Answer:
longer for less massive stars.
Explanation:
A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.
It is typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He).
Some of the examples of stars are Canopus, Sun (closest to the Earth), Betelgeus, Antares, Vega etc.
Generally, the time taken for the collapse of an interstellar cloud fragment to the period (time) when a main-sequence star is given birth to, is usually longer for less massive stars.
This ultimately implies that, stars that are not so massive or big in size are transformed from interstellar cloud fragment to a main-sequence star is lesser.
Time taken from, when in interstellar cloud fragment first begins collapsing until it gives birth to a main-sequence star is longer for less massive stars.
What is interstellar cloud fragment?The process of Interstellar cloud fragment results the development into the stars.
The formation of stars is takes place within the highly dense concentration of the interstellar gases and the dust. These interstellar gases and the dust forms the molecular or interstellar clouds.
Here, in these molecular clouds the gases are present like helium, CO, hydrogen in major amount. Now this clouds started to collapse due to their own weight and density, and the formation of star is started.
Here, the H₂ atoms, which are fused forms helium atoms, to their core to form the main-sequence star.
In general, the time it takes from when in interstellar cloud fragment first begins collapsing until it gives birth to a main-sequence star. The time required to form the main-sequence star is lesser than the star which has big size.
Hence, the time taken from, when in interstellar cloud fragment first begins collapsing until it gives birth to a main-sequence star is longer for less massive stars.
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Astronomy Question
Who thinks Venus is the hottest planet instead of the lava planet?
say aye or nay
Answer:
AYE
Explanation:
Water has a heat capacity of 4.184 J/g °C. If 50 g of water has a temperature of 30ºC and a piece of hot copper is added to the water causing the temperature to increase to 70ºC. What is the amount of heat absorbed by the water?
The amount of heat absorbed by the water will be 8368 J.
What are heat gain and heat loss?Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.
It is given by the formula as ;
[tex]\rm Q= mcdt[/tex]
The given data in the problem is;
Equilibrium temperature = 30°C.
mass of water = 50 g ,
Temperature change = 70ºC
The specific heat of water =4.184 J//g °C
The amount of heat absorbed by the water is;
[tex]\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J[/tex]
Hence, the amount of heat absorbed by the water will be 8368 J.
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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to greatest:
a. Hollow Cylinder
b. Solid Cylinder
c. Hollow Sphere
d. Solid Sphere
Answer:
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
Explanation:
To answer this question, let's analyze the problem. Let's use conservation of energy
Starting point. Highest point
Em₀ = U = m g h
Final point. To get off the ramp
Em_f = K = ½ mv² + ½ I w²
notice that we include the kinetic energy of translation and rotation
energy is conserved
Em₀ = Em_f
mgh = ½ m v² +1/2 I w²
angular and linear velocity are related
v = w r
w = v / r
we substitute
mg h = ½ v² (m + I / r²)
v² = 2 gh [tex]\frac{m}{m+ \frac{I}{r^2} }[/tex]
v² = 2gh [tex]\frac{1}{1 + \frac{I}{m r^2} }[/tex]
this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)
v² = v₀² + 2 a L
where L is the length of the plane
v² = 2 a L
a = v² / 2L
we substitute
a = [tex]g \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }[/tex]
let's use trigonometry
sin θ = h / L
we substitute
a = g sin θ \ \frac{h}{L} \ \frac{1}{1+ \frac{I}{m r^2 } }
the moment of inertia of each object is tabulated, let's find the acceleration of each object
a) Hollow cylinder
I = m r²
we look for the acerleracion
a₁ = g sin θ [tex]\frac{1}{1 + \frac{mr^2 }{m r^2 } }[/tex]1/1 + mr² / mr² =
a₁ = g sin θ ½
b) solid cylinder
I = ½ m r²
a₂ = g sin θ [tex]\frac{1}{1 + \frac{1}{2} \frac{mr^2}{mr^2} }[/tex] = g sin θ [tex]\frac{1}{1+ \frac{1}{2} }[/tex]
