Answer:
clouds are created when water vapor and invisible gas turns into water droplets and these water droplets form tiny particles like dust and float in the air.
you have cumulus
cirrus
cirrostraus
Explanation:
how do i find out the maximum speed and things? also, if you can give the answer and tell me why that would be amazing!!
Answer:
See the answers and the explanation below.
Explanation:
To solve these questions we must understand that speed is the relationship between the displacement and the time that the displacement lasts.
1. 300 [km], in 5 [hr]
d = displacement [km]
[tex]v = d/t[/tex]
where:
v = velocity [km/h]
t = time [hr]
[tex]v = 300/5\\v = 60 [km/h][/tex]
In one hour : [tex]d = 60*1\\d = 60 [km][/tex]
b. Car B 100 [km] in 2 [hr]
[tex]v =100/2\\v = 50 [km/h]\\[/tex]
C. The car A has the greatest average speed.
A car accelerates at 2 m/s2. Assuming the car starts from rest, how much time does it need to accelerate to a speed of 16 m/s?
Answer:
8 seconds
Explanation:
as per 1st equation of motion
v=u+at,
u= initial velocity, v= final velocity , t=time, a=acceleration
since it starts from rest , u=0
v=16m/s
a=2m/s^2
16=0+2t
16=2t
t=8
Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. If this body is released from rest in a horizontal position, what is the angular speed of the meter stick as it swings through its lowest position
Answer:
5.35 rad/s
Explanation:
From the question, we are toldthat an Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick.
Solving this question, the potential energy of the particles must equal to the Kinectic energy i.e
P.E=K.E
Mgh= m½Iω²-------------eqn(*)
Where M= mass of the particles
g= acceleration due to gravity= 9.81m/s^2
ω= angular speed =?
h= height of the particles in the stick on the metre stick= ( 50cm + 80cm)= (0.5m + 0.8m)= 130cm=1.3m
If we substitute the values into eqn(*) we have
m×9.81× (1.3m)= 1/2× m×[ (0.5m)² + [(0.8m)²]× ω²
m(12.74m²/s²)= 1/2× m× (0.25+0.64)× ω
m(12.74m²/s²)= 1/2× m× 0.89× ω²
We can cancel out "m"
12.74= 1/2×0.89 × ω²
12.74×2= 0.89ω²
25.48= 0.89ω²
ω²= 28.629
ω= √28.629
ω=5.35 rad/s
Hence, the angular speed of the meter stick as it swings through its lowest position is 5.35 rad/s
Determine e when I = 0.50 A and R = 12 W.
Answer:
The correct answer is "24 V".
Explanation:
The given values are:
Current,
I = 0.50 A
Resistance,
R = 12 W
As we know,
⇒ [tex]I = 0.5\times (\frac{E}{2R})[/tex]
On substituting the given values, we get
⇒ [tex]0.5= (\frac{E}{4\times 12} )[/tex]
⇒ [tex]0.5= (\frac{E}{48} )[/tex]
⇒ [tex]E=24 \ V[/tex]
a 15 kg block of substance with specific heat capacity 840J/kg is heated by 15 c.assume its volume change to be negligible by how much its internal energy increases
Answer:
Q = 189000 [J]
Explanation:
The internal energy or heat can be calculated by means of the following expression.
[tex]Q=m*c_{p}*DT[/tex]
where:
Q = internal energy or heat [J]
m = mass = 15 [kg]
Cp = 840 [J/kg*°C]
DT = temperature change = 15 [°C]
[tex]Q = 15*840*15\\Q = 189000 [J][/tex]
My tiger trap needs 735 N of force on top of it to activate. What is the lightest tiger I can trap?
