Answer:
Skateboard
Explanation:
Acceleration is change in velocity over time.
a = Δv / Δt
The airplane's acceleration is:
a = (1005 km/h − 1000 km/h) / 10 s
a = 0.5 km/h/s
The skateboard's acceleration is:
a = (5 km/h − 0 km/h) / 1 s
a = 5 km/h/s
A 1-kilogram object is thrown horizontally and a 2-kilogram object is dropped vertically at
the same instant and from the same point above the ground. Which of the following will
be the same for both objects at any given instant?
Acceleration is the change in the velocity of the object with time. Acceleration is the same for both objects so both objects will reach the ground at the same time.
How do you find out the common value for both objects?Given that both the object dropped at the same time. So the gravitational acceleration of both the objects will be the same which is nearly 10 m/s.
The force at object 1 is given below.
[tex]F_1 = m_1\times g = m_1\times a_1[/tex]
Where m1 is the mass of object 1 and a1 is the acceleration of object 1.
[tex]F _1 = 1\times 10 = 10 \;\rm N[/tex]
The acceleration of object 1 is calculated as below.
[tex]a_1 = \dfrac{F_1}{m_1}[/tex]
[tex]a_1=\dfrac{10}{1} = 10 \;\rm m/s^2[/tex]
The force at object 2 is given below.
[tex]F_2 = m_2\times g = m_2\times a_2[/tex]
Where m2 is the mass of object 2 and a2 is the acceleration of object 2.
[tex]F_2 = 2\times 10 =20 \;\rm N[/tex]
The acceleration of object 2 is calculated as below.
[tex]a_2 = \dfrac{F_2}{m_2}[/tex]
[tex]a_2=\dfrac{20}{2} = 10 \;\rm m/s^2[/tex]
Hence we can conclude that the acceleration of both the objects is the same at every point so they will reach the ground at the same time.
To know more about the acceleration, follow the link given below.
https://brainly.com/question/12134554.
State and explain Gauss's law in
electrostatics and use it to determine the electric field
intensity due to a line charge.
To find:-
To find the electric field due to a point charge using Gauss's Law .Answer:-
Gauss's law in electrostatics states that the electric flux through a closed surface is equal to 1/ε0 times the charge enclosed by that surface, that is,
[tex]\implies \displaystyle \phi_E = \oint \vec{E} .\vec{ds} \\[/tex]
Also ,
[tex]\implies\displaystyle \oint \vec{E}.\vec{ds} =\dfrac{q}{\epsilon_0} \\[/tex]
Linear density of charge:-
It ise defined as charge per unit length. Mathematically,
[tex]\implies\displaystyle \lambda =\dfrac{q}{\ell} \\[/tex]
where ,
[tex] q[/tex] is the charge .[tex]\ell[/tex] is the length of the conductor.For figure see attachment.
Now let us assume that the linear density of charge is [tex]\lambda[/tex] . There is a point P near the wire at a distance of " r " from it , where we are interested in calculating the electrical field. Imagine a coaxial Gaussian cylinder of length [tex]\ell[/tex] , and radius " r " such that the point P lines on its surface.
Calculating Electrical flux through the cylinder:-
[tex]\implies\displaystyle \phi_E = 0 + 0 + \oint E.ds \\[/tex]
[tex]\implies\displaystyle\phi_E = \oint E.ds \ \cos0^o \\[/tex]
[tex]\implies\displaystyle \phi_E = \oint Eds \\[/tex]
[tex]\implies\displaystyle \phi_E = E\oint ds \\[/tex]
[tex]\implies\displaystyle \phi_E = E (2\pi r\ell) \ \ dots (1) \\[/tex]
From Gauss's law:-
[tex]\implies\displaystyle \phi_E =\dfrac{q}{\epsilon_0} \\[/tex]
[tex]\implies\displaystyle \phi_E = \dfrac{\lambda \ell}{\epsilon_0}\ \dots (2) \\[/tex]
Therefore , from equation (1) and (2) , we have;
[tex]\implies\displaystyle E \times 2\pi r \ell =\dfrac{\lambda\ell}{\epsilon_0}\\[/tex]
[tex]\implies\displaystyle \underline{\underline{\green{ E =\dfrac{1}{4\pi \epsilon_0}\dfrac{2\lambda}{r}}}}\\[/tex]
And we are done!