a₂ = g sin θ ⅔
c) hollow sphere
I = 2/3 m r²
a₃ = g sin θ [tex]\frac{1}{1 + \frac{2}{3} }[/tex]
a₃ = g sin θ [tex]\frac{3}{5}[/tex]
d) solid sphere
I = 2/5 m r²
a₄ = g sin θ [tex]\frac{1 }{1 + \frac{2}{5} }[/tex]
a₄ = g sin θ [tex]\frac{5}{7}[/tex]
We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)
a) a₁ = g sin θ ½ = g sin θ [tex]\frac{105}{210}[/tex]
b) a₂ = g sinθ ⅔ = g sin θ [tex]\frac{140}{210}[/tex]
c) a₃ = g sin θ [tex]\frac{3}{5}[/tex]= g sin θ [tex]\frac{126}{210}[/tex]
d) a₄ = g sin θ [tex]\frac{5}{7}[/tex] = g sin θ [tex]\frac{150}{210}[/tex]
the order of acceleration from lower to higher is
a₁ <a₃ <a₂ <a₄
acceleration are
hollow cylinder < hollow sphere < solid cylinder < solid sphere
If three forces F1 = 20N, 300NE, F2 = 50N along W, F3 = 40N 500NW act on a body. Find the resultant in magnitude and direction.
Answer:
110N north west
Explanation:
.........
If three forces F1 = 20N, 300NE, F2 = 50N along W, F3 = 40N 500NW act on a body. The resultant in magnitude and direction is 110N north west
What are the types of force ?Force is a quantitative parameter and it is an interaction between two physical bodies such as an object and its environment, there are different types of forces in nature.
it can be defined as pushing or pulling of any object resulting from the object’s interaction or movement, without force the objects can not move
If an object in moving state the it will be either static or motion, the position of the object will only be changed if it is pushed or pulled and The external push or pull upon the object called as Force.
The contact force types are Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force
Non-Contact forces are occur from a distance such as Electromagnetic Force, Gravitational Force, Nuclear Force
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when drawing electric field lines ___________ charges have vectors point away/out and______ charges have vectors point toward/in.
Answer:
Positive, Negative
Explanation:
The image I've attached shows that vectors point into the negative source and vectors point away from the positive source.
what is 60mph (miles per hour) in meters per second? ( A mile is 5280ft)
please someone help me
Answer:
60mph=26.8224meters per second
Explanation:
If the length of the standing wave below is 2 meters, what is the wavelength of the standing
wave? *
Answer:
fffffgggggggggggggghhh
A –5 μC charge is placed 2 mm from a +3 μC charge. Find the force between the two charges?
Answer:
-33750 N
Explanation:
Use coulomb's law: [tex]\frac{k(q1)(q2)}{r^{2} } = \frac{(9x10^{9})(-5x10^{-6})(3x10^{-6})}{0.002^{2} } = -33750 N[/tex]
A 1.0μF capacitor with an initial stored energy of 0.50 J is discharged through a 1.0MΩ resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time t, (c) the potential difference V C across the capacitor, (d) the potential difference V R across the resistor, and (e) the rate at which thermal energy is produced in the resistor.
Answer:
A) q_o = 0.001 C
B) I = 0.001•e^(-t)
C) V_c = 1000e^(-t)
D) V_r = 1000e^(-t)
E) P = e^(-2t) watts
Explanation:
A) We are given;
Initial stored energy; U_o = 0.5 J
Capacitance; C = 1.0μF = 1 × 10^(-6) F
To find the charge, we will use the formula for energy in capacitors which is given by;
U = q²/2C
Thus, since we are dealing with initial energy, U is U_o and q is q_o
Making q the subject, we have;
q_o = √2CU_o
q_o = √(2 × 1 × 10^(-6) × 0.5)
q_o = 0.001 C
B) The charge as a function of time is expressed as;
q = q_o•e^(-t/RC)
Now the current is gotten by differentiating the charge function. Thus;
I = (q_o/RC)•e^(-t/RC)
Where;
R is Resistance = 1.0MΩ = 1 × 10^(6) Ω
C is capacitance = 1 × 10^(-6) F
(q_o/RC) is the initial current = 0.001/(1 × 10^(6) × 1 × 10^(-6))
(q_o/RC) = 0.001 A
Thus;
I = 0.001•e^(-t/(1 × 10^(6) × 1 × 10^(-6)))
I = 0.001•e^(-t)
C) Formula for potential difference across the capacitor is;
V_c = IR
I = 0.001•e^(-t)
R = 1 × 10^(6) Ω
Thus;
V_c = 1 × 10^(6) × 0.001•e^(-t)
V_c = 1000e^(-t)
D) Potential difference across the resistor will be the same as that across the capacitor because the resistor is connected in parallel to the capacitor.