7203 kg
0.013 kg
75 kg
725 kg
Answer:
75kg
Explanation:
[tex]F=mg[/tex]
[tex]m=\frac{F}{g}[/tex]
[tex]m=\frac{735}{9.8}[/tex]
[tex]m=75kg[/tex]
Therefore, the answer is the third option 75 kg
Match each method of transferring electric charge with the correct description
friction
transfer of electric charge without direct
contact
induction
transfer of electric charge by rubbing
conduction
transfer of electric charge by direct contact
The matching of each method of transferring electric charge with the correct description should be explained below.
Matching of transferring electric charge?The friction means the transfer of electric charge via rubbing. The conduction means the transfer of electric charge via the direct contact.
Also, the induction means the transfer of the electric charge without the direct contact.
In this way it should be matched.
Learn more about electric here: https://brainly.com/question/23056096
Answer:
this is the answer that is correct
A sled of mass 50 kg is pulled along a snow-covered, flat ground. The static friction coefficient is 0.3 and the kinetic friction coefficient is 0.1.
Draw a FBD of all the forces acting on the sled.
What is the weight of the sled?
What force is needed to start the sled moving?
What force is needed to keep the sled moving at a constant velocity?
Answer:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled.
b) The weight of the sled is 490.35 newtons.
c) A force of 147.105 newtons is needed to start the sled moving.
d) A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
Explanation:
a) We kindly invite you to see below the Free Body Diagram of the forces acting on the sled. All forces are listed:
[tex]F[/tex] - External force exerted on the sled, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force from the ground on the mass, measured in newtons.
[tex]W[/tex] - Weight, measured in newtons.
b) The weight of the sled is determined by the following formula:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]m = 50\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the weight of the sled is:
[tex]W = (50\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 490.35\,N[/tex]
The weight of the sled is 490.35 newtons.
c) The minimum force needed to start the sled moving on the horizontal ground is:
[tex]F_{min,s} = \mu_{s}\cdot W[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]W[/tex] - Weight of the sled, measured in newtons.
If we know that [tex]\mu_{s} = 0.3[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to start the sled moving is:
[tex]F_{min,s} = 0.3\cdot (490.35\,N)[/tex]
[tex]F_{min,s} = 147.105\,N[/tex]
A force of 147.105 newtons is needed to start the sled moving.
d) The minimum force needed to keep the sled moving at constant velocity is:
[tex]F_{min,k} = \mu_{k}\cdot W[/tex] (3)
Where [tex]\mu_{k}[/tex] is the kinetic coefficient of friction, dimensionless.
If we know that [tex]\mu_{k} = 0.1[/tex] and [tex]W = 490.35\,N[/tex], then the force needed to keep the sled moving at a constant velocity is:
[tex]F_{min,k} = 0.1\cdot (490.35\,N)[/tex]
[tex]F_{min,k} = 49.035\,N[/tex]
A force of 49.035 newtons is needed to keep the sled moving at a constant velocity.
A pendulum clock uses a simple pendulum as its timing device. The clock is correct at noon. The next day, when the clock reads noon, the actual time is 11:50 a.m. What is the fractional change in pendulum length that must be made so the clock runs at the correct rate
Answer: the length of the pendulum should be 1.4% longer
Explanation:
Given that;
when its noon, the clock reads 11:50 am,
i.e we have 10 minutes delay ⇒ 10min × 60 = 600secs
we know that in simple pendulum
T = 2π√(l/g)
d means delay and c means correct;
24hrs = 86400 secs
Now
Td/Tc = (86400-600) / 86400 = 0.993 = [2π√(ld/gc)] / [2π√(ld/gc)] = √(ld/lc)
so ld/lc = 0.98616
lc = 1.014 ld
lc/l.d = -1 + 1.014 = 0.014 × 100 = 1.4% longer
therefore the length of the pendulum should be 1.4% longer
What is the kinetic energy of a 1500 kg vehicle traveling at a velocity of 8 m/s?
