The answer is C) kilogram
The SI unit in the given options is kilogram, the correct option is C.
What is SI unit?The most common unit of measurement is the standard unit. These units are standardized, which means that everyone understands an object's size, weight, and other properties.
Standard Units of Measurement have fixed values that cannot be changed.
The International System of Units, abbreviated SI in all languages and sometimes pleonastically as the SI system, is the modern version of the metric system and the world's most widely used measurement system.
The SI is a coherent system of units of measurement that begins with seven base units: the second (symbol s, the unit of time), the metre (m, length), the kilogram (kg, mass), the ampere (A, electric current), and so on.
Thus, the correct option is C.
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Lab report on velocity of sound
The current in a resistor is 2.0 A, and its power is 78 W. What is the voltage?
Answer:
39 volts
Explanation:
Use the equation [tex]P=VI[/tex]
[tex]78=V(2)[/tex]
[tex]V=39[/tex]
Someone with a force of 900 N does not float in a freshwater pool. To prevent them from sinking, 20 N must be applied in an upward fashion. What is their volume and average density?
Explanation:
The buoyant force [tex]F_B[/tex] is defined as
[tex]F_B = \rho_wgV[/tex]
where [tex]\rho_w[/tex] is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is [tex]997\:\text{kg/m}^3.[/tex] Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:
[tex]F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}[/tex]
Plugging in the values, we get
[tex]V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3[/tex]
Recall that the weight of an object in terms of its density and volume is given by
[tex]W = \rho gV[/tex]
Using the value for the volume above, we can solve for the density of the object as follows:
[tex]\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}[/tex]
[tex]\:\:\:\:\:= 44,798\:\text{kg/m}^3[/tex]
A 2.0 kg particle moving along the z-axis experiences the
force shown in (Figure 1). The particle's velocity is
3.0 m/s at x = 0 m.
A) At what point on the x axis does the particle have a turning point?
At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.
Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.
If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
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The amount of work done in example B is:
Answer:
Explanation:
20 n is an unknown amount
If that is supposed to be 20 N(ewtons)
then W = Fd = 20(15) = 300 J
Answer: it will be 300 newton meters
Explanation:
Swim swim swim swim swim swim swim swim swim swim swim swim.
A 400 g ball swings in a vertical circle at the end of
a 1.5-m-long string. When the ball is at the bottom
of the circle, the tension in the string is 10 N.
A) what is the speed of the ball at that point?
Answer:
a = 25 m/s2
Explanation:
A = f/m
A = Speed/Acceleration
F =‘Force
M = Mass
This question has two parts. First, answer part A. Then, answer part B. Part A: Which statement best summarizes the theme of the text? O A. You do not always recognize what is most valuable. O B. Keep your friends close, but your enemies closer. W O C. Fine possessions do not make a fine person. h 0 D. The best things come in small packages. Part B: Which evidence from the text best supports your answer in part A? O A. "The other animals of the forest bowed to him, and they often spoke of his antlers with admiration." B. "The Stag was so engrossed that he did not notice that a Lion had crept up alongside him." C. ". . . but the branches of some of the trees hung low, and vines curled around them." D. "The long legs that I hated would have saved me, but the antlers that I loved have led to my destruction!"
Answer: for part A Its A the ANSWER for part B its D
Explanation:
Answer:
a and b
Explanation:
What is a list of all the states of matter?
Answer:
3
Explanation:
state of matter are solid
liquid and
gases
Answer:
3
Explanation:
state of a matter are solid liquid and gas
A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg of the wateris in the liquid state. What is the pressure in the vessel, and what is the volume of the vessel
In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
[tex]x=\frac{m_{steam}}{m_{total}}[/tex]
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
[tex]x=\frac{2kg}{10kg} =0.20[/tex]
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
[tex]v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg[/tex]
This means that the volume of the container will be:
[tex]V=10kg*0.4727m^3/kg\\\\V=4.73m^3[/tex]
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A machine lifts a load of 100N through a vertical distance of 2m in seconds. What is the work done by machine?