Thus;
V_r = V_c = 1000e^(-t)
E) rate at which thermal energy is produced is basically the power.
Thus;
P = (V_r)²/R
P = (1000²e^(-2t))/1 × 10^(6)
P = e^(-2t) watts
A student rubbed two balloons with a piece of wool. What will happen when the balloons are
brought near each other?
Attract
Repel
Answer:
# ICSE board exam 2021
Explanation:
# ICSE board exam 2021
# ICSE board exam 2021
# ICSE board exam 2021
what happens to the work done when a force is doubled and the distance moved remain the same?
Answer:
Work done gets doubled.
Explanation:
The work done by a force is given by :
W = Fd
Where
F is force and d is distance move
If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.
At the time when force is doubled and distance moved remain the same so Work done gets doubled.
The following information should be considered:
The work done by a force is given by :
W = Fd
Where
F is force and d is distance move
In the case when the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.
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Which of the following are vectors? *
2 points
Distance
Displacement
Speed
Time
Velocity
Answer:
Displacement and Velocity
What is the specific heat of a substance that absorbs 1600 joules of heat when a sample
of 18 g of the substance increases in temperature from 20 °C to 31°C? Round your answer
to two decimal places if necessary.
Answer:
8.08 J/g °C
Explanation:
Q=m*Cp*ΔT-->
Cp=Q/(m*ΔT) -->
Cp=1600/[18*(31-20)]-->
Cp=8.08 J/g °C
molecules , like hormones , are made up of which of the following
A. cells
B. Atoms
C. Tissues
D. Organs
Answer:
atoms
Explanation:
Hormones are derived from amino acids or lipids. Amine hormones originate from the amino acids tryptophan or tyrosine. Larger amino acid hormones include peptides and protein hormones. Steroid hormones are derived from cholesterol.
A current loop ABCDA consists of a metal rod. a resistor R, and a pair of conducting rails separated by a distance d. The rod has a weight mg and it is falling with an instantaneous speed v. There is a constant magnetic field B which is perpendicular to the paper and directed into the paper. Find the direction of the induced current through the resistor It. A to B B to A 0 What is the magnitude of the induced current? If B = 3.6T, d = 7m, m = 4.7kg, R = 8.2 ohm, and g = 9.8 m/s2 . find the terminal velocity. (When the terminal velocity is readied, there is no net force on the rod, so the magnetic force is equal and opposite to the weight of the rod.) Answer in units of m/s.
