Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
An equation for the period of a planet is 4 pie² r³/Gm where T is in secs, r is in meters, G is in m³/kgs² m is in kg, show that the equation is dimensionally correct.
Answer:
[tex]\displaystyle T = \sqrt{\frac{4\, \pi^{2} \, r^{3}}{G \cdot m}}[/tex].
The unit of both sides of this equation are [tex]\rm s[/tex].
Explanation:
The unit of the left-hand side is [tex]\rm s[/tex], same as the unit of [tex]T[/tex].
The following makes use of the fact that for any non-zero value [tex]x[/tex], the power [tex]x^{-1}[/tex] is equivalent to [tex]\displaystyle \frac{1}{x}[/tex].
On the right-hand side of this equation:
[tex]\pi[/tex] has no unit.The unit of [tex]r[/tex] is [tex]\rm m[/tex].The unit of [tex]G[/tex] is [tex]\displaystyle \rm \frac{m^{3}}{kg \cdot s^{2}}[/tex], which is equivalent to [tex]\rm m^{3} \cdot kg^{-1} \cdot s^{-2}[/tex].The unit of [tex]m[/tex] is [tex]\rm kg[/tex].[tex]\begin{aligned}& \rm \sqrt{\frac{(m)^{3}}{(m^{3} \cdot kg^{-1} \cdot s^{-2}) \cdot (kg)}} \\ &= \rm \sqrt{\frac{m^{3}}{m^{3} \cdot s^{-2}}} = \sqrt{s^{2}} = s\end{aligned}[/tex].
Hence, the unit on the right-hand side of this equation is also [tex]\rm s[/tex].
A 2kg ball is rolled along the floor for 0.8 m at a constant speed of 6 m/s. What is the work done by gravity?
A, 0
B, 16 J
C, 72 J
D, 450 J
E, 90 J
=F×s×cosa=2×g×0,8×cos90°= 0
The work done by gravity on a ball of 2 kg which is moving with a constant speed of 6 meter per second is zero. Thus, the correct option is A.
What is Work?Work is the energy transfer to or from an object through the application of force along with the displacement. For a constant force aligned with the direction of motion, the work done is equal to the product of the force strength which is applied and the distance traveled by the object.
Work = Force × Displacement
Force = Mass × Acceleration
Acceleration of the ball is zero as it is moving with a constant speed. Therefore, the work done by the gravity is zero.
Therefore, the correct option is A.
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define nortons theorem
Answer:
In direct-current circuit theory, Norton's theorem is a simplification that can be applied to networks made of linear time-invariant resistances, voltage sources, and current sources. At a pair of terminals of the network, it can be replaced by a current source and a single resistor in parallel.
15 . A scientist who studies the whole environment as a working unit .
Botanist
Chemist
Ecologist
Entomologist
Answer:
Ecologist.
Your answer is Ecologist.
(Ecologist) is a scientist who studies the whole environment as a working unit.
Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. What is the effective "spring constant" of this simple harmonic motion?
Express your answer to three significant digits and include the appropriate units.
We have that the spring constant is mathematically given as
[tex]k=2.37*10^{11}N/m[/tex]
Generally, the equation for angular velocity is mathematically given by
[tex]\omega=\sqrt{k}{m}[/tex]
Where
k=spring constant
And
[tex]\omega =\frac{2\pi}{T}[/tex]
Therefore
[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]
Hence giving spring constant k
[tex]k=m((\frac{2 \pi}{T})^2[/tex]
Generally
Mass of earth [tex]m=5.97*10^{24}[/tex]
Period for on complete resolution of Earth around the Sun
[tex]T=365 days[/tex]
[tex]T=365*24*3600[/tex]
Therefore
[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]
[tex]k=2.37*10^{11}N/m[/tex]
In conclusion
The effective spring constant of this simple harmonic motion is
[tex]k=2.37*10^{11}N/m[/tex]
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The source of sound moves away from the listener.The listener has the impression that the source is lower in pitch. Why?
When the source is moving away from the observer the velocity of the source is added to the speed of light. This increases the value of the denominator, decreasing the value of the observed frequency. Frequency corresponds to pitch or tone; a lower observed frequency will result in a lower observed pitch.