Answer:
48,000 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]k = \frac{1}{2} \times 1500 \times {8}^{2} \\ = 750 \times 64[/tex]
We have the final answer as
48,000 JHope this helps you
In compressed state, a spring 3.00 meters long hanging from the ceiling has a potential energy of 2.0 joules. If the entire potential energy is used to release a ball of mass
Answer:
0.068kg
Explanation:
In compressed state, a spring 3.00 meters long hanging from the ceiling has a potential energy of 2.0 joules. If the entire potential energy is used to release a ball of mass
Potential energy is expressed using the formula
PE = mgh
m is the mass
g is the acceleration due to gravity = 9.8m/s²
h is the height = 3.0m
PE = 2.0Joules
Required
Mass of the ball
Substitute the given values into the formula as shown;
2.0 = m(9.8)(3)
2 = 29.4m
m = 2/29.4
m = 0.068kg
Hence the mass of the ball is 0.068kg
An earth satellite travels in a circular orbit at 20,000 mph if the radius of the orbit is 4,300 mi what angular velocity is generated?
Answer:
0.00129rad/s
Explanation:
The angular velocity is expressed as;
v = wr
w is the angular velocity
r is the radius
Given
v = 20,000 mph
r = 4300mi
Get w;
w = v/r
w = 20000* 0.44704/4300*1609.34
w = 8940.8/6,920,162
w = 0.00129rad/s
Hence the angular velocity generated is 0.00129rad/s
A constant force of magnitude 23 N acts on an object for 3.1 s. What is the magnitude of the impulse
Answer:
71.3 NsExplanation:
The impulse of an object can be found by using the formula
impulse = force × time
From the question we have
impulse= 23 × 3.1
We have the final answer as
71.3 NsHope this helps you
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
s = 40m - 4m = 36m
W = F × s
= 10N × 36m = 360J
A bit of explanation :W = Work (J)
F = Force / weight (N)
s = distance (m)
Work done in physics is the product of force and displacement. The displacement for the object is 36 m and force acts on it is 10 N. Then the work done is 360 J.
What is work done?Work done is the dot product of force acting on a body and the resultant displacement. When a force applied on an object results in a displacement from its position, the force is said to be work done.
Work done is a vector quantity thus, characterised by a magnitude and direction. The common unit of work done is joule.
Given that force applied on the weight = 10 N
displacement occurred = 40 m - 4 m = 36 m
Work done = F . ds
ds = 36 m and f = 10 N
Then W = 10 N × 36m
= 360 J.
Therefore, the work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth is 360 J.
To find more on work done, refer here:
https://brainly.com/question/13662169
#SPJ2
What is the significance of Nucleotides in Chromosomes?
Answer:
it comprises of the DNA/RNA bipolymer molecules
In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.
Answer:
in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.
Explanation:
Given request solutions
L=500N a,M.A=? a,M.A=L/E =5/3
E=300N b,V.R=? b,V.R=2
c,efficiency =? c,£=M.A/V.R=5/6
£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS
I THINK I HELPED
How can you measure the strength of any electric field?
Answer:
The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q.
Explanation:
Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.
A 1500 kg car sits on a 3.5° inclined hill. Find the force of friction required to keep it from
sliding down the hill. The coefficient of static friction is μ=0.45
The same 1500 kg car is coasting at 50 km/h when it encounters a (friction
free!) hill that drops 14 m vertically. It then travels through 30.0 m of mud with an effective kinetic friction coefficient of 0.25. Determine the speed of the car after it emerges from the mud (in km/h).
Answer:
a. 6602.7 N b. 64.44 km/h
Explanation:
a. Find the force of friction required to keep it from sliding down the hill.
The frictional force f equals the component of the weight of the car, W perpendicular to the inclined hill = Wcosθ times the coefficient of static friction, μ = 0.45.
