Answer:
a) The laser device
b) The tape
Explanation:
First, there is a need to understand what accuracy and precision mean.
Accuracy is the closeness of a measurement to its true (pre-determined) value.
Precision is the closeness of repeated measurements to each other.
Since 1 feet = 12 inches, then, 5 feet and 5.2 inches would be equivalent to 65.2 inches. This value represents the true value of my height.
The tape measured the height as 65 3/8, which is equivalent to 65.375 inches.
The laser device measured the height as 65.31.
Error = true value - measured value
Absolute error from the tape = 65.2 - 65.375
= -0.175 inches
Absolute error from laser device = 65.2 - 65.31
= -0.11
a) The magnitude of error from the tape is more than that of the laser device. Hence, the laser device is said to be more accurate.
b) Even though there were just single readings from both instruments, the tape can be read to the nearest 1/8 of an inch and as such, can give more precisive measurements than the laser device.
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t = 0.0. Knowing that automobile A has a constant acceleration of 0.8 m/s? and automobile B has a constant deceleration of 0.4 m/s2. Automobile A will overtake B after traveling a distance SA: A B. Side view
Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t = [tex]\frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150} }{2}[/tex]
t = [tex]\frac{12.5 \ \pm 27.5}{2}[/tex]
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m
1. Consider a 1000 kg car rounding a curve on a flat road of radius 50 m at a speed of
50 km/h (14 m/s).
a. Will the car make the turn if the pavement is dry and the coefficient of static
friction is 0.60?
Answer:
The car will make the turn perfectly
Explanation:
Given that the centripetal force= mv^2/r
M= mass of the car
v = speed of the car
r= radius
Hence;
F = 1000 × (14)^2/50
F= 3920 N
The frictional force = μmg
μ = coefficient of static friction
m= mass
g = acceleration due to gravity
Frictional force= 0.6 × 1000× 10
Frictional force = 6000 N
The car will not skid off the curve because the frictional force is greater than the centripetal force.
A penny of mass 3.10 g rests on a small 20.0 g block supported by a spinning disk with radius of 12.0 cm. The coefficients of friction between block and disk are 0.850 (static) and 0.575 (kinetic) while those for the penny and block are 0.395 (kinetic) and 0.495 (static). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk
Answer:
do this Q yourself because i havent read the chapter
The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.
How to solveThis is calculated using the coefficient of static friction between the penny and block, which is 0.495.
The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.
This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.
The angular velocity of the disk is then calculated from this force and the radius of the disk.
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A source of emf is connected by wires to a resistor and electrons flow in the circuit the wire diameter is teh same throughout teh circuit compared to the drift speed of the electrons before entering the source of emf, the drift speed ot eh electrons afte rleaving the source of emf is: ___________
a. faster.
b. slower.
c. the same.
d. either A or B depending on circumstances.
e. any of A, B , or C depending on circumstances.
A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then
n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / n ≈ 14.0 g/mol
Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type of wave is this
"Longitudinal wave" is the wave where the difference between the coils increases as well as decreases.
Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.
Thus the above answer is correct.
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Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time and distance?
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[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]
The answer is...
C. Work and time.
[tex]HopeThisHelps!![/tex]
[tex]AnimeVines[/tex]
An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?
We have that the angle is
[tex]\theta=32.53[/tex]
From the Question we are told that
E⃗ =5.00×105ı^N/C
Generally the equation for Tension is mathematically given
[tex]W=Tcos\theta[/tex]
Where
[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]
[tex]\theta=32.53[/tex]
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Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?
When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.