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
Determine the density in kg \cm of solid whose Made is 1080 and whose dimension in cm are length=3 ,width=4,and height=3
Answer:
d = 30kg/cm³
Explanation:
d = m/v
d = 1080kg/(3cm*4cm*3cm)
d = 30kg/cm³
a beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a focal length of 40 cm
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
Difference between scissors and nut cracker
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from the center of the wire. (in Nm2/C, keep 3 significant figures)
Answer:
[tex]E=35921.96N/C[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=0.321mm[/tex]
Charge Density [tex]\mu=0.100[/tex]
Distance [tex]d= 5.00 cm[/tex]
Generally the equation for electric field is mathematically given by
[tex]E=\frac{mu}{2\pi E_0r}[/tex]
[tex]E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}[/tex]
[tex]E=35921.96N/C[/tex]
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
When the drag force on an object falling through the air equals the force of gravity, the object has reached
terminal force.
terminal acceleration,
terminal illness.
terminal velocity
cyclist always bends when moving the direction opposite to the wind. Give reasons
Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B
Explanation:
Let [tex]\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}[/tex] and [tex]\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}[/tex]
The sum of the two vectors is
[tex]\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}[/tex]
[tex] = 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}[/tex]
The difference between the two vectors can be written as
[tex]\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}[/tex]
[tex]= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}[/tex]
The cells lie odjacent to the sieve tubes
Answer:
Almost always adjacent to nucleus containing companion cells, which have been produced as sister cells with the sieve elements from the same mother cell.A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular
Answer:
option A
Explanation:
simple harmonic motion
Answer:
random motion I think not sure
The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced
Answer:
The total surface area is "90 km²".
Explanation:
Given:
Power from solar installations,
= 27 GW
Other natural installations,
= 740 GW
Intensity,
[tex]\frac{F}{At}=\frac{P}{A}=1000 \ W/m^2[/tex]
%n,
= 30%
Now,
⇒ %n = [tex]\frac{out.}{Inp.}\times 100[/tex]
then,
⇒ [tex]Inp.=\frac{27}{30}\times 100[/tex]
[tex]=90 \ GW[/tex]
As we know,
⇒ [tex]I=\frac{P}{A}[/tex]
by substituting the values, we get
[tex]1000=\frac{90\times 10^9}{A}[/tex]
[tex]A = \frac{90\times 10^9}{10^3}[/tex]
[tex]=90\times 10^6[/tex]
[tex]=90 \ km^2[/tex]
A bicycle tire with a volume of 0.00210 m^3 is filled to its recommended absolute pressure of 495 kPa on a cold winter day when the tire's temperature is -14°C. The cyclist then brings his bicycle into a hot laundry room at 32°C.
a. If the tire warms up while its volume remains constant, will the pressure increase be greater than, less than, or equal to the manufacturer's stated 10% overpressure limit?
b. Find the absolute pressure in the tire when it warms to 32 degrees Celcius at constant volume.
(A) The pressure will be greater than 10% overpressure limit.
(B) The final pressure will be "582.915 kPa".
Given:
Volume,
[tex]V = 0.0021 \ m^3[/tex]Initial pressure,
[tex]P_o= 495 \ kPa[/tex]Initial temperature,
[tex]T_o = -14^{\circ} C[/tex][tex]= 259 \ K[/tex]
Final temperature,
[tex]T = 32^{\circ} C[/tex](B)
Number of moles,
→ [tex]n = (\frac{P_o V}{RT_o} )[/tex]
then,
The final absolute pressure,
→ [tex]P = \frac{nRT}{V}[/tex]
[tex]= (\frac{P_o V}{RT_o} )(\frac{RT}{V} )[/tex]
[tex]=(\frac{T}{T_o} )P_o[/tex]
[tex]= (\frac{305}{259} )\times 495[/tex]
[tex]= 582.915 \ kPa[/tex]
Thus the above approach is correct.
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The 52-g arrow is launched so that it hits and embeds in a 1.50 kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.47 m higher than the block's starting point. How fast was the arrow moving before it joined the block? What mechanical work must you do to lift a uniform log that is 3.1 m long and has a mass of 100 kg from the horizontal to a vertical position?