Answer:
a) [tex]\vec{d} =B-A[/tex]
b) [tex]I=\frac{B*v*d*\sin 90 \textdegree}{R}[/tex]
c) [tex]v \approx 0.6m/s[/tex]
Explanation:
From the question we are told that
Magnetic field strength [tex]B=3.6T[/tex]
Distance traveled [tex]d=7m[/tex]
Mass [tex]m=4.7 kg[/tex]
Resistance [tex]r=8.2 ohm[/tex]
Gravitational acceleration [tex]g=9.8m/s^2[/tex]
[tex]\theta =90 \textdegree[/tex] Because of perpendicularity
a)
Generally the direction of the current will be given as
[tex]\vec{d} =B-A[/tex]
Because it opposes increases of magnetic flux
b)
Generally the equation for induced EMF [tex]E[/tex] is mathematically given as
[tex]E=B*v*d*\sin \theta[/tex]
[tex]E=B*v*d*\sin 90 \textdegree[/tex]
Generally the equation for induced current [tex]I[/tex] is mathematically given as
[tex]I=E/R[/tex]
[tex]I=\frac{B*v*d*\sin 90 \textdegree}{R}[/tex]
c)
Generally the the equation for force F at terminal speed is mathematically given as
[tex]F=mg[/tex]
[tex]mg=B*I*d*\sin 90 \textdegree[/tex]
[tex]mg=B*(\frac{B * v * d }{R}) *d[/tex]
[tex]v=\frac{m*g*R}{B^2*D^2}[/tex]
[tex]v=\frac{4.7*9.8*8.2}{3.6^2*7^2}[/tex]
[tex]v=0.59475m/s[/tex]
[tex]v \approx 0.6m/s[/tex]
Two rubber bullets (each of the same mass) are fired at the same velocity towards two different blocks of equal mass. One block is made of clay and the bullet gets stuck in it, the clay block bullet begins to move in the direction the bullet was fired. The other block is made of aluminum and the bullet bounces off the block, the aluminum block also begins to move in the direction the bullet was fired. Which block (clay or aluminum) will move with greater velocity after being struck by the bullet
Answer:
Aluminum board will move with a higher velocity
Explanation:
The velocity of the block will be higher when the impulse imparted by the bullet is higher.
In case of bullet bouncing off, the impulse imparted on the aluminum board is high and hence, it will move with a high velocity as compared to that of the clay board.
As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equation dVdt=kV , where t is measured in years. The volume of the glacier is 400km3 at time t=0 . At the moment when the volume of the glacier is 300km3 , the volume is decreasing at the rate of 15km3 per year. What is the volume V in terms of time
Solve the differential equation:
dV/dt = k V → 1/V dV/dt = k
→ d/dt [ln(V)] = k
→ ln(V) = k t + C
→ V (t )= exp(k t + C ) = C exp(k t ) = C e ᵏᵗ
At t = 0, the glacier has volume 400 km³ of ice, so
V (0) = 400 → C e⁰ = C = 400
Find when the glacier's volume is 300 km³:
V (t ) = 400 e ᵏᵗ = 300 → e ᵏᵗ = 3/4
→ k t = ln(3/4)
→ t = 1/k ln(3/4)
At this time, the volume is decreasing at a rate of 15 km³/yr, so
V ' (t ) = C k e ᵏᵗ → V ' (1/k ln(3/4)) = 400 k exp(k × 1/k ln(3/4)) = -15
→ 3/4 k = -3/80
→ k = -1/20
Then the volume V (t ) of the glacier at time t is
V (t ) = 400 exp(-1/20 t )
The volume in terms of time will be "V(t) = 400 exp (-[tex]\frac{1}{20}[/tex] t)".
Differential equation and VolumeAccording to the question,
Glacier's volume, V(0) = 400 km³
→ [tex]\frac{dV}{dt}[/tex] = kV
[tex]\frac{1}{V}[/tex] [tex]\frac{dV}{dt}[/tex] = k
ln(V) = kt + C
Now,
V (t) = exp (kt + C)
= C exp (kt)
= C[tex]e^{kt}[/tex]
When volume of glacier be "300 km³",
V(t) = 400 [tex]e^{kt}[/tex] = 300
[tex]e^{kt}[/tex] = [tex]\frac{3}{4}[/tex]
By taking log,
kt = ln([tex]\frac{3}{4}[/tex])
t = [tex]\frac{1}{k}[/tex] ln([tex]\frac{3}{4}[/tex])
When volume decrease at 15 km³/yr, then
→ V'(t) = Ck[tex]e^{kt}[/tex]
= 400 k exp (k × [tex]\frac{1}{k}[/tex] kn ([tex]\frac{3}{4}[/tex]))
= -15
Now,
[tex]\frac{3}{4}[/tex] k = - [tex]\frac{3}{80}[/tex]
By applying cross-multiplication,
k = - [tex]\frac{1}{20}[/tex]
Thus the response above is correct.
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