Explanation:
In this chapter, we will study the important concepts of kinetic energy and the
closely related concept of work and power.
A- Kinetic Energy
Kinetic energy is a physical quantity, which is associated with the moving objects
and defined as:
K = ½ mv2
If the body is stationary (v=0), its kinetic energy is zero. The SI unit of kinetic
energy is kg.m2
/s2
or Joule (J), where 1 J=1 kg.m2
/s2
. Kinetic energy is a scalar
quantity.
B- Work
The work is defined as the ability to perform a force along a certain displacement.
There are different types of work as follows:
1- Work done by a constant force
The work done by the constant force F is given by the scalar product of the force F
and the displacement d.
WF = F.d = Fd cosθ
where θ is the angle between the force and displacement. The above equation means
that the work is the product of the displacement magnitude by component of the
force parallel to the displacement. Therefore, work is a scalar quantity (only
magnitude, no direction) and can be positive, negative, or zero. The SI unit of work
is (N.m) or joule (J) where 1 N.m = 1 J.
Special cases and remarks:
• If the angle between the force and displacement is zero (parallel), the work is
WF = F d (maximum work)
For the vertical part, W = (200 N) * (10 m) * cos (0 deg) = 2000 J. For the horizontal part, W = (50 N) * (35 m) * cos (0 deg) = 1750 J. The total work done is 3750 J (the sum of the two parts).
Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 475 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)
Why can't we feel all the action-reaction forces around us?
Answer:
This happens when the mass of one of the objects is very large and does not move.
Explanation:
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.7 cm from its equilibrium position?
A)The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
1. force constant k
2. total energy E
3. mass m
4. maximum velocity vmax
5. amplitude A
6. potential energy U at x
7. kinetic energy K at x
8. position x from equilibrium
B)What is the kinetic energy of the object on the spring when the spring is compressed 4.7 cm from its equilibrium position?
C)What is the potential energy U of the toy when the spring is compressed 4.7 cm from its equilibrium position?
Hi there!
Part A:
The only quantities explicitly given to us are:
3. mass (m)
4. Maximum velocity (vmax)
5. Amplitude (A)
8. Position x from equilibrium
Part B:
To solve, we must begin by calculating the force constant, 'k'.
We can use the following relationship:
[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]
We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:
[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]
[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]
Now, we can find the velocity at 4.7cm (0.047m) using the equation:
[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]
Plug this value into the equation for kinetic energy:
[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]
Part C:
The potential energy of a spring is given as:
[tex]U = \frac{1}{2}kx^2[/tex]
Find 'k' using the derived quantity above:
[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]
Now, calculate potential energy:
[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]
Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.150 rev/srev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.
What is the radius of the circle in which Jack travels? Treat him as a point mass.
Answer:
Explanation:
At the top of the arc, 3/4 of the acceleration of gravity is use to supply the necessary centripetal acceleration.
0.75g = ω²R
R = 0.75g/ω²
R = 0.75(9.81) / (0.15 rev/s)(2π rad/rev)²
R = 8.283006...
R = 8.28 m
One of the great challenges of cosmology today is to ---
A
determine the amount of matter in the Universe
B
find intelligent signals emanating from outer space
С
look backward in time to before the Big Bang
D
locate wormholes to help define the structure of the Universe
Answer:
Steady-state theory, in cosmology, a view that the universe is always expanding but maintaining a constant average density, with matter being continuously created to form new stars and galaxies at the same rate that old ones become unobservable as a consequence of their increasing distance and velocity of recession.
Cosmic inflation is a faster-than-light expansion of the universe that spawned many others. ... Cosmic inflation solves these problems at a stroke. In its earliest instants, the universe expanded faster than light (light's speed limit only applies to things within the universe).
How fast would a(n) 75 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 390 m/s
Answer:
Explanation:
½(75)v² = ½(0.008)390²
v² = (0.008)390²/75
v² = 16.224
v = 4.027...
v = 4.0 m/s
According to the table what was the hikers total displacement the graph has 4km 6km 4km 6km
Answer:
0
Explanation:
0 is the answer
How does friction help us in walking.