A metric unit of volume equal to a milliliter
In the early 1900s, Robert Millikan used small charged droplets of oil, suspended in an electric field, to make the first quantitative measurements of the electron’s charge. A 0.71-um-diameter droplet of oil, having a charge of , is suspended in midair between two horizontal plates of a parallel-plate capacitor. The upward electric force on the droplet is exactly balanced by the downward force of gravity. The oil has a density of 860 kg/m^3 , and the capacitor plates are 4.5 mm apart.
Part A
What must the potential difference between the plates be to hold the droplet in equilibrium?
Express your answer to two significant figures and include the appropriate units
ΔV=
Answer:
The potential difference is [tex]\Delta V = 44.40 \ V [/tex]
Explanation:
From the question we are told that
The diameter of the droplet of oil is [tex]d = 0.71 \mu m = 0.71 *10^{-6} \ m[/tex]
The density of the oil is [tex]\rho = 860 kg/m^3[/tex]
The distance of separation of the capacitor plate is [tex]l = 4.5 \ mm = 0.0045 \ m[/tex]
Generally the radius of the droplet is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{0.71 *10^{-6} }{2}[/tex]
=> [tex]r = 3.55 *0^{-7} \ m[/tex]
Generally the mass of the oil droplet is mathematically represented as
[tex]m = \rho * V[/tex]
Here V is the volume of the oil droplet which is mathematically represented as
[tex]V = \frac{4}{3} * 3.142 * (3.55 *0^{-7} )^3[/tex]
[tex]V = 1.874 *10^{-19} \ m^3[/tex]
So
[tex]m = 860 * 1.874 *10^{-19} [/tex]
=> [tex]m = 1.611 *10^{-16} \ kg [/tex]
Generally the electric force acting on the droplet is mathematically represented as
[tex]F = E * q[/tex]
Here q is the charge on an electron with value [tex]q = 1.60*10^{-19}\ C[/tex]
This force is equivalent to the weight of the droplet which is mathematically represented as
[tex]W = mg[/tex]
So
[tex]E * q = m * g[/tex]
Here E is the electric field which is mathematically represented as
[tex]E = \frac{\Delta V}{l}[/tex]
[tex]\frac{\Delta V}{l} * q = m * g[/tex]
=> [tex]\Delta V = \frac{m * g * l }{q}[/tex]
=> [tex]\Delta V = \frac{1.611 *10^{-16} * 9.8 * 0.0045 }{1.60*10^{-19}}[/tex]
=> [tex]\Delta V = 44.40 \ V [/tex]
A man is pulling a box with the force of 10N. There is nobody else on the other side of the box. What is the total amount of force on the box? *
If the length of an ax is 20 cm and the width is 5 cm what is the MA of a knife?
O 2
0 4
O 10
O6
Answer:
4
Explanation:
From the question given above, the following data were obtained from:
Length of an axe = 20 cm
Width = 5 cm
Mechanical advantage (MA) =.?
The mechanica advantage is simply defined as the ratio of the length to the width of the axe. Mathematically, it is expressed as:
Mechanical advantage = length /width
With the above formula, we can obtain the mechanical advantage of the axe as follow:
Length of an axe = 20 cm
Width = 5 cm
Mechanical advantage (MA) =.?
MA = 20/5
MA = 4
Therefore, the mechanical advantage (MA) of the axe is 4
Which of the following is the primary force that shapes the Earth's landscape?
A pigeon takes off from where it was initially on the ground, near where home base is located at Regions Field (close to 1st Ave. South and 14th street in Birmingham), and the bird flies at what you can consider a constant speed of 18 m/s along a direct straight line to Birmingham International Airport, where he lands on the roof of one of the main terminals there. 5.E. How many seconds does it take the bird to make this trip
Answer:
1.84 secondsExplanation:
Given
initial velocity u = 0m/s
final velocity v = 18m/s
To get the time it takes the bird to make this trip, we will use the equation of motion
v = u+gt
g is the acceleration due to gravity = 9.81m/s²
t is the time
Substitute into the formula:
[tex]18 = 0+(9.81)t\\18 = 9.81t\\9.81t = 18\\t = \frac{18}{9.81}\\ t = 1.84secs[/tex]
Hence it took the bird 1.84seconds to make this trip
Which of the following graphs shows the motion of an object that starts to travel forward, stops for several seconds, then returns to its original position?
