Answer:
Her average speed is: 100 m/min
Explanation:
Recall that the formula for average speed is given by:
Speed = Distance / time
Then in our case, this is
Speed = 300 m / 3 min = 100 m/min
Objects 1 and 2 attract each other with a gravitational force
of 18.0 units. If the mass of Object 1 is halved AND the
mass of object 2 is tripled, then the new gravitational force
will be units.
Answer:
the answer is 72.0 units:) thank me later!!!!!!!
In an element's square on the periodic table, the number with the greatest numerical value represents the
number of neutrons.
O number of electrons.
88
atomic nunner.
O atomic mass.
Answer:
0 atomic mass because it widens as the numerical value represents
Answer: ATOMIC MASS
Explanation:
I got it right
Power is measured in unit of Joules per second or
a) seconds
b) hertz
c) joules
d) watts
e) newtons
A force of 6.7 N acts on a 30 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.
Answer:
(a) 0.748 J
(b) 2.245 J
(c) 3.74 J
(d) 4.482 W
Explanation:
(a) Work done W = Force × distance
W = F×d,
Where d = 1/2(at²)
Therefore,
W =1/2(F×at²)................ Equation 1
Where a = acceleration, t = time.
But,
a = F/m...................... Equation 2
Where m = mass.
Substitute equation 1 into equation 2
W = 1/2(F²t²/m)................. Equation 3
Given: F = 6.7 N, t = 1 s, m = 30 kg
Substitute into equation 3
W₁ = 1/2(6.7²×1²/30)
W = 0.748 J.
(b) Similarly,
The work done in the second seconds is
Where t₂ = 2 s
W₂ = 1/2(F²t₂²/m)- W₁
W = 1/2(6.7²×2²/30)-0.748
W = 2.245 J
(c) The work done in the third seconds is
Where t₃ = 3 s
W₃ = 1/2(F²t₃²/m)-(W₂+W₃)
W = 1/2(6.7²×3²/30)-(2.993)
W = 3.74 J.
(d) P = Fv ............... Equation 4
Where v = velocity.
and,
v = at..................... Equation 5
Substitute equation 5 into equation 4
P = Fat................... Equation 6
Given: F = 6.7 N, a = 6.7/30 = 0.223 m/s², t = 3 s
Substitute into equation 6
P = 6.7×0.223×3
P = 4.482 W.
A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calculate the angular acceleration
Answer:
2 rad/s²
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 1 Kg
Radius (r) = 0.2 m
Angular speed (w) = 20 rad/s
Time (t) = 10 s
Angular acceleration (a) =?
Angular acceleration is defined as:
Angular acceleration (a) =Angular speed (w) / time
a = w/t
With the above formula, we can obtain the angular acceleration of the sphere as follow:
Angular speed (w) = 20 rad/s
Time (t) = 10 s
Angular acceleration (a) =?
a = w/t
a = 20 / 10
a = 2 rad/s²
Thus, the angular acceleration of the sphere is 2 rad/s²
Three charges, each separated by 100 m from adjacent charges, are located along a horizontal line: a -3.00 C charge on the left, a 2.00 C charge in the middle, and a 1.00 C charge on the right. What is the resultant force on the 1.00 C charge due to the other two
Answer:
F= 11.25*10⁵ N to the right.
Explanation:
Assuming that the three charges can be treated like point charges, they must obey Coulomb's Law.Due to the linearity of this Law, we can use superposition in order to find the resultant force on the 1.00 C charge due to the other two.First, we find the force that the -3.00 C charge (located 200 m to the left) exerts on the 1.00 C, as follows:[tex]F_{13} = \frac{K*q_{1}*q_{2}}{r_{13} ^{2} } = \frac{9e9*(-3.00C)(1.00C)}{(200m)^{2}} = -6.75e5 N (1)[/tex]
Then, in the same way, we can find the force that the 2.00 C exerts on the 1.00 C charge, located 100 m away to the left:[tex]F_{23} = \frac{K*q_{3}*q_{2}}{r_{23} ^{2} } = \frac{9e9*(2.00C)(1.00C)}{(100m)^{2}} = 18e5 N (2)[/tex]
Since both vectors are on the same line, their sum is directly the algebraic sum, as follows:F₃ = F₁₃ + F₂₃ = -6.75*10⁵ N + 18.00*10⁵ N = 11.25*10⁵ N to the right, assuming this direction as positive.A person standing on a building ledge throws a ball vertically from a launch position 55 m above the ground. It takes 2.0 s for the ball to hit the ground.
