Answer:
Measurements allow people to find their way to new places. Measurements such as miles or kilometers are used by GPS systems to give directions. Time measurements help to create schedules so tasks get done on time. Measurements are used in food as well. Ingredients in recipes have to be measured to make the dish correctly. Serving sizes are a measurement that keep people healthy by showing how much of each food you should eat.
Which item will be shipped third?
—-
Answer:
I know it's groceries
Explanation:
electronics ship before clothing
electronics ship after groceries
urgent items are first so
order:
1.) A/Electronics
2.) Clothing/B
3.) Groceries(since groceries aren't urgent)
thing is it's C or D I'm leaning to D since it says it ships last but i dont know so if I'm wrong sorry.
Is it true that as we gain mass the force of gravity on us decreases
Answer:
No. As we gain mass the force of gravity on us does not decrease
The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.
Answer:
ΔT = 0.017 °C
Explanation:
According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:
Change in Internal Energy = (0.85)(Kinetic Energy)
[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]
where,
ΔT = increase in temperature = ?
v = speed of block = 4 m/s
C = specific heat capacity of copper = 389 J/kg.°C
Therefore,
[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]
ΔT = 0.017 °C
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced
Answer:
The total surface area is "90 km²".
Explanation:
Given:
Power from solar installations,
= 27 GW
Other natural installations,
= 740 GW
Intensity,
[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]
%n,
= 30%
Now,
⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]
then,
⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]
[tex]=90 \ GW[/tex]
As we know,
⇒ [tex]I=\frac{P}{A}[/tex]
by substituting the values, we get
[tex]1000=\frac{90\times 10^9}{A}[/tex]
[tex]A = \frac{90\times 10^9}{10^3}[/tex]
[tex]=90\times 10^6[/tex]
[tex]=90 \ km^2[/tex]
Which of the units of the following physical quantities are derived
Answer:
where is the attachment
Explanation:
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
The cells lie odjacent to the sieve tubes
Answer:
Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.Ibrah open a bottle of perfume infront of the room. After few minutes the smell of perfume reach the whole room. Explain why this happens
a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
What star is known as the "cold planet"?
Explanation:
OGLE-2005-BLG-390Lb.
PSR B1620-26 b. Surface Temperature: 72 Kelvin. ...
Neptune. Surface Temperature: 72 Kelvin. ...
Uranus. Surface Temperature: 76 Kelvin. ...
Saturn. Surface Temperature: 134 Kelvin. ...
Jupiter. Image Courtesy: NASA. ...
OGLE-2016-BLG-1195Lb. Surface Temperature: Unknown
A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?
A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J
=F×s×cosa=2×g×0,8×cos90°= 0
The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.
What is Work?Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.
Work = Force × Displacement
Force = Mass × Acceleration
Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.
Therefore, the correct option is A.
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The masses of two heavenly bodies are 2×10‘16’ and 4×10 ‘22’ kg respectively and the distance between than is 30000km. find the gravitational force between them ? ans. 2.668× 10-9N
[tex]F = 5.93×10^{13}\:\text{N}[/tex]
Explanation:
Given:
[tex]m_1= 2×10^{16}\:\text{kg}[/tex]
[tex]m_2= 4×10^{22}\:\text{kg}[/tex]
[tex]r = 30000\:\text{km} = 3×10^7\:\text{m}[/tex]
Using Newton's universal law of gravitation, we can write
[tex]F = G\dfrac{m_1m_2}{r^2}[/tex]
[tex]\:\:\:\:=(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)\dfrac{(2×10^{16}\:\text{kg})(4×10^{22}\:\text{kg})}{(3×10^7\:\text{m})^2}[/tex]
[tex]\:\:\:\:= 5.93×10^{13}\:\text{N}[/tex]
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
Which sequence shows the chain of energy transfers that create surface currents on the ocean?
Answer:
The correct answer is A. The sun is the energy source of the surface currents in the ocean
The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .
What is surface current ?Surface currents are currents that are located in the upper feet of the ocean , they are simply how water moves from one place to another . Pattern of surface current are determined by wind direction .
Surface currents are formed by global wind system that are fueled by energy from the sun . Because of heating effect of sun , the earth's atmosphere gets warmed up . As we know , warm air is lighter then cool air , it rises up and create low pressure near the equator because of this wind causes surface currents the ocean .
hence , The energy transfer starts from solar energy , then wind energy and finally wind energy is the cause of surface current .
learn more about surface current
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How can I solve this?
You have three capacitors of values 40 F, 10 F and 50 F. What would their equivalent capacitance (in F) be if they were connected in parallel with each other? Enter your answer as a number only, to one decimal place.
Explanation:
The equivalent capacitance of capacitors in parallel can be determined as
[tex]C_{eq} = C_1 + C_2 + C_3[/tex]
[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
This question is incomplete, the complete question is;
A bat flies towards a moth at 7.1 m/s while the moth is flying towards the bat at 4.4 m/s. The bat emits a sound wave of 51.7 kHz.
After enjoying a tasty meal of the first moth, the bat goes after another moth. Flying with the same speed and emitting the same frequency, this time the bat detects a reflected frequency of 55.5 kHz. How fast is the second moth moving
Answer:
the second moth is moving at 5.062 m/s
Explanation:
Given the data in the question;
Using doppler's effect
[tex]f_{moth[/tex] = f₀( [tex]v_{s[/tex] ± [tex]v_{observer[/tex] / [tex]v_{s[/tex] ± [tex]v_{source[/tex] )
f₁ = f₀( ([tex]v_{s[/tex] + v₂) / ( [tex]v_{s[/tex] - v₁ ) )
frequency reflected from the moth,
Now, moth is the source and the bat is the receiver
f₂ = f₁( ([tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) )
hence, f = f₀[ ( ( [tex]v_{s[/tex] + v₁ ) / ( [tex]v_{s[/tex] - v₂ ) ) ( ( [tex]v_{s[/tex] + u₂ ) / ( [tex]v_{s[/tex] - u₁ ) )
we know that, the velocity of sound [tex]v_{s[/tex] = 343 m/s.
given that v₁ and v₂ { velocity of bat } = 7.1 m/s, f₀ = 51.7 kHz and f = 55.5 kHz.
we substitute
55.5 = 51.7[ ( ( 343 + 7.1 ) / ( 343 - 7.1 ) ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ ( 350.1 / 335.9 ) ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 51.7[ 1.04227 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 = 53.885359 ( ( 343 + u ) / ( 343 - u ) ) ]
55.5 / 53.885359 = ( 343 + u ) / ( 343 - u )
1.02996 = ( 343 + u₂ ) / ( 343 - u )
( 343 + u₂ ) = 1.02996( 343 - u )
343 + u = 353.27628 - 1.02996u
u + 1.02996u = 353.27628 - 343
2.02996u = 10.27628
u = 10.27628 / 2.02996
u = 5.062 m/s
Therefore, the second moth is moving at 5.062 m/s
OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular
Answer:
option A
Explanation:
simple harmonic motion
Answer:
random motion I think not sure
Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)
Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance
Answer:
杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音
Explanation:
莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd
Answer:
B velocity
Explanation:
An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J
Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top of it as shown. The block w2 is attached to avertical wall by a string 6m long. If the coefficient of friction between all surface is 0.25 and the system is in equilibrium find the magnitude of the horizontal force applied to the lower block
The horizontal force applied to the block is approximately 1,420.84 N
The known parameters;
The mass of the block, w₁ = 400 kg
The orientation of the surface on which the block rest, w₁ = Horizontal
The mass of the block placed on top of the 400 kg block, w₂ = 100 kg
The length of the string to which the block w₂ is attached, l = 6 m
The coefficient of friction between the surface, μ = 0.25
The state of the system of blocks and applied force = Equilibrium
Strategy;
Calculate the forces acting on the blocks and string
The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N
The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N
Let T represent the tension in the string
The upward force from the string = T × sin(θ)
sin(θ) = √(6² - 5²)/6
Therefore;
The upward force from the string = T×√(6² - 5²)/6
The frictional force = (W₂ - The upward force from the string) × μ
The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25
The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)
∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]
[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]
The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N
Therefore;
The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68
The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]
The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P
Where;
P = The horizontal force applied to the block
P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84
The horizontal force applied to the block, P ≈ 1,420.84 N
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A boat is able to move at 7.6 m/s in still water. If the boat is placed on the south shore of a river (water current of 3.4 m/s [SE]), and the captain wants to head straight across to the north shore:
a) In what direction should the captain point the boat?
b) Calculate the time it will take to cross (the river is 212.0 m from the south to the north shore).
Answer:
I don't get it rewrite please
A boy walks from point C to point D which is 50 m apart. Then, he walks back to point C. what is his displacement of his whole journey ?
A.25 m
B.75 m
C.50 m
D.0 m
Answer: D. 0 m
Explanation:
Concept:
Here, we need to know the concept of displacement.
Displacement is defined to be the change in position of an object.
The difference between displacement and distance is the total movement of an object without any regard to direction, while displacement is the pure change of position.
If you are still confused, please refer to the attachment below for a graphical explanation.
Solve:
STEP ONE: the boy walks from point C to point D (a distance of 50 m)
C ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ D
50 m
STEP TWO: the boy walks from point D to point C (a distance of 50 m)
D ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ C
50 m
STEP THREE: find the displacement
The boy started with point C
The boy ended with point C
He did not change his position throughout the journey.
Therefore, his displacement is 0 m.
Hope this helps!! :)
Please let me know if you have any questions
Give examples of motion in which the directions of the velocity and acceleration vectors are (a) opposite, (b) the same, and (c) mutually perpendicular
Answer:
a) When moving body applies brake then velocity and acceleration would be in opposite direction
b) When body starts to increase velocity then velocity and acceleration would be in same direction
c) When body is circulating then velocity and acceleration would be perpendicular to each other
Explanation:
a) When body applies brake then its velocity starts decreasing, in this case its acceleration would try to stop the moving body. So direction of velocity would be same as direction of motion of body but direction of acceleration would be in opposite direction
b) When body starts to increase velocity, its acceleration would make the body to move faster. So direction of velocity would be the direction of motion of body and acceleration would also be in same direction
c) When body moves in circular path then its acceleration would be towards centre of circle and velocity would try to snap the body out of circle to straight line which in tangent to circle.