Answer:
Pls can you send the following graphs
To a stationary observer, a man jogs east at 2 m/s and a woman jogs west at
3 m/s. From the woman's frame of reference, what is the man's velocity?
Answer:
5 m/s east
Explanation:
i just took the test and it was correct
Answer:
Just took the test. 5 m/s east.
Explanation:
What happens to the acceleration of a ball if the force increases while the mass remains the same?
Answer:
if the force increases while the mass of the ball remains the same, the acceleration increases.
Explanation:
According to Newton's second law of motion, the acceleration of an object produced by a net force is directly proportional to the magnitude of the net force Fn and inversely proportional to the mass of the object m.
[tex]\displaystyle a=\frac{Fn}{m}[/tex]
Suppose the net force doubles to 2Fn. The new acceleration will be:
[tex]\displaystyle a'=\frac{2Fn}{m}=2\frac{Fn}{m}=2a[/tex]
The acceleration doubles too. Thus, if the force increases while the mass of the ball remains the same, the acceleration increases.
how does a fly function in a energy pyramid
Answer:
A pyramid of energy shows the relative amount of energy available at each trophic level of a food chain or web. On average, about 10 percent of the energy available within one trophic level is transferred to the next level.
Explanation:
A planet similar to the Earth has a radius 7.4 × 106 m and has an acceleration of gravity of 10 m/s 2 on the planet’s surface. The planet rotates about its axis with a period of 25 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight, what is the new period? Answer in units of h.
Part 2:
By what factor would the speed of the object be increased when the planet is rotating at the higher speed?
Answer:
1.5 hr
16.7
Explanation:
Zero apparent weight means there's no normal force.
Sum the forces in the centripetal direction.
∑F = ma
mg = mv²/r
v = √(gr)
v = √(7.4×10⁶ m × 10 m/s²)
v = 8602 m/s
The circumference of the equator is:
C = 2πr
C = 2π (7.4×10⁶ m)
C = 4.65×10⁷ m
So the period is:
T = C / v
T = (4.65×10⁷ m) / (8602 m/s)
T = 5405 s
T = 1.5 hr
The initial speed is:
v = C / T
v = (4.65×10⁷ m) / (25 h × 3600 s/h)
v = 517 m/s
The speed increases by a factor of:
8602 m/s / 517 m/s
16.7
The new period of the planet at the given radius is 1.5 h.
The speed of the planet increased by a factor of 16.7.
The given parameters;
radius of the Earth, R = 7.4 x 10⁶ macceleration due gravity, g = 10 m/s²period, T = 25 hThe speed of the planet at the given radius is calculated as follows;
[tex]mg = \frac{mv^2}{r} \\\\gr = v^2\\\\v= \sqrt{rg} \\\\v = \sqrt{7.4 \times 10^6 \times 10} \\\\v = 8602.3 \ m/s[/tex]
The new period of the planet is calculated as follows;
[tex]T = \frac{2\pi R}{v} \\\\T = \frac{2 \pi \times 7.4 \times 10^6}{8602.3} \\\\T =5,405.72 s \\\\T = \frac{5405.72}{3600} = 1.5 \ hr[/tex]
The initial speed of the planet is calculated as follows;
[tex]v_i = \frac{2\pi R}{T} \\\\v_i = \frac{2\pi \times 7.4\times 10^6}{25 \times 3600} \\\\v_i =516.68 \ m/s[/tex]
The increase in the speed of the planet is calculated as;
[tex]= \frac{v}{v_i} \\\\= \frac{8602.3}{516.68} \\\\= 16.7[/tex]
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Doctors were furious. Their patients didn't care about regular medicine anymore. All they wanted was a wave of Dr. Mesmer's wand. The doctors griped — and groused — and fussed — and fumed. What is most likely the meaning of the word furious?
