Answer:
7.69km/hr due east
Explanation:
Given parameters:
Displacement due west = 3km
Displacement due east = 2km
Time taken = 8 minutes
Unknown:
Average velocity = ?
Solution:
Average velocity is the rate of change of displacement with time.
Velocity = [tex]\frac{displacement}{time }[/tex]
Displacement is deals with the start and the end;
Here, displacement = 3km - 2km = 1km due east
Time = 8min
60min = 1hr
8min = [tex]\frac{8}{60}[/tex] = 0.13hr
Average velocity = [tex]\frac{1km}{0.13hr}[/tex] = 7.69km/hr due east
The
of an object shows how fast the object is moving at a specific
Answer:
Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.
40 POINTS!
A 2,200 kg SUV is traveling at 25 m/s. What is the magnitude of its momentum?
A. 55,000 kg·m/s
B. 550 kg·m/s
C. 2,200 kg·m/s
D. 88 kg·m/s
Answer:55,000 kg•m/s
Explanation:
A student determines the density ρ of steel by taking measurements from a steel wire
Mass- 6.2 +-0.1g
Length- 25.0 +-0.1m
Diameter- 2.00 +-0.01mm
He uses the equation ρ= 4m/πd^2l
What is the percentage uncertainty in his calculated value of density ?
Answer:
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
Explanation:
We can estimate the absolute uncertainty by the definition of total differential. That is:
[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)
Where:
[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].
[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.
[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.
[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.
[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.
[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.
[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.
Partial derivatives are, respectively:
[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)
[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)
[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)
And we expand (1) as follows:
[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]
[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)
If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:
[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]
[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]
And the expected density is:
[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)
[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]
[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]
The percentage uncertainty in his calculated value of density is:
[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)
If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:
[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]
[tex]\%e = \pm 0.713\,\%[/tex]
The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].
Grade 8 Science Admin. May 2018 Released
37 A physics teacher performed a demonstration for a science class by pulling a crate across the
floor and measuring the force with a spring scale. While she pulled, a student measured the
acceleration of the crate with a handheld electronic device. The results of three trials are
shown below
Motion
F = 900 N
Actual Acceleration
of Crate
Trial
1
140 kg
Acceleration
(m/s)
0.36
0.34
0.38
0.36
2
3
Average
The teacher asked the class to calculate the acceleration of the crate based on the crate's
mass and the force she applied. What conclusion can be made about the difference between
the calculated acceleration and the actual acceleration that occurred in the trials?
A Another force in the direction of the motion produced a lower acceleration than calculated.
B An opposing force caused by friction produced a lower acceleration than calculated.
C Another force in the direction of the motion produced a higher acceleration than
calculated.
D An opposing force caused by friction produced a higher acceleration than calculated,
Answer:
B. An opposing force caused by friction produced a lower acceleration than calculated
Explanation:
Answer:
An opposing force caused by friction produced a lower acceleration than calculated.
Explanation:
A dart is thrown horizontally at a target's center that is 5.00\,\text m5.00m5, point, 00, start text, m, end text away. The dart hits the target 0.150\,\text m0.150m0, point, 150, start text, m, end text below the target's center.What was the initial horizontal velocity of the dart?
Answer:
Explanation:
Given
the range = 5.00m (distance moved in horizontal direction)
Height = 0.150m
Required
Initial velocity of the dart
Using the formula for calculating range of a projectile;
R = U√2H/g
5 = U√2(0.15)/9.8
5 = U√0.0306
5 = 0.1749U
U = 5/0.1749
U = 28.59m/s
Hence the initial horizontal velocity of the dart is 28.59m/s
Answer:
28.59
Explanation:
khan academy
Gravitational potential energy is potential energy based on what?
a. temperature
b.height
c. length
d. volume
Answer:
Height
Explanation:
the more height the more gravitational energy
what do humming birds eat? and how do they get their food?
can a chemical change be undone?
Answer:
no
Explanation:
because it will bind together in the molecular level
5,000 joules of thermal energy were applied to 1-kg aluminum bar. What was the temperature increase?
Answer:
ΔT = 4.9°C
Explanation:
The thermal energy of the bar can be given as follows:
Thermal Energy = mCΔT
where,
m = mass of bar = 1 kg
C = specific heat capacity of aluminum = 1020 J/kg.°C
ΔT = Change in Temperature = ?
Therefore,
5000 J = (1 kg)(1020 J/kg.°C)ΔT
ΔT = (5000 J)/(1020 J/°C)
ΔT = 4.9°C
A stone is thrown vertically upwards. It takes 2 seconds to reach it highest point the acceleration due to gravity is 10m/s. With what velocity must it be thrown ?
Answer:
The velocity the stone must be thrown with is 20 m/s.
Explanation:
Given;
time taken for the stone to reach it highest point, t = 2 s
acceleration due to gravity, g = 10 m/s²
The maximum height reached by the stone at the given time is calculated as;
[tex]h = \frac{1}{2}gt^2\\\\h = \frac{1}{2}*10*2^2\\\\h = 20 \ m[/tex]
Determine the initial of the stone;
v² = u² + 2gh
where;
v is the final velocity of the stone at maximum height = 0
u is the initial velocity of the stone = ?
0 = u² - 2(10 x 20)
0 = u² - 400
u² = 400
u = √400
u = 20 m/s
Therefore, the velocity the stone must be thrown with is 20 m/s.
For a substance to change phases, the amount of Internal energy must change. Water exists in three phases: liquid, solid (ce), and gas (water vapor). Which of the following lists the
phases in order of increasing total energy?
O
gas, liquid, solid
solid, gas, liquid
solid, liquid, gas
oliquid, gas, solld
Convert 318 meters per second to kilometers per hour
Answer:
1144.8
Explanation:
1 meter per second is equal to 3.6 kilometers per hour
Please help I will give brainliest
Answer:
ask questions that the findings bring up
Explanation:
hope this helps :)
Answer:
ask questions that the findings bring up
!!ill give brainliest!
Explain why !
PLEASE HELP ME QUICKLY!!!!!!!!!!!!!!
Answer:
I think thats the correct answer
Explanation:
Hand washing is important especially in this time of pandemic
!!please help !!!!! I’ll give brainliest
Answer:
45 degrees
Explanation:
a projectile travels the farthest when it is launches at the angle of 45 degree.
A box sitting still on the ground by itself has a Normal Force of 700N, what is the mass? (gravity’s acceleration is 9.80 m/s²)
Answer:
Mass, m = 71.42 kg
Explanation:
Given that,
Normal force acting on a box, F = 700 N
We need to find the mass of a box. Let it is m. Normal force acting on an object is balanced by its weight such that,
F = mg
Where m is the mass of the box
[tex]m=\dfrac{F}{g}\\\\m=\dfrac{700\ N}{9.8\ m/s^2}\\\\m=71.42\ kg[/tex]
So, the mass of the box is 71.42 kg.
A system has both thermal energy and chemical potential energy. According
to the law of conservation of energy, which statement describes the total
energy of this system?
O A. It will decrease over time as chemical potential energy is
transformed into thermal energy.
B. It will decrease as thermal energy is transferred out of the system.
C. It will remain constant because the total amount of thermal energy
plus chemical potential energy stays the same.
OD. It will increase as thermal energy is created from chemical
potential energy
Answer:
C. It will remain constant because the total amount of thermal energy plus chemical potential energy stays the same.
Explanation:
According to the law of conservation of energy, the total energy of this system will remain constant because the total amount of thermal energy plus chemical potential energy stays the same.
The law of conservation of energy states that "energy is neither created nor destroyed in system but transformed from one form to another". Total energy in a system will always be conserved. It will remain constant.Newton's first law of motion states than an object's motion will not change
unless?
Answer:
it is hit by an external force
Explanation:
A 7.40-kg object initially has 347 J of gravitational potential energy. Then an elevator lifts the object a distance of 20.6 m above its previous position. How much work did the elevator perform on the object?
Answer:
W = 1493.9 J = 1.49 KJ
Explanation:
The work done by the elevator on the object will be equal to the gain in is potential energy:
W = ΔP.E
W = mgΔh
where,
W = Work = ?
m = mass of object = 7.4 kg
g = 9.8 m/s²
Δh = gain in height = 20.6 m
Therefore,
W = (7.4 kg)(9.8 m/s²)(20.6 m)
W = 1493.9 J = 1.49 KJ
how we will solve this question?
36.45cL=______=μL
Answer:
0.364
I believe... Good luck!
A stone is thrown at an angle of 34.0° above the horizontal from the top edge of a cliff with an initial speed of 18.3 m/s. A stopwatch measures the stone's trajectory time from the top of the cliff to the bottom at 2.6 s. What is the height of the cliff?
Answer:
Explanation:
Using the formula for calculating the maximum height in projectile
H = u²sin²theta/2g
u is the initial velocity = 18.3m/s
g is the acceleration due to gravity = 9.8m/s²
theta = 34°
Substitute
H = u²sin²theta/2g
H = 18.3²(sin34)²/2(9.8)
H = 334.89(0.5592)²/19.62
H = 5.337m
Hence the height of the cliff is 5.337m
What must follow the division of the nucleus to complete the process of binary fission
Answer:
Replication of the DNA must occur. Segregation of the "original" and its "replica" follow. Cytokinesis ends the cell division process. Whether the cell was eukaryotic or prokaryotic, these basic events must occur.
An archer stands on the ground and fires an arrow at a target. A second archer stands at the top of a building and holds an arrow in his hand. Which arrow has more potential energy?
In the same scenario described in question 7, which arrow has more kinetic energy? Explain.
Answer:
The arrow of the second archer standing on a high building will have more potential energy.
The arrow of the first archer standing on ground will have more kinetic energy.
Explanation:
POTENTIAL ENERGY:
The potential energy of an object depends upon its height, as given in the formula:
P.E = mgh
Hence, the arrow with the greater height will have more potential energy.
Therefore, the arrow of the second archer standing on a high building will have more potential energy.
KINETIC ENERGY:
The kinetic energy of an object depends upon its speed, as given in the formula:
K.E = (1/2)mv²
Hence, the arrow with the greater speed will have more kinetic energy energy. Since, the second arrow is stationary, it will have zero kinetic energy. But, the first arrow will have some K.E due to its speed.
Therefore, the arrow of the first archer standing on ground will have more kinetic energy.
(a) The second archer standing at the top of the building will have more potential energy.
(b) The first archer on the ground level will have the more kinetic energy.
The potential energy of an object is the energy possessed by the object by virtue its position above above the ground.
P.E = mgh
where;
h is the height above the groundThus, the second archer standing at the top of the building will have more potential energy.
(b) The kinetic energy of an object is the energy possessed by the object due to its motion.
K.E = ¹/₂mv²
where;
v is the speed of the objectAn object in motion, has zero velocity at maximum height but maximum velocity on the ground level.
Thus, the first archer on the ground level will have the more kinetic energy.
Learn more here:https://brainly.com/question/21866017
41. A statue weighs 1,000N and exerts a pressure of 20,000 Pa. How big is
the base of the statue in square meters?
please help
Answer:
The answer is 0.05 m²Explanation:
The area of the base of the statue can be found by using the formula
[tex]a = \frac{f}{p} \\ [/tex]
f is the force
p is the pressure
From the question we have
[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]
We have the final answer as
0.05 m²Hope this helps you
An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of the volume is occupied by liquid water and the rest by vapor. Now an electric resistor placed in the tank is turned on, and the tank is observed to contain saturated water vapor after 20 min. Determine:
(a) the volume of the tank
(b) the final temperature
(c) the electric power rating of the resistor
Solution:
Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg
Temperature = 200 °C
And 25% of the volume by liquid water is steam.
State 1
[tex]$m=\frac{V}{v}$[/tex]
[tex]$m=m_f+m_g$[/tex]
[tex]$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$[/tex]
[tex]$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$[/tex] (taking the value of [tex]$v_g$[/tex] and [tex]$v_g$[/tex] at 200°C )
[tex]$V=6.304 \times 10^{-3}$[/tex]
Now quality of vapor
[tex]$x=\frac{m_g}{m}$[/tex]
[tex]$=3.377 \times 10^{-3}$[/tex]
Internal energy at state 1 can be found out by
[tex]$u_1=u_f+xu_{fg}$[/tex]
[tex]$=850.65+3.377\times10^{-3}\times 1744.65$[/tex]
= 856.54 kJ/kg
After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor [tex]$v_2=v_g \text { and }\ x=1$[/tex]
Tank is rigid, so volume of tank is constant.
[tex]$v_g=v_2=\frac{V}{m}$[/tex]
[tex]$v_g=\frac{6.304\times 10^{-3}}{1.4}$[/tex]
[tex]$v_g=4.502 \times 10^{-3} \ m^3 /kg$[/tex]
Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature
[tex]$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$[/tex]
= 369.11° C
Internal energy at state 2
[tex]$u_2=2154.9 \ kJ/kg$[/tex]
Now power rating of the resistor
[tex]$P=\frac{m(u_2-u_1)}{t}$[/tex]
[tex]$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$[/tex]
= 1.51 kW
Help on this pls. !!!!!
Answer:
C
Explanation:
Which two staments explain how a cell's parts help it get nutrients
Answer:
I think it's A and D.
You throw an orange out a window at a height of 12.00 meters
upwards at an angle of 32° to the horizontal at a velocity of 3.5
m/s. What is the maximum height that the orange will reach?
Answer:
12.18
Explanation:
The maximum height that the orange will reach is 12.18 m.
What is projectile?When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion.
You throw an orange out a window at a height of 12.00 meters upwards at an angle of 32° to the horizontal at a velocity of 3.5 m/s.
If u is the initial speed, the maximum height of the ball from the point of throwing is
H = u² sin²θ / 2g
Put the values, we get
H = 3.5² sin²32° / (2x9.81)
H = 0.1753 m
So, the maximum height calculated from the ground is
12 + 0.18 = 12.18 m
Thus ,the maximum height that the orange will reach is 12.18 m.
Learn more about projectile.
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