Now, given that:
the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.after hitting the wall, the final velocity will be (v) = 0 km/hAssumptions:
Suppose we make an assumption that the distance traveled during the collision of the car with the brick wall (S) = 1 mThat the car's acceleration is also constant.∴
For a motion under constant acceleration, we can apply the kinematic equation:
[tex]\mathsf{v^2 = u^2 + 2as}[/tex]
where;
v = final velocity u = initial velocitya = accelerations = distanceFrom the above equation, making acceleration (a) the subject of the formula:
[tex]\mathsf{v^2 - u^2 =2as }[/tex]
[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]
The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.
since 1 km/h = 0.2778 m/s
100 km/h = 27.78 m/s
[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]
[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]
a = - 385.86 m/s²
Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;
v = u + at
where;
v = 0
-u = at
[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]
[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]
t = 0.07 seconds
An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.
Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.
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A denser object will usually have a ____ index of
refraction.
Answer:
A denser object will usually have a high index of
refraction.
A denser object will usually have a high index of refraction.
What is index of refraction?The refractive index is the ratio of the speed of light in vacuum and speed of light in any medium.
n = c/v
The density is greater for the denser medium (water, oil, mercury, etc) then the rarer medium (any gas or air).
When a light ray travels through denser medium, its velocity is reduced and the refracted ray bends towards normal.
As, the index of refraction is inversely proportional to the velocity of light in the medium, index of refraction will be high for denser object.
Thus, denser object have high index of refraction.
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A solid object is made of two materials, one material having density of 2 000 kg/m3 and the other having density of 6 000 kg/m3. If the object contains equal masses of the materials, what is its average density
Answer:
[tex]\rho_{avg}=4000kg/m^3[/tex]
Explanation:
From the question we are told that:
Density of Material 1 [tex]\rho_1=2000kg/m^3[/tex]
Density of Material 2 [tex]\rho_2=6000kg/m^3[/tex]
Generally the equation for Average density is mathematically given by
[tex]\rho_{avg}=frac{\rho _1+rho _2}{2}[/tex]
[tex]\rho_{avg}=\frac{2000+6000}{2}[/tex]
[tex]\rho_{avg}=4000kg/m^3[/tex]
At a distance of 14,000 km from the center of Planet Z-99, the acceleration due to gravity is 32 m/s2. What is the acceleration due to gravity at a point 28,000 km from the center of this planet
A body of mass m feels a gravitational force due to the planet of
F = GmM/R ² = ma
where
• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant
• M is the mass of the planet
• R is the distance between the body and the planet's center
• a is the acceleration due to gravity
Solving for a gives
a = GM/R ²
Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of
GM/(2R)² = 1/4 × GM/R ² = a/4
so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².
What are the major sources of energy utilized during a 100 meter race, a 1000 meter race, and a marathon
Answer:
The energy from food and then from plants and then from sun.
As sun is the ultimate source of energy.
Explanation:
Distance = 100 m, 1000m, marathon
As the distance is covered by the person, so the muscular energy is used and thus the energy comes form out food.
As we know that the energy can neither be created nor be destroyed it can transform from one form to another.
So, the energy form the food which we consume is converted into the kinetic energy as we run.
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?
Answer:
The electric potential energy is 6.72 x 10^-11 J.
Explanation:
Potential difference, V = 4.2 x 10^8 V
charge of electron, q = - 1.6 x 10^-19 C
Let the potential energy is U.
U = q V
U = 1.6 x 10^-19 x 4.2 x 10^8
U = 6.72 x 10^-11 J
In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank
Answer:
v₂ = 15.24 m / s
Explanation:
This is an exercise in fluid mechanics
Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
They indicate the pressure in the factory P₁ = 140000 Pa, the velocity
v₁ = 5.5 m / s and the initial height is zero y₁ = 0
the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m
P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²
let's calculate
140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²
138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²
v₂² = 2 (138000 /ρ - 58.8 + 15.125)
v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]
In the exercise they do not indicate what type of liquid is being used, suppose it is water with
ρ = 1000 kg / m³
v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]
v₂ = 15.24 m / s
If the Velocity of the body
is increased to 3v, determine the kinetic energy
ANSWER; KE=5mv^2 so it is proportional to v^2.
Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.