With what initial speed was the ball thrown?
Answer:
Vo = 17.69 [m/s]
Explanation:
To solve this problem we must use two equations of kinematics.
[tex]v_{f}^{2} =v_{o}^{2} +2*g*h\\v_{f}=v_{o}+g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity [m/s]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 55 [m]
t = time = 2 [s]
Now we replace the gravity acceleration into the second equation:
[tex]v_{f}=v_{o}+9.81*2\\v_{f}=v_{o}+19.62[/tex]
And then into the first equation:
[tex](v_{o}+19.62)^{2}=v_{o}^{2}+2*9.81*55\\v_{o} ^{2}+2*v_{o}*19.62+384.94=v_{o}^{2} + 1079.1\\39.24*v_{0}=694.16\\v_{o}=17.69[m/s][/tex]
The initial speed at which the ball is thrown upward is 17.7 m/s.
According to the question the initial position of the ball, y = 55 m, and the final position of the ball is y' = 0 m. We have assumed upward direction as positive direction and downward as negative direction.
so the total displacement:
d = y'-y = 0 - 55
d = -55 m
now applying the second equation of motion:
d = ut - (1/2)gt²
where t = 2s ( given ) and g = 9.8 m/s².
-55 = 2u - 0.5×9.8×4
-55 = 2u - 19.6
u = -17.7 m/s
the negative sign indicated that the initial velocity is opposite to the direction of displacement.
This means the initial velocity is upward as it should be.
Therefore, the initial speed of the ball is 17.7m/s
Learn more:
https://brainly.com/question/24018491
what is physical quantity
Answer:
physical quantity is a property of a material or system that can be qualified by measurement
[tex] \huge\mathfrak{ANSWER}[/tex]
the quantities by meals of which we describe the law of physics are known as physical quantities.
in order to measure a physical quantity we as you my certain magnitude of this quantity a standard and call its unit of that quantity to express the measurement of any physical quantity two thing must be mentioned.
the unit in which the quantity is measured.the numerical value which do not the magnitude of that quantity in term of the choosen unit.For Example -:
when we say that the length of a road is 20 m its mean that the unit of length is matter and the length of rod 20 time magnitude of metre.in other number the numerical value of physical quantities inversely proportional to its unit if N1 and N2 be the numerical value of a physical quantity into different unit u1 and u2 then[tex] \bold{ n_1 (u_1) = n_2(u_2)}[/tex]
A sports car accelerates from rest to 42 m/s in 6.2 s, what is its acceleration? (6.8 m/s) with proof please
Answer:
Explanation:
As the car Started from rest means that Initial Velocity "Vi = 0 m/s" and final Velocity is given "Vf = 42m/s". Time is given "t = 6.2s"
Acceleration is required a =?
Use Formula;; a = [tex]\frac{Vf-Vi}{t}[/tex]
a = [tex]\frac{42-0}{6.2}[/tex]
a = 6.7741 m/s² ≈ 6.8 m/s²
Mark me as brainliest if you got it...
If the evaporator outlet temperature on an r410A system is 50f and the evaporator superheat is 10f, what is the evaporating pressure of the refrigerant in the system
Answer:
So, the evaporating pressure of the R410A = 118 psig
Explanation:
Solution:
For R410A system:
Data Given:
Evaporator Outlet Temperature = 50°F
Evaporator Superheat = 10°F
Required:
Evaporating Pressure in the system = ?
For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.
Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.
Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat
Evaporator inlet Temperature = 50°F - 10°F
Evaporator inlet Temperature = 40°F
Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.