A quietly curious B quite thrilled C mildly annoyed D extremely angry
Answer:
am pretty sure the answer is d
A coil is wrapped with 206 turns of wire on the perimeter of a circular frame (of radius 31 cm). Each turn has the same area, equal to that of the frame. A uniform magnetic field is directed perpendicular to the plane of the coil. This field changes at a constant rate from 23 mT to 54 mT in 24 ms. What is the magnitude of the induced average E in the coil, over the time interval 24 ms during which the field changes? Answer in units of V.
Answer:
E = -80.33 V
Explanation:
For this exercise we use Faraday's law of induction
E = - N d[tex]\phi _{B}[/tex] / dt
where the magnetic flux is
\phi _{B} = B A cos θ
In the exercise they indicate that the magnetic field is perpendicular to the coil, therefore the normal of the coil area and the field has the same direction, consequently the angle is zero. Since the area is constant
E = - N A dB / dt
let's look for each term
a =π r²
they tell us that the variation in the field is linear
dB / dT = (B_f-B₀) / t
we substitute
E = - N (π r²) (B_f -B₀) / t
let's calculate
E = - 206 π 0.31² (54 - 23) 10⁻³ / 24 10⁻³
E = -80.33 V
Which of the following represents an example of thermal energy?
O a
Saxophone
Ob
Oven
с
Powerlines
Od
Windmill
Answer:
The answer of this question is windmill
A 30.0 g object fell from a height y = 2.94 m. It’s velocity when y is zero is approximately equal to (in m/s) (a) 0 (b) 19.6 (c) 29.4 (d) 8.64
Analysing the question:
We are given:
mass of object (m) = 0.03 kg
height of object (h) = 2.94 m
acceleration due to gravity (a) = 10 m/s²
initial velocity (u) = 0 m/s [the object fell, we are NOT given any initial velocity]
final velocity (v) = v m/s
Solving for Final velocity:
from the third equation of motion:
v² - u² = 2ah
replacing the variables
(v)² - (0)² = 2(10)(2.94)
v² = 58.8
v = 7.67 m/s
15. An astronaut on the moon has a 110 kg crate and a 230 kg crate. How do the forces required to liftyihe crates straight up on the moon compare with the forces required to lift them on Earth?
Explanation:
Mass of carte 1 is 110 kg and that of crate 2 is 230 kg
Force required to lift crates is equal to its crate i.e. W = mg
On Moon, a = 1.625 m/s²
Weight of crate 1, W = 110 kg × 1.625 m/s² = 178.75 N
Weight of crate 2, W = 230 kg × 1.625 m/s² = 373.75 N
On Earth, g = 9.8 m/s²
Weight of crate 1, W = 110 kg × 9.8 m/s² = 1078 N
Weight of crate 2, W = 230 kg × 9.8 m/s² = 2254 N
Hence, this is the required solution.
The forces required to lift the crates straight up on the Moon is lesser than the forces required to lift them on Earth.
Given the following data:
Mass A = 110 kgMass B = 230 kgScientific data:
Acceleration due to gravity on Earth = 9.8 [tex]m/s^2[/tex]Acceleration due to gravity on Moon = 1.6 [tex]m/s^2[/tex]The formula for weight.Mathematically, the weight of an object is given by the formula;
[tex]Weight = mg[/tex]
Where;
m is the mass of the object.g is the acceleration due to gravity.For the weight on Moon:[tex]Weight \;A= 110 \times 1.6[/tex]
Weight A = 176 Newton.
[tex]Weight \;B=230 \times 1.6[/tex]
Weight B = 368 Newton.
For the weight on Earth:[tex]Weight \;A= 110 \times 9.8[/tex]
Weight A = 1,078 Newton.
[tex]Weight \;B=230 \times 9.8[/tex]
Weight B = 2,254 Newton.
In conclusion, the forces required to lift the crates straight up on the Moon is lesser than the forces required to lift them on Earth.
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Phoebe wants to measure the volume of a small, irregularly
shaped rock. Which of the following instruments would be best
for Phoebe to use?
Answer:
A graduated cylinder
Explanation:
What is the equivalent temperature in kelvin if you have a metal at 50°F?
A.
50 K
B.