6. Two astronauts of equal mass step from the top of a rocket on Venus. One slides down a 10- degree ramp while the other slides down a 75-degree ramp. If all friction is ignored, which astronaut reaches the surface of Venus with the lower kinetic energy
Answer:
Final kinetic energies of both astronauts will be the same.
Explanation:
If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.
Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.
Similarly, the final kinetic energies of astronauts will also be the same.
A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil. The average emf induced in the coil is
Answer:
Explanation:
From the question we are told that:
Number of turns [tex]N=10[/tex]
Area [tex]a=0.23m^2[/tex]
Magnetic field [tex]B=0.947T[/tex]
Generally the equation for maximum flux is mathematically given by
[tex]\phi=NBa[/tex]
[tex]\phi=10*0.047*0.23[/tex]
[tex]\phi=0.1081wbi[/tex]
Therefore induced emf
[tex]e= \frac{d\phi}{dt}[/tex]
Since
[tex]t=0[/tex]
Therefore
[tex]e=0[/tex]
80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting temperature.
Answer:
the resulting temperature is 23.37 ⁰C
Explanation:
Given;
mass of the iron, m₁ = 80 g = 0.08 kg
mass of the water, m₂ = 200 g = 0.2 kg
mass of the iron vessel, m₃ = 50 g = 0.05 kg
initial temperature of the iron, t₁ = 100 ⁰C
initial temperature of the water, t₂ = 20 ⁰C
specific heat capacity of iron, c₁ = 462 J/kg⁰C
specific heat capacity of water, c₂ = 4,200 J/kg⁰C
let the temperature of the resulting mixture = T
Apply the principle of conservation of energy;
heat lost by the hot iron = heat gained by the water
[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]
Therefore, the resulting temperature is 23.37 ⁰C
A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?
Answer:
(a) the velocity of the shirt is 2.14 m/s
(b) the velocity of the shirt is 5.3 m/s
Explanation:
Given;
initial velocity of the shirt, u = 5.3 m/s
height of the platform above the ground, h = 4.00 m
(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]
(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]
The top of a swimming pool is at ground level. If the pool is 2.60 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions?
a. The pool is completely filled with water.
______m below ground level
b. The pool is filled halfway with water.
______m below ground level
Answer:
a) [tex]d_g=1.95m[/tex]
b) [tex]d_g'=2.3m[/tex]
Explanation:
From the question we are told that:
Depth [tex]d=2.60[/tex]
a)
Generally the equation for distance to ground is mathematically given by
[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]
Where
[tex]n=\frac{4}{3}[/tex]
Therefore
[tex]d_g=\frac{d}{n}[/tex]
[tex]d_g=\frac{2.6}{4/3}[/tex]
[tex]d_g=1.95m[/tex]
b)
For when The pool is filled halfway with water
[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]
[tex]d_g'=\frac{1.3}{4/3}[/tex]
[tex]d_g'=0.98m[/tex]
Therefore
[tex]d_g'=(1.3+0.98)[/tex]
[tex]d_g'=2.3m[/tex]
dujevduxjehhsusheheh
m=100g
F-?
Answer:
Force = mass × acceleration
[tex]F =(100 \times 1000) \times 10 \\ = 1 \times {10}^{6} \: newtons[/tex]
A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the operating frequency, the resistance R is equal to the inductive reactance XL. The technician notices that when the plate separation of the parallel-plate capacitor is reduced to one-half its original value, the current in the circuit doubles. Determine the initial capacitive reactance in terms of the resistance R.
Answer:
Xc = (0.467 - 0.427j)R
Explanation:
Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is
Z = √[R² + (XL - XC)²]
Since the inductive reactance XL equals the resistance R, we have that
Z = √[R² + (XL - XC)²]
Z = √[R² + (R - XC)²]
Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]
Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus
Z' = √[R² + (R - XC')²]
Z' = √[R² + (R - 2XC)²]
The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]
Since the current doubles, I' = 2I.
V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]
1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]
√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]
squaring both sides, we have
[R² + (R - XC)²] = 4[R² + (R - 2XC)²]
expanding the brackets, we have
[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]
[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]
2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²
collecting like terms, we have
16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²
14RXC - 15XC² = 6R²
15XC² - 14RXC + 6R² = 0
Using the quadratic formula to find XC, we have
[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]
Since it is capacitive, we take the negative part.
So, Xc = (0.467 - 0.427j)R
monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how many total number of maximuima can be seen
Answer:
The total number of maxima that can be seen is 11
Explanation:
Given the data in the question
wavelength λ = 500 nm = 5 × 10⁻⁷ m
if the third order maximum is 32
i.e m = 3 and θ = 32°
Now, we know that condition for diffraction maximum is as follows;
d × sinθ = m × λ
so we substitute in our given values
d × sin( 32° ) = 3 × 5 × 10⁻⁷ m
d × sin( 32° ) = 1.5 × 10⁻⁶ m
d = [ 1.5 × 10⁻⁶ m ] / sin( 32° )
d = 2.83 × 10⁻⁶ m
Now, maxima n when θ = 90° will be;
sin( 90° ) = nλ / d
1 = nλ / d
d = nλ
n = d / λ
we substitute
n = [ 2.83 × 10⁻⁶ m ] / [ 5 × 10⁻⁷ m ]
n = 5.66
so 5 is the max value
hence, total maxima value is;
⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11
Therefore, total number of maxima that can be seen is 11
A car starting at rest accelerates at 3m/seconds square How far has the car travelled after 4s?
Answer:
24 meters
Explanation:
Find the final velocity. 12m/s
d=[final-initial]/2×time
D=(6m/s)×4=24 m/s
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________
Answer:
The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".
Explanation:
As the outer spherical shell is conducting, so there is no electric field in side from
⇒ [tex]r_b_1 < r < r_b_2[/tex].
So the electric potential at all points inside the conducting shell that from
⇒ [tex]r_b_1<r<r_b_2[/tex]
and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:
⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]
Thus the above is the right solution.
A cylindrical water tank has a height of 20cm and a radius of 14cm. If it is filled to 2/5 of its capacity, calculate.
I. Quantity of water in the tank
II. Quantity of water left to fill the tank to its capacity.
Answer:
4.926 L Y 7.389 L
Explanation:
first you calculate the tank volume
V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]
then you convert to liters
[tex]12315 cm^{3}[/tex] = 12.315 l
then you calculate the liters of water
2/5(12.35 l) = 4.926 l
finally we calculate the amount without water
12.315 l - 4.926 l = 7.389 l
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Which quantities below of a solid object on this planet are NOT the same as on Earth?
Choose all
possible answers.
Weight
Mass
Volume
Density
Acceleration when it falls vertically.
Color
Answer:
Weight, acceleration when it falls vertically, are not same as that of earth.
Explanation:
Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.
So, the weight of object is not same as that on earth.
The mass is defined as the amount of matter contained in the object.
So, the mass of the object is same as that of earth.
The volume of the object is defined as the space occupied by the object.
So, the volume of the object is same as that of earth.
The density is defined as the ratio of mass of the object to its volume.
So, the density of the object is same as that of earth.
The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.
So, the acceleration is not same as that of earth.
The color of the object is its characteristic.
It is same as that of earth.
Can a conductor be given limitless charge
Answer:
No
Explanation:
You could try to give it enough to fill all valence electrons in all of the atoms in the conductor, but practically this could not be achieved.
Consider a neutron star with a mass equal to the sun, a radius of 19 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star
Answer:
120 km/s
Explanation:
Given data :
Radius of the star is r = 19 km
Rotational time period of the star is T = 1 s
Therefore, we know that the velocity of the star is given by :
[tex]$V=\frac{2\pi r}{T}$[/tex]
[tex]$V=\frac{2 \times 3.14 \times 19\times 10^3}{1}$[/tex]
V = 119380.52 m/s
Therefore, the velocity of the point on the equator of the star is = 120 km/s