So, the evaporating pressure of the R410A = 118 psig
Which is the BEST scientific explanation for temperature? *
A)how hot something is
B)how much kinetic energy an object has.
C)how far away molecules are from each other.
D)the average kinetic energy of the molecules.
Answer:
B
Explanation:
bc all the other answers seem way to simple and boring
Objects 1 and 2 attract each other with a gravitational force
of 72.0 units. If the distance separating Objects 1 and 2 is
changed to four times the original value (i.e., quadrupled),
then the new gravitational force will be units.
Answer:
The new gravitational force will be of 4.5 units
Explanation:
Recall that the formula for the gravitational force between two objects of masses m1 and m2 separated by a distance d is given by:
[tex]F_g=G\,\frac{m1*m2}{d^2}[/tex]
in our case, we are told that such gives 72 units of force:
[tex]F_g=G\,\frac{m1*m2}{d^2} =72[/tex]
Then we change the distance between the objects to 4 times the original (4 * d), such will produce a new gravitational force Fg':
[tex]F_g'=G\,\frac{m1*m2}{(4*d)^2} =G\,\frac{m1*m2}{16*d^2} = \frac{1}{16} *G\,\frac{m1*m2}{d^2}=\frac{1}{16} *\,72=4.5[/tex]
Therefore the new gravitational force would be of 4.5 units
Use the following information about earth and its moon to determine the distance between them
answer is the SECOND one on Edge
3.8 x 10^8m
the chart would’ve been helpful though.
Answer:
B
Explanation:
edge2020
A helium-neon laser (λ = 633 nm) illuminates a single slit and is observed on a screen 1.50 m behind the slit. The distance between the first and second minima in the diffraction pattern is 3.75 mm .
Required:
What is the width (in mm) of the slit?
Answer:
d = 0.25 mm
Explanation:
The position that the Minimal fall in single silt diffraction pattern can be written as
y=λmD/d ..........eqn(1)
(y2 - y1 )= difference of y that exist between first and second minima
D= distance of slit from the screen=1.50m
d= width of the slit
λ= 633 nm = 633×10^-9m
y= position of pth minimal
eqn(1) can be written as
(y2 - y1)= λmD/d .........eqn(2)
If we substitute the given values we have
(y2 - y1 )= [ (2×633×10^-9 × 1.50) /d] -[ (1×633×10^-9 × 1.50) /d]
Simplyfying
(y2 - y1)=(9.495×10^-7)/d
But The distance between the first and second minima in the diffraction pattern = 3.75, which implies that (y2 - y1) = 3.75mm=
✓ we can substitute (y2 - y1)= 3.75mm=3.75 ×10^-3 into expression above
3.75 ×10^-3 =(9.495×10^-7)/d
The we can make "d" subject of the formula
d=(9.495×10^-7)/ (3.75 ×10^-3)
d=0.00025m
d=0.25×10^-3 m
d=0.25 mm
Hence, the width (in mm) of the slit is 0.25mm
You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 KPa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 KPa .
Find the net downward force on the tank's flat bottom, of area 2.15 m2 , exerted by the water and air inside the tank and the air outside the tank.
Answer: F = 6262.2 kN
Explanation: Pressure is defined as force per area. But pressure varies according to the depth of a fluid: in air, it decreases the higher the altitude, while in water, it increases the deeper you go.
So, at the bottom of the tank, besides the pressure of air inside the tank and air outside the tank, there is pressure of water due to its depth.
Pressure due to the depth is calculated as
[tex]P=h.\rho.g[/tex]
h is the depth in m
ρ is density of the fluid, in this case is water, so ρ = 997 kg/m³
g is acceleration due to gravity, which, in this case, is 3.71 m/s²
Then, pressure at the bottom of the tank due to variation in depth is
[tex]P=14.4(997)(3.71)[/tex]
P = 53263.73 Pa or 53.26 kPa
Assuming positive referential is downward, all pressures at the bottom point down, so total or resultant pressure is:
[tex]P_{r}=P_{1}+P_{2}+P_{3}[/tex]
[tex]P_{r}=150+88+53.26[/tex]
[tex]P_{r}=[/tex] 291.26 kPa
At last, pressure is force per area:
[tex]P=\frac{F}{A}[/tex]
[tex]F=P.A[/tex]
[tex]F_{r}=P_{r}.A[/tex]
[tex]F_{r}=291.26.10^{3}(2.15)[/tex]
[tex]F_{r}=[/tex] 626209 N or 626.2 kN
At the cylindrical tank's flat bottom, net force has magnitude 626.2 kN.