What is most likely the amount of energy available at a trophic level of primary consumers if the amount of energy available to secondary consumers in that food web is 200 kilocalories?
0 kilocalories
20 kilocalories
200 kilocalories
2,000 kilocalories
Answer:
200 kilocalories
Explanation:
A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct
Answer:
The race car will crash into the outer wall
Explanation:
max fr = μsN = 1.1 mg = 11 m
mv2/R = m(100)2/(400) = 25 m > fr
can someone please help no.1,2,3
Answer:
12345678910i don't know
a coconut falls from the top of a tree and takes 3.5 seconds to reach the ground. How tall is the tree?
Hello!
To solve, we can begin by using the kinematic equation:
[tex]d = v_it + \frac{1}{2}at^2[/tex]
Where:
vi = initial velocity (m/s)
t = time (s)
a = acceleration (in this case, due to gravity. g = 9.8 m/s²)
Since the object falls from rest, the initial velocity is 0 m/s.
[tex]d = \frac{1}{2}at^2[/tex]
Plug in the given values:
[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]
Please help. I'm mot sure what I need to do first...
Answer:
0.80 kN
Explanation:
Hope you understood it
The_____ scale is called an absolute temperature scale, and its zero point is called absolute zero.
Kelvin
Fahrenheit
Celsius
The Celcius Scale is called an absolute Temperature scale,and it's zero point called absolute zero
_______________
A constant horizontal force of 30.0 N is exerted by a string attached to a 5.0 kg block being pulled across a tabletop. The block also experiences a frictional force of 5.0 N due to contact with the table. What is the horizontal acceleration of the block?
Answer:
A 5.00- kg block is placed on top of a 10.0 -kg block (Fig. P5.68). A horizontal force of 45.0 N is applied to the 10.0-kg block, and the 5.00- kg block is tied to the wall. The coefficient of kinetic friction between all surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks.
(b) Determine the tension in the string and the magnitude of the acceleration of the 10.0-kg block.
The horizontal acceleration of the block is 5 m/s².
To calculate the horizontal acceleration on the block, we use the formula below.
Formula:
ma = (F-F')............... Equation 1Where:
m = mass of the blocka = Horizontal acceleration of the blockF = Horizontal force exerted on the stringF' = Frictional forceMake "a" the subject of the equation.
a = (F-F')/m............... Equation 2Substitute these values into equation 2
F = 30 NF' = 5 Nm = 5.0 kgSubstitute these values into equation 2
a = (30-5)/5a = 25/5a = 5 m/s²Hence the horizontal acceleration of the block is 5 m/s².
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an observer sees two spaceships flying apart with speed .99c. What is the speed of one spaceship as viewed by the other
Answer:
V2 = (V1 - u) / (1 - V1 u / c^2)
V1 = speed of ship in observer frame = .99 c to right
u = speed of frame 2 = -.99 c to left relative to observer
V2 = speed of V1 relative to V2
V2 = (.99 - (-.99 ) / (1 - .99 (-.99)) c
V2 = 1.98 / (1 + .99^2) c = .99995 c
how does the structure of compounds determines the properties of the compounds?
Answer:
The chemical structure of the molecule is responsible for each of these characteristics. The chemical structure is comprised of the bonding angle, the kind of bonds, the size of the molecule, and the interactions that occur among the molecules. Even little changes in the chemical structure of a molecule may have a significant impact on the characteristics of the substance.
Explanation:
Hope it helps:)
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.
Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.
The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 20 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
1: m = 20 kg, r = 0.25·R
2: m = 10 kg, r = 1.0·R
3: m = 10 kg, r = 0.25·R
4: m = 15 kg, r = 0.75·R
5: m = 10 kg, r = 0.5·R
6: m = 40 kg, r = 0.25·R
According to the principle of conservation of angular momentum, we have;
[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]
The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]
Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of m·r² increases, the value of [tex]\omega _f[/tex] decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =
10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].
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