Answer:
[tex]v_1=87.40m/s[/tex]
Explanation:
From the question we are told that:
Mass of arrow [tex]m=52g[/tex]
Mass of rock [tex]m_r=1.50kg[/tex]
Height [tex]h=0.47m[/tex]
Generally the equation for Velocity is mathematically given by
[tex]v = \sqrt{(2gh)}[/tex]
[tex]v=\sqrt{(2 * 9.8m/s² * 0.47m) }[/tex]
[tex]v= 3.035m/s[/tex]
Generally the equation for conservation of momentum is mathematically given by
[tex]m_1v_1=m_2v_2[/tex]
[tex]0.052kg * v = 1.5 * 3.03m/s[/tex]
[tex]v_1=87.40m/s[/tex]
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
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A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
Which of the following choices is not an example of climate?
0000
San Diego has mild, warm temperatures and sea breezes year-round.
Anchorage has short, cool summers and long, snowy winters.
It will be 78° on Friday in Clovis.
Florida is tropical, with a significant rainy season.
Answer:
Florida is tropical, with a significant rainy seson
1. A 2.7-kg copper block is given an initial speed of 4.0 m/s on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs 85% of its initial kinetic energy as internal energy, calculate its increase in temperature.
Answer:
ΔT = 0.017 °C
Explanation:
According to the given condition, the change in internal energy of the block must be equal to 85% of its kinetic energy:
Change in Internal Energy = (0.85)(Kinetic Energy)
[tex]mC\Delta T = (0.85)\frac{1}{2}mv^2\\\\C\Delta T = (0.425)v^2\\\\\Delta T = \frac{0.425v^2}{C}[/tex]
where,
ΔT = increase in temperature = ?
v = speed of block = 4 m/s
C = specific heat capacity of copper = 389 J/kg.°C
Therefore,
[tex]\Delta T = \frac{(0.425)(4\ m/s)^2}{389}\\\\[/tex]
ΔT = 0.017 °C
Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.
Answer:
PlROCA
Explanation:
A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.
∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a
==> a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²
To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².
The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:
Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)
F[tex]_{horizontal}[/tex] = 150 N × cos(60°)
F[tex]_{horizontal}[/tex] = 150 N × 0.5
F[tex]_{horizontal}[/tex] = 75 N
Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)
F[tex]_{vertical}[/tex] = 150 N × sin(60°)
F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)
F[tex]_{vertical}[/tex] ≈ 129.9 N
Now, let's calculate the net force in the horizontal direction:
Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]
F[tex]_{net horizontal}[/tex] = 75 N - 15 N
F[tex]_{net horizontal}[/tex] = 60 N
Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:
F[tex]_{net horizontal}[/tex] = m × a
60 N = 20 kg × a
Now, solve for acceleration (a):
a = 60 N / 20 kg
a = 3 m/s²
So, the acceleration of the box is 3 m/s².
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What is an internal resistance?
Explanation:
some thing inside a resistor
The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle
Answer:
4.38 m/s
Explanation:
An ideal spring is hung vertically from the ceiling. When a 2.0-kg mass hangs at rest from it the spring is extended 6.0 cm from its relaxed length. A downward external force is now applied to the mass to extend the spring an additional 10 cm. While the spring is being extended by the force, the work done by the spring is:
a. -3.6 J
b. -3.3 F
c. -3.4 times 10^-5 J
d. 3.3 J
e. 3.6 J
Answer:
b) - 3.3 J
Explanation:
Given;
mass, m = 2 kg
initial extension of the spring, x = 6 cm = 0.06 m
The weight of the mass on the spring;
W = mg
where;
g is acceleration due to gravity = 9.81 m/s²
W = 2 x 9.81
W = 19.62 N
The spring constant is calculated as;
W = kx
k = W/x
k = 19.62 / 0.06
k = 327 N/m
The work done by the spring when it is extended to an additional 10 cm;
work done = force x distance
distance = extension, x = 10 cm = 0.1 m
The work done by the spring opposes the applied force by acting in opposite direction to the force.
W = - Fx
W = - (kx) x
W = - kx²
W = - (327) x (0.1)²
W = - 3.27 J
W ≅ - 3.3 J
Therefore, the work done by the spring by opposing the applied force is -3.3 J