A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 50 percent. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate be higher or lower than this value
Answer: 600MW
Explanation:
First and foremost, we have to calculate the rate of heat that is supposed to the power plant. This will be:
= 600 / 50%
= 600 / 0.5
= 1200 MW
The rate of heat transfer to the river water will then be:
= 1200 - 600
= 600MW
Therefore, the actual heat transfer rate is thesame as this value as we got 600MW.
convert 7.6 meters to milemeters
1 m = 1000 mm
So, 7.6 m = 7.6 × 1000 mm = 7600 mm
Place the lunar phases in the correct order. !QUICK! (I WILL GIVE BRAINLEST!!!!!)
Answer:
Here:
Explanation:
These eight phases are, in order, new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, third quarter and waning crescent. The cycle repeats once a month (every 29.5 days).
Calculate the increase in the internal energy when 1000 g of boiling water at 212° F is converted into steam at the same
temperature. Atmospheric pressure is 1.013 X 10 Pa and the latent heat of vaporization is 2.260 MJ kg
-1
Answer:
The increase in the internal energy is 1.840 x 10⁶ J.
Explanation:
Given;
mass of the water, m = 1000 g = 1 kg
temperature of the boiling water, t = 212 ° F = 100 ° C
latent heat of vaporization, L = 2.260 MJ/kg = 2.260 x 10⁶ J/kg
The internal energy of the boiling water is calculated as;
Q₁ = mcΔθ
where;
c is specific heat capacity of water, = 4200 J/kg.⁰C
Δθ is change in temperature = 100 ° C
Q₁ = 1 x 4200 x 100
Q₁ = 420,000 J
The internal energy of the vaporized steam is calculated as;
Q₂ = mL
Q₂ = 1 x 2.260 x 10⁶
Q₂ = 2,260,000 J
The increase in the internal energy is calculated as;
ΔQ = Q₂ - Q₁
ΔQ = 2,260,000 J - 420,000 J
ΔQ = 1,840,000 J
ΔQ = 1.840 x 10⁶ J
Therefore, the increase in the internal energy is 1.840 x 10⁶ J.
A caterpillar crawls 20 feet in 5 minutes. What is the caterpillars crawling speed?
Equation needed:
How does friction impact work? (that's the whole question)
The force of water flowing over a dam spillway is to be measured in a geometrically similar small-scale laboratory model. The width of the prototype is 30 m and the width of the model is 2 m. If the force on the model spillway is measured to 10 kN, the expected force on the prototype spillway for dynamically similar conditions is:
Answer:
Jay shree krishna j to the same time kr rhe h u have UCC fm rob and studing a few minutes ago
I need help with these four problems.. anyone??
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the: A. propagation speed of the radiation. B. wavelength of t
Answer:
D.number of photons emitted.
Explanation:
These are the options for the question
A.propagation speed of the radiation.
B.wavelength of the radiation.
C.polarization of photons emitted.
D.number of photons emitted.
.
Electromagnetic energy of any radiation is proportional to the photons present. And we know that intensity is ratio of energy and unit time.
Hence, The intensity of the radiation emitted by the oxygen sensor is directly proportional to the number of photons emitted
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the number of photons emitted. The correct answer is D.
The intensity of radiation refers to the amount of energy carried by the radiation per unit of time and unit of area. In the context of the oxygen sensor, the intensity of the radiation emitted by the sensor is directly proportional to the number of photons emitted.
The more photons emitted, the higher the intensity of the radiation. The intensity is not directly related to the propagation speed, wavelength, or polarization of the photons emitted, but rather the quantity or number of photons being emitted.