150 K
C.
283 K
D.
300 K
Answer:
The answer is C 283 kelvin
Answer: C.
283 K
Explanation: edmentum
An airplane flying at 119 m/s is accelerated uniformly at the rate of 0.5 m/s2 for 10
seconds. What is its final speed?
Answer:
124m/s
Explanation:
v=u+at
v=119+(0.5×10)
v=119+5
v=124m/s
You move 6 meters east and move 8 meters south what’s the distance and displacement
The distance travelled in total is 14 m. But the resultant displacement is the square root of the sum of squares of each distances covered in different directions, that is 10 m.
What is displacement ?Displacement is a physical quantity that, that measures how far we travelled from a mean position. The change in position is dependent on the magnitude and direction as well. Hence, displacement is a vector quantity.
The distance covered by an object is total length or sum magnitudes of the length it covered. It is a scalar quantity. Here, the distance covered is 6 + 8 = 14 m.
The resultant displacement here, is calculated as follows:
ds = √(6² + 8²) = √100 = 10 m
Therefore, the displacement of us will be 10 m.
Find more on displacement :
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A body starts with a velocity of 20 m/s, accelerates uniformly for a time t, moves with constant velocity for the same time, and finally decelerates to rest, again taking a time t. Given that the final deceleration is three times the original acceleration, and the total distance travelled is 140 m, find t.
Answer:
t=0.14s
Explanation:
t=[tex] \frac{s}{v} [/tex]t=[tex] \frac{20}{140} [/tex]Where s= speedV=distance And t=timeA wind tunnel draws atmospheric air at 20°C and 101.3 kPa by a large fan located near the exit of the tunnel. If the air velocity in the tunnel is 80 m/s, determine the pressure in the tunnel.
Answer:
The pressure in the tunnel is 97.45 kPa
Explanation:
Given;
atmospheric pressure, P₁ = 101.3 kPa
air velocity in the tunnel, v₂ = 80 m/s
density of air at 20°C, ρ = 1.204 kg/m³
pressure in the tunnel, P₂ = ?
Apply Bernoulli's equation;
P₁ + ¹/₂ρv₁² + ρgz₁ = P₂ + ¹/₂ρv₂² + ρgz₂
v₁ = 0
z₁ = z₂ = 0
P₁ = P₂ + ¹/₂ρv₂²
P₂ = P₁ - ¹/₂ρv₂²
P₂ = 101.3 kPa - (0.5 x 1.204 x 80²)
P₂ = 101.3 kPa - 3.8528 kPa
P₂ = 97.45 kPa
Therefore, the pressure in the tunnel is 97.45 kPa
Why does a sound come out when you blow into a bottle?
When does the jet of air convert to vibrations?
How do waves transfer energy? Be detailed.
Answer:
Electromagnetic waves transfer energy through vibrations of electric and in magnetic fields. For sound waves the energy is transferred through the vibration of air particles or particles of a solid. Waves also carry energy, the amount of energy they carry depends on their frequency and their amplitude. The more energy and the higher frequency amplitude, the more energy waves are ^‿^
A freight train car has a mass of 2,000 kilograms and an acceleration of 1.8 m/s/s. What is the average force behind that train car?
Answer:
F = 3600 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force [N]
m = mass = 2000 [kg]
a = acceleration = 1.8 [m/s^2]
Now replacing:
F = 2000*1.8
F = 3600 [N]
If you walk North for 40 feet and then walk East for 30 feet, what is your total distance
travelled? What is your displacement?
Answer:
The person travelled a total distance of (40+30) feet = 70 feet
Displacement= 50feet (im not positive on the displancement)
displacemeny is the shortest distance between two pionts
Explanation:
My bill at Office Depot came to $92.35 before my teacher discount. Once the clerk applies
my 15% teacher discount, how much will I have to pay?
$106.20
$13.85
$78.50
$92.35
Answer: $78.50 or C.
Explanation:
$92.35 x 15% or .15 is $78.50
which can occur in a physical change
Answer:
A physical change involves a change in physical properties. Examples of physical properties include melting, transition to a gas, change of strength, change of durability, changes to crystal form, textural change, shape, size, color, volume and density.
Explanation:
