Answer:
c
Explanation:
Answer:
the answer is c i just did it
Explanation:
3) An explorer walks 13 km due east, then 18 km north, and finally 3 km west.
a) What is the total distance walked?
b) What is the resulting displacement of the explorer from the starting point?
Answer: 34 km, 21 km 61 degrees north of east
Explanation: distance = 13 + 3 + 18 = 34
displacement = 13 - 3 = 10
10^2 + 18^2 = 424
find the square root of 424 ( 20.5 rounded to 21 )
The total distance walked is 34 km and the resulting displacement is 20.6 km.
a) The total distance is gotten by summing up all the distance.
Total distance = Distance moved east + distance moved north + distance moved west
Total distance = 13 km + 18 km + 3 km = 34 km
b) The displacement is the distance from the beginning point to end point.
Displacement² = 18² + (13 - 3)² = 18² + 10²
Displacement² = 424
Displacement = 20.6 km
Therefore the total distance walked is 34 km and the resulting displacement is 20.6 km.
Find out more at: https://brainly.com/question/12874363
How many elements in oxygen gas? PLEASE ANSWER!
Answer:
8
Explanation:
Answer:
Chemical Properties of Oxygen
At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.
Explanation:
Earth's atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1800 protons per second. What would be the electric current in amperes intercepted by a 41 × 107 km2 area on the planet
Answer:
Electric current in amperes = 1.1808 A
Explanation:
Given:
Intercept protons rate = 1800 protons per second
Area = 41 × 10⁷ km²
Find:
Electric current in amperes
Computation:
Current density = Intercept protons rate × 1.6 × 10⁻¹⁹
Current density = 1800 × 1.6 × 10⁻¹⁹
Current density = 2.88 × 10⁻¹⁶
1 km² = 10⁶m²
So,
Electric current in amperes = 2.88 × 10⁻¹⁶ × 41 × 10⁷ × 10⁶
Electric current in amperes = 1.1808 A
An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump
Answer:
The mechanical efficiency of the pump is 91.8 %
Explanation:
Given;
input power, p = 44 kw
density of oil, ρ = 860 kg/m³
motor efficiency, η = 90 %
inlet diameter, d₁ = 8 cm
outlet diameter, d₂ = 12 cm
volume flow rate, V = 0.1 m³/s
pressure rise, P = 500kPa
output power = motor efficiency x input power
output power = 0.9 x 44 = 39.6 kW
Thus, the mechanical input power = 39.6 kW
The mechanical output power is given by change in mechanical energy;
[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]
The mechanical efficiency is given by
η = mechanical output power / mechanical input power
η = 36.347 / 39.6
η = 0.918
η = 91.8 %
Therefore, the mechanical efficiency of the pump is 91.8 %
Forces of 6lbs and 4lbs act on an object. What is the maximum and minimum net force on the object? Explain your answer.
i cant understand your question define briefly
A woman walks 150 min 45 seconds. What was her average speed?
Answer:
0.3
Explanation: when you average out the number sthat is the answer you get
An unstrained horizontal spring has a length of 0.34 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.024 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. g
Answer:
a
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
b
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
Explanation:
Generally the force exerted on the string is mathematically represented as
[tex]F = k * e[/tex]
substituting values 180 N/m for k and 0.024 m for e
[tex]F = 180 * 0.024[/tex]
[tex]F = 4.32 \ N[/tex]
This force can also equivalent to the electrostatic force between the charges i.e
[tex]F = k * \frac{q^2}{ r^2}[/tex]
substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have
[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]
[tex]q = \sqrt{1.929 *10^{-10}}[/tex]
[tex]q = 1.389 *10^{-5} \ C [/tex]
Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible algebraic signs of the charges are
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
For any object in projectile motion, the vertical velocity is independent of gravity?
Answer:
the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity
Explanation:
In projectile launching, the movement is separated into two movements, one on the x-axis and the other on the y-axis, related through time.
In horizontal movement, the speed is constant, because there is no acceleration in this axis, the effect of air friction is almost always eliminated.
In the other movement on the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity.
This value of the eceleration of gravity is constant for small distances
compared to the radius of the Earth, for higher altitudes an expansion in beings of the distance is used, giving a linear dependence.
Help me please! You will get the brainliest!
Answer:
0.5 m.
