \times {10}^{ - 4} \: kg \\ \\ \bf \: E=mc^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > E=7.00 \times {10}^{ - 4} \times ({3 \times {10}^{8} })^{2} \\ \\ = > \green{ \boxed{ E = 6.3 \times {10}^{13} \: J}}[/tex]
Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading in rad of a flashlight beam that is originally 5.90 cm in diameter with an average wavelength of 610 nm.
Answer:
The answer is "[tex]1.2566 \times 10^{-5}\,rad[/tex]".
Explanation:
As per the Rayleigh Criterion the minimum angular spreading, for a circular aperture:
[tex]\theta_{\mathrm{min}}\approx \sin\theta=1.22\,\frac{\lambda}{d}[/tex]
[tex]\theta_{\mathrm{min}}=\mathrm{1.22\,\frac{\left( 610\,nm \right)}{\left( 5.90\,cm \right)}=1.22\,\frac{\left( 610\times10^{-9}\,m \right)}{\left( 5.90\times10^{-2}\,m \right)}}[/tex]
[tex]=1.22\times 103.389 \times 10^{-7}\\\\=1.22\times 1.03 \times 10^{-5}\\\\=\mathrm{1.2566 \times 10^{-5}\,rad}[/tex]
A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.5 kg person stands on the platform at a distance of 1.05 m from the center, and a 28.3 kg dog sits on the platform near the person 1.43 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
I_syst = 278.41477 kg.m²
Explanation:
Mass of platform; m1 = 117 kg
Radius; r = 1.61 m
Moment of inertia here is;
I1 = m1•r²/2
I1 = 117 × 1.61²/2
I1 = 151.63785 kg.m²
Mass of person; m2 = 62.5 kg
Distance of person from centre; r = 1.05 m
Moment of inertia here is;
I2 = m2•r²
I2 = 62.5 × 1.05²
I2 = 68.90625 kg.m²
Mass of dog; m3 = 28.3 kg
Distance of Dog from centre; r = 1.43 m
I3 = 28.3 × 1.43²
I3 = 57.87067 kg.m²
Thus,moment of inertia of the system;
I_syst = I1 + I2 + I3
I_syst = 151.63785 + 68.90625 + 57.87067
I_syst = 278.41477 kg.m²
How does 'g' vary from place to place?
Explanation:
The acceleration g varies by about 1/2 of 1 percent with position on Earth's surface, from about 9.78 metres per second per second at the Equator to approximately 9.83 metres per second per second at the poles.
The refractive index of glass is 1.52
and that of air is 1.00. Draw a labelled diagram to show how a light ray bends when it travels from glass to air.
Explanation:
..upper side is glass ..
a motorcycle covers 500 meters in 25 seconds. calculate the average velocity
Vtb xe máy là:
v=s : t= 500 : 25=20(m/s)
which physical property of the gas molecules gives the measurement of temperature
The temperature of a gas molecule is measured by the average translational kinetic energy
Answer:
The temperature of a gas is a measure of the average translational kinetic energy of the molecules. In a hot gas, the molecules move faster than in a cold gas; the mass remains the same, but the kinetic energy, and hence the temperature, is greater because of the increased velocity of the molecules
Explanation:
This is also from Go0gle because my explanation would've been an essay long .
but in shorter version if the gas molecules move fast it's hot an if it moves slow its cold hope this helps .
Write a balanced nuclear reaction for one complete cycle
Answer:
H + H ------> He + energy. He + He -----> H + H + He.
Explanation:
Hydrogen having one proton and no neutron fuse with hydrogen having one proton and one neutron forming helium atom with the release of photon. After that two helium atoms combine together forming two hydrogen atoms having one proton each whereas one helium atom having two protons and two neutrons present in their nucleus so the end product of this reaction is hydrogen atoms and helium.
A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied by the club is 115 N. What is the speed of the ball as it leaves the tee
Answer:
v = 50.5 m/s
Explanation:
F = (m)(^v/^t)
115N = (0.04551kg)(v/(0.020s))
2,526.917161 m/s² = v/(0.020s)
v = 50.53834322 m/s
v = 50.5 m/s
Phương trình chuyển động thẳng đều của một chất điểm có dạng: x = 2t – 10. (x: km, t: h). Quãng đường đi được của chất điểm sau 2h là bao nhiêu?
Answer:
Distance cover in 2 hour = 6 kilometer
Explanation:
Given equation:
x = 2t - 10
where
x = kilometer
t = hour
Find:
Distance cover in 2 hour
Computation:
T = 2
So,
x = 2t - 10
x = 2(2) - 10
x = 4 - 10
x = -6
Distance cover in 2 hour = 6 kilometer
Answer:
The distance is 6 km.
Explanation:
The equation of uniform linear motion of a particle has the form: x = 2t – 10. (x: km, t: h). What is the distance traveled by the particle after 2 hours?
x = 2t - 10
distance traveled after t = 2 hours
Substitute t = 2 in the given expression
x = 2 x 2 - 10
x = 4 - 10
x = - 6 km
So, the distance is 6 km.