Therefore, The intensity of the radiation emitted by the oxygen sensor is directly proportional to the number of photons emitted. The correct answer is D.
To know more about the radiation emitted:
https://brainly.com/question/3745068
#SPJ6
The complete question is:
The intensity of the radiation emitted by the oxygen sensor is directly proportional to the
A. propagation speed of the radiation.
B. wavelength of the radiation.
C. polarization of photons emitted.
D. number of photons emitted.
Which ray diagram demonstrates the phenomenon of absorption?
tum
Answer:
hjbyvtf ghbj
Explanation:
uhgbnm,likjh
a mechanic uses a hydraulic lift to raise a 1,200 kg car 0.50 m off the ground . how much work does the lift do on the car
Answer:
The work done is 5880 J
Explanation:
Recall the formula for work as force times distance, therefore in this case the force is that oppose to the gravitational force on the car(of magnitude m * g), and the distance is 0.5 meters.
Then:
Work = 1200 * 9.8 * 0.5 J = 5880 J
The Moon has a mass of 7.36 1022 kg and a radius of 1.74 106 m. (a) What is the acceleration due to gravity on the Moon
Answer:
1.622m/s²
Explanation:
Given the following
mass of 7.36 × 10²² kg
Radius r = 1.74×10^6m
Gravitational constant G = 6.67×10^-11
Acceleration due to gravity us expressed as:
g = GM/r²
Substitute the given values into the formula
g = 6.67×10^-11×7.36 × 10²²/(1.74×10^6)²
g = 49.0912×10¹¹/3.0276×10¹²
g = 16.22×10^{11-12}
g = 16.22×10^-1
g = 1.622m/s²
Hence the acceleration due to gravity on the Moon is 1.622m/s²
Two isotopes of carbon, carbon-12 and carbon-13, have masses of 1.993 10^-26 kg and 2.159 10^-26 kg, respectively. These two isotopes are singly ionized ( e) and each is given a speed of 7.50 10^5 m/s. The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.9000 T.
Required:
Determine the spatial separation between the two isotopes after they have traveled through a half-circle.
Answer:
0.0174 m
Explanation:
Given that:
mass of carbon 12 [tex]m_{c \ 12} = 1.993 \times 10^{-26} \ kg[/tex]
mass of carbon 13 [tex]m_{c \ 13} = 2.159 \times 10^{-26} \ kg[/tex]
Speed V = [tex]7.50 \times 10^ 5 \ m/s[/tex]
[tex]q = 1.6 \times 10^{-19 } \ C[/tex]
B = 0.9000 T
[tex]R_1 = \dfrac{ m_{c \ 12} \ v}{ qB}[/tex]
[tex]R_1 = \dfrac{ 1.993 \times 10^{-26} \ 7.50 \times 10^5}{ 1.6 \times 10^{-19} \times 0.90000}[/tex]
[tex]R_1 =0.1038 \ m[/tex]
[tex]R_2 = \dfrac{ m_{c \ 13} \ v}{ qB}[/tex]
[tex]R_2= \dfrac{2.159 \times 10^{-26} \ 7.50 \times 10^5}{ 1.6 \times 10^{-19} \times 0.90000}[/tex]
[tex]R_2 = 0.1125 \ m[/tex]
The spatial separation (D) = [tex]2R_2 - 2R_1[/tex]
[tex]D = 2(0.1125 \ m) - 2(0.1038 \ m)[/tex]
D = 0.0174 m
4. What is the mass of the block
of iron illustrated below?
Ty in advance
Henry is researching the efficacy of a new drug for treating agoraphobia (fear of public places). His assistant, who is unaware of the study’s purpose, randomly assigns half of his 200 subjects a 30 day supply of the drug (Fearnot) and the other half a 30 day supply of a placebo (i.e., sugar pill). A month later the subjects are given a questionnaire to rate their fear of being in public. Henry finds that there is no difference between the two groups in their level of fear. How many subjects participated in this study?