Explanation:
From the question given above, the following data were obtained:
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) travalled by the book =?
We can obtain the distance travelled by the book by applying the formula of work done. This is illustrated below:
Work done (W) = Force (F) × Distance (s)
W = F × s
Force (F) applied = 5 N
Work done (W) = 2.5 J
Distance (s) travalled by the book =?
2.5 = 5 × s
Divide both side by 5
s = 2.5/5
s = 0.5 m
From the above calculation, the book will travel 0.5 m when a force of 5 N is applied.
When a gas undergoes an isothermal process, there is
(A) no heat added to the gas.
(B) no work done by (or on) the gas.
(C) no change in the temperature of the gas.
(D) no change in the volume of the gas.
(E) no change in the pressure of the gas.
Answer:
(C) no change in the temperature of the gas.
Explanation:
isothermal process of a gas, can result to compression of the gas or expansion of the gas at a constant temperature (ΔT = 0).
For isothermal gas expansion, work is done in reducing the pressure of the gas by increasing its volume at a constant temperature. The change in the internal energy of this process is zero (ΔU = 0).
For isothermal gas compression, work is done in decreasing the volume of the gas by increasing its pressure at a constant temperature. The change in the internal energy of this process is also zero (ΔU = 0).
Therefore, when a gas undergoes an isothermal process, there is no change in the temperature of the gas.
sam exerts a force of 65 N on a lawn mower with a mass of 25 kg. which formula can be used to calculate the acceleration of a lawn mower
Answer:
F = ma - with this you will get 2.6 m/s^2
Explanation:
Use this formula to get the acceleration...
where F is force, M is mass and A is acceleration
by using this we get...
65 = 25 * a
so, a = 65/25
Therefore, the acceleration is 2.6 m/s^2
Hope that helped :)
The acceleration of a lawn mower will be 2.6 m/s². It is the ratio of force and the mass.
What is acceleration?The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The given data in the problem is;
The force act on a lawn mower,[tex]\rm F = 65 \ N[/tex]
Mass of lawn mower,[tex]\rm m = 25 kg[/tex]
The acceleration of a lawn mower is,[tex]\rm a = ?[/tex]
Acceleration, is found as the ratio of force and the mass.
[tex]\rm F= ma \\\\ a = \frac{F}{m} \\\\ a= \frac{65}{25} \\\\ a = 2.6 \ m/s^2[/tex]
Hence, the acceleration of a lawn mower will be 2.6 m/s².
To learn more about acceleration, refer to the link;
https://brainly.com/question/2437624
#SPJ2
Physics Question (attached)
Answer:
The answer is "[tex]\bold{10.6 \ \ \frac{m}{s^2}}[/tex]"
Explanation:
The Formula of acceleration:
[tex]\bold{a= \frac{v-v_0}{t}}[/tex]
Given value:
[tex]t= 50\\v= 856\\v_0= 324[/tex]
Calculating the acceleration:
[tex]\to a = \frac{(856 - 324)}{50}\\\\[/tex]
[tex]= \frac{(532)}{50}\\\\= \frac{(106.4)}{10}\\\\= 10.64 \ \ \frac{ m}{s^2}[/tex]
An excited squirrel starts from rest and accelerates towards a nut at 1.0 m/s2 for 2.0 seconds. What is the squirrel’s displacement?
Answer:
[tex]x=8m[/tex]
Explanation:
Hello,
In this case, the most suitable kinematic equation for the calculation of the squirrel's displacement is:
[tex]x=v_ot+\frac{1}{2}at^2[/tex]
Whereas the initial velocity is zero as it starts from rest, thus, we obtain:
[tex]x=\frac{1}{2}*1.0\frac{m}{s^2}*(2.0s)^2\\ \\x=8m[/tex]
Best regards.
What is an example of intellectual development?
Answer:
learning new activities like riding a bike and playing sports. and also having the ability to collect information and remember them (memory). <3
The brakes of a car are applied, causing it to slow down at a rate of 10 ft/s2. Knowing that the car stops in 300 ft, determine
(a) how fast the car was traveling immediately before the brakes were applied,
(b) the time required for the car to stop.
Answer:
(a) The velocity of the car before the brakes were applied is 77.46 ft/s
(b) The time required for the car to stop is 7.8 s
Explanation:
Given;
acceleration of the car, a = 10 ft/s²
distance traveled by the car, d = 300 ft
(a) the velocity of the car before the brakes were applied is given;
v² = u² + 2ad
v² = 0 + 2(10 x 300)
v² = 6000
v = √6000
v = 77.46 ft/s
(b) the time required for the car to stop
d = ut + ¹/₂at²
d = 0 + ¹/₂at²
d = ¹/₂at²
t² = 2d / a
t = √ ( 2d / a)
t = √ ( 2 x 300 / 10)
t = 7.8 s
Therefore, the time required for the car to stop is 7.8 s
The speed of the car immediately before the brakes were applied was 77.46 ft/s and it took 7.75 s for the car to stop.
Using the formula:
v² = u² + 2as
Where S is the distance = 300 ft, u is the initial velocity, a is the acceleration = -10 ft/s², v is the final velocity = 0 ft/s (stops)
0² = u² + 2(-10)(300)
0 = u² - 6000
u² = 6000
u = 77.46 ft/s
b)
v = u + at
0 = 77.46 - 10t
t = 7.75 s
The speed of the car immediately before the brakes were applied was 77.46 ft/s and it took 7.75 s for the car to stop.
Find out more at: https://brainly.com/question/3715235
Physics question URGENT PLEASE!!!
[tex]\left(95\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{231}{1.00}\dfrac{\mathrm{in}^3}{\rm gal}\right)\left(\dfrac{1.0}{12}\dfrac{\rm ft}{\rm in}\right)^3\left(\dfrac{1.0}{60}\dfrac{\rm min}{\rm s}\right)\approx0.211661\dfrac{\mathrm{ft}^3}{\rm s}[/tex]
Multiplying 95 gal by 231 converts it to in^3. Dividing by 12^3 converts in^3 to ft^3. Dividing by 60 converts min to s.
A car travels 20 meters east
in 1.0 second and then travels
10 meters west in 20 seconds. The
displacement of the car at the end
of this 3.0-second interval is-
Answer:
Alright the other day.
If a 500kg elephant is sliding across a frictionless patch of ice, how much force is needed to keep the elephant from slowing down?
Answer:
4905NExplanation:
The force needed to keep the elephant from slowing down is expressed as shown according to Newtons second law of motion.
Force = mass * acceleration due to gravity
Given
Mass of elephant = 500kg
acceleration due to gravity = 9.81m/s²
Force = 500*9.81
Force = 4905N
Hence the force needed to keep the elephant from slowing down is 4905N
g uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work is needed to stretch the spring from 35 cm to 37 cm? (Round your answer to two decimal places.) .02 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)
Answer:
W= 0.319992 J and distance is 3.75 cm
Explanation:
The energy needed to stretch the spring from 30 cm to 45 cm = 3 J
Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.
So first find the work done to stretch the spring from 35 cm to 45 cm.
Work done, w = (1/2) kx^2
3 = (1/2)k(0.45 – 0.30)^2
k = 266.66 N/m
now, x1 = 0.35 – 0.30 = 0.05m
x2 = 0.37 – 0.30 = 0.07m
Now the amount of work done to stretch from 35cm to 37.
w = (1/2) k (x2^2 – x1^2)
w = (1/2) (266.66) (0.07^2 – 0.05^2 )
w= 0.319992 J
(b). Given F = 10 N
F = kx
x = F / k
x = 10 / 266.66
x = 0.0375m
x = 3.75 cm
Thus, distance is 3.75cm
According to Newton's second law, what is the acceleration of the released cart?
In order to have good physical fitness, you __________. A. should be able to complete daily tasks without being physically taxed B. must be able to run a 4-minute mile C. need to work out at the gym four times a week D. don't need to worry about eating healthy foods
Answer:
A) should be able to complete daily tasks without being physically taxed
Explanation:
B) you dont need to run that fast to have good physical fitness
C) you dont need to workout four times a week
D) you should always eat healthy as diet contributes to health
In order to have good physical fitness, you should be able to complete daily tasks without being physically taxed.
What is physical fitness?Physical fitness is the ability to be fit to do work without getting tired easily.
Exercise and workouts are necessary to build and improve physical fitness.
Exercises required include jogging and running.
Therefore, in order to have good physical fitness, you should be able to complete daily tasks without being physically taxed.