What is an effect of continental drift?
Answer: An effect of continental drift is causing tectonic plates resting upon the convecting mantle to move which results in natural disasters like earthquakes, volcanic eruptions, and more.
matter can enter and exit which of the following systems?
A. isolated only
B. open only
C. both open and isolated
D. Both closed and isolated
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone
Answer:
The centripetal acceleration of the stone is 5 m/s²
Explanation:
The length of the string to which the stone is attached, r = 1 m
The speed with which the string is rotated, v = 5 m/s
The centripetal acceleration, [tex]a_c[/tex], is given as follows;
[tex]a_c = \dfrac{v^2}{r}[/tex]
Therefore, the centripetal acceleration of the stone found as follows;
[tex]a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2[/tex]
The centripetal acceleration of the stone, [tex]a_c[/tex] = 5 m/s².
can someone pls help me calculate this?
Answer:
10N is the answerrrerreer
Answer:
f= 100N
Explanation: F=m×(v₀-vf/t)
=0.05ₓ(200-0/0.1)
=0.05ₓ2000
=100N
please mark as brainliest
A long distance runner running a 5km track is pacing himself by running 4.5km/h at 9km/h and the rest at 12.5km/h9
Complete Question:
A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/hr and the rest at 12.5km/hr. What is the average speed?
Answer:
Average speed = 9.7333 km/h
Explanation:
Let the total distance be divided into A and B.Given the following data;
Total distance = 5 kmDistance A = 4.5 kmSpeed A = 9.5 km/hrSpeed B = 12.5 km/hrTo find the average speed;
First of all, we would determine the time taken to cover distance A in speed A by using the formula;
[tex] Time \ A = \frac {Distance \; A}{Speed \; A} [/tex]
Substituting the values into the formula, we have;
[tex] Time \ A = \frac {4.5}{9.5} [/tex]
Time A = 0.4737 hours
Total distance = distance A + distance B
5 = 4.5 + distance B
Distance B = 5 - 4.5
Distance B = 0.5 Km
Next, we would determine the time to cover distance B in speed B;
[tex] Time \ B = \frac {0.5}{12.5} [/tex]
Time A = 0.04 hours
Total time = time A + time B
Total time = 0.4737 + 0.04
Total time = 0.5137 hours
Now, we would solve for the average speed;
Mathematically, the average speed of an object is given by the formula;
[tex] Average \; speed = \frac {total \; distance}{total \; time} [/tex]
[tex] Average \; speed = \frac {5}{0.5137} [/tex]
Average speed = 9.7333 km/h
A car travels first 10 km in 20 minutes and another 10 km in 30 minutes. What is the average speed of the car in m/s?
Total distance = 10 km + 10 km = 20 km
1 km = 1000 m
20km x 1000 = 20,000 m
Total time = 20 min. + 30 min. = 50 minutes
Average speed = Distance / time
Average speed = 20,000/50 min
Average speed = 400 m/s
explain how force and surface area affect the applied pressure.
Answer:
force and surface area are two factors affecting pressure on solids
more the force you apply, more will be the pressure
pressure and force are directly proportional meaning if Force is greater, pressure will also be greater
more the surface area of the solid less will be the pressure
surface area and pressure are inversely proportional meaning if surface are is big, pressure will be less, surface area small, pressure will be greater
Answer and Explanation:
We have a basic equation: Pressure = Force/Area.
So for example:
Increase pressure - increase the force or reduce the area the force acts on.
Decrease pressure - decrease the force, or increase the area the force acts on
The force per unit area is pressure. The force on the object is spread over the surface area. The area where the force is applied is divided by the equation for pressure.
Cheers,
A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degrees while the angle of refraction is 24 degrees. What is the index of refraction of amber?
(index of refraction of water is 1.33)
maize is a monocotyledonous seed and pea is a dicotyledonous seed why? give short and the suitable answer I will mark you as a brainelist
Answer:
A dicot is a flowering plant that has one seed leaves. The monocot plants have a single cotyledon. Maize only has one cotyledon in their seed, so it's a monocot. Seeds having two Cotyles are mainly called a Dicot. A pea is a dicotyledonous plant, the seed (the pea itself) has two halves, cotyledons, hence dicot being 2.
Explanation:
One or more of the cotyledons are the first to appear from a germinating seed. Based on the number of cotyledons, botanists classify flowering plants (angiosperms) into :
a) plants with one embryonic leaf, termed monocotyledonous (monocots).
b) plants with two embryonic leaves, termed dicotyledonous (dicots).
Helpful Link:
https://www.vedantu.com/question-answer/in-pea-caster-and-maize-the-number-of-cotyledons-class-11-biology-cbse-5f626a17e5bde9062ff6d2a3
; (b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.
Explanation:
(b) A uniform beam 150cm long weighs 3.5kg and
supported on knife-edges at its ends. The beam
supports a weight 7kg at a distance 30cm from
one end. Find the reactions of the supports.