Learn more about physical fitness at: https://brainly.com/question/1744272
#SPJ2
Using the equation for for Newton's Second law, F=ma, solve the following problem. You have been given an object with a mass of 6g and an acceleration of 2 m/s2, what is the force?
Group of answer choices
A. 12N
B. 3N
C. 8N
D. 120N
Answer:
F = 0.012 N
Explanation:
Given that,
Mass of the object, m = 6 g
Acceleration, a = 2 m/s²
1 kg = 1000 grams
6 g = 0.006 kg
Force, F = ma
So,
[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]
So, the force is 0.012 N.
Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the x-y plane), the definition of acceleration is
Answer:
a = √ (a_t² + a_c²)
a_t = dv / dt , a_c = v² / r
Explanation:
In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.
a = aₓ i ^ + a_y j ^
Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains
a = √ (a_t² + a_c²)
where the tangential acceleration is
a_t = dv / dt
the centripetal acceleration is
a_c = v² / r
A model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
What is the airplanes resultant speed (magnitude of vector)
What is the airplanes heading (direction of velocity vector
Answer:
1.) 19.21 m/s
2.) 57 degree
Explanation: Given that a model airplane is flying north with a velocity of of 15m/s. A strong wind blowing east at 12m/s
The airplanes resultant speed can be calculated by using pythagorean theorem
R = sqrt ( 15^2 + 12^2 )
R = sqrt( 225 + 144 )
R = sqrt( 369 )
R = 19.21 m/s
The magnitude of the resultant vector is 19.21 m/s
The direction of velocity vector will be:
Tan Ø = 15 /12
Tan Ø = 1.25
Ø = tan^-1(1.25)
Ø = 57.04
Ø = 57 degree.
Therefore, the airplanes is heading 57 degree in the horizontal direction of velocity vector
What does a diagonal upward sloping line on a position time graph mean?
diagonal line means the velocity is constant.
4. What is the velocity of an object that doesn't move?
It depends on the object b. it depends on the speed c. it depends on the height
O mis
help
Answer:
Acceleration /Speed
Explanation:
An objects Velocity can be determined by acceleration,
Please pay attention in your middle school class, speed and velocity quiz.
Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of length 0.500 mm . What is the electric potential energy UUU of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Answer:
The value is [tex]U = 0.06 \ J [/tex]
Explanation:
From the question we are told that
The value of charge on each three point charge is
[tex]q_1 = q_2 = q_3 =q= 1.05 \mu C = 1.05 *10^{-6} \ C[/tex]
The length of the sides of the equilateral triangle is [tex]r = 0.500 \ [/tex]
Generally the total potential energy is mathematically represented as
[tex]U = k * [ \frac{q_1 * q_2}{r} + \frac{q_2 * q_3}{r} + \frac{q_3 * q_1}{r} ][/tex]
=> [tex]U = k * 3 * \frac{q^2}{r} [/tex]
Here k is coulomb constant with value [tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
=> [tex]U = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 } [/tex]
[tex]U = 0.06 \ J [/tex]
A river flows from south to north at 5.6 km/hr. On the west bank of this river, a boat launches and travels perpendicular to the current with a velocity of 7.7 km/hr due east. If the river is 1.6 km wide at this point, how far downstream does the boat land on the east bank of the river relative to the point it started at on the west bank?
Answer:
Downstream displacement = 1.1648 km
Explanation:
Given:
Speed of flow = 5.6 km/h
Speed of current upstream = 7.7 km/h
Increase in size = 1.6 km wide
Computation:
Total time to cross = Increase in size / Speed of current upstream
Total time to cross = 1.6 / 7.7
Total time to cross = 0.208 h
Downstream displacement = Speed of flow × Total time to cross
Downstream displacement = 5.6 × 0.208
Downstream displacement = 1.1648 km
Give an example of a situation in which you would describe an
object's position in
a. one dimension.
b. two dimensions.
three dimensions.
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.394 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
The distance traveled by the second block is 0.197 m
Explanation:
Given;
mass of the small block, m₁ = 0.2 kg
distance traveled by the block, d₁ =0.394 m
time of travel, t = 2 s
mass of the second block, m₂ = 0.4 kg
distance traveled by the second block, d₂ = ?
The work done per unit time on the inclined plane is given by;
[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]
Therefore, the distance traveled by the second block is 0.197 m