Since f = μN = μWcosθ = μmgcosθ where m = mass of car = 1500 kg, g = acceleration due to gravity = 9.8 m/s² and θ = angle of incline of hill = 3.5°
So, f = μmgcosθ
= 0.45 × 1500 kg × 9.8 m/s²cos3.5°
= 6615cos3.5°
= 6602.7 N
b. Determine the speed of the car after it emerges from the mud (in km/h)
Since the car drops a vertical height of 14 m, its potential energy decreases by mgh and its kinetic energy increases by mgh where m =mass of car and h = height drop = 14 m. So its kinetic energy increase is ΔK = mgh = 1500 kg × 9.8 m/s² × 14 m = 205800 J
Since it has an initial velocity of u = 50 km/h = 50 km/h 1000m/3600 s = 13.89 m/s, its initial kinetic energy is K = 1/2mu² = 1/2 × 1500 kg × (13.89 m/s)² = 144699.08 J.
Its new kinetic energy after the drop is thus K' = K + ΔK = 144699.08 J + 205800 J = 350499.08 J
Let v be its velocity after the drop, since K' = 1/2mv²,
v = √(2K'/m) = √(2 × 350499.08 J/1500 kg) = √(700998.16 J/1500 kg) = √(467.332 J/kg) = 21.62 m/s
Now, from work kinetic energy principles, the kinetic energy change in the car is the work done on car by friction
So, ΔK" = -fd = -μmgd
Let v' be the velocity of the car after emerging from the mud and moving a distance d = 30.0 m.
So, 1/2m(v'² - v²) = -μmgd
v'² - v² = -2μgd
v'² = v² - 2μgd
Substituting the values of the variables, we have
v'² = (21.62 m/s)² - 2 × 0.25 × 9.8 m/s² × 30.0 m
v'² = 467.42 m²/s² - 147 m²/s²
v'² = 320.42 m²/s²
taking square root of both sides, we have
v' = √(320.42 m²/s²)
= 17.9 m/s
Converting v to km/h we have v' = 17.9 m/s × 3600 s/h × 1 km/1000 m = 64.44 km/h.
So, the car emerges from the mud with a speed of 64.44 km/h
Investigator Campbell has bullets that were collected from the crime scene and puts them under her comparison microscope. What other item would she need to examine?
The victim's clothing with the gunshot hole
Test bullets fired from the suspected gun
A bullet that has been fired into a wall or hard surface
The barrel of the suspected gun
Answer:
Test bullets fired from the suspected gun
Explanation:
This is the correct answer!! I took the test!!
At a state championship High School football game, the intensity level of the shout of a single person in the stands at the center of the field is 48.1 dB. What would be the intensity level at the center of the field if all 4841 fans at the game shouted from roughly the same distance away from that center point
Answer:
The value is [tex]\beta_f = 84.95 \ dB[/tex]
Explanation:
From the question we are told that
The intensity level of the shout of a single person is [tex]\beta = 48.1 \ dB[/tex]
The number of fans is [tex]n = 4841[/tex]
Gnerally intensity level is mathematically represented as
[tex]\beta = 10 log * \frac{I}{I_o }[/tex]
Here [tex]I_o[/tex] is the minimum intensity of sound human ear can pick and the value is
[tex]I_o = 1 * 10^{-12} \ W/m ^2[/tex]
when [tex]\beta = 48.1 \ dB[/tex]
[tex]48.1 = 10 log * \frac{I}{ 1 * 10^{-12}}[/tex]
=> [tex]4.81 = log ( \frac{ I}{ 1 * 10^{-12}} )[/tex]
taking antilog of both sides
[tex]64565.42 = \frac{I}{ 1 *10^{-12}}[/tex]
=> [tex]I = 6.457 *10^{-8} \ W/m^2[/tex]
Generally the intensity for the whole fans is mathematically represented as
[tex]I_f = n * I[/tex]
=> [tex]I_f = 4841 * 6.457 *10^{-8 }[/tex]
=> [tex]I_f = 0.0003126 \ W/m^2[/tex]
Gnerally the intensity level for the whole fans is mathematically represented as
[tex]\beta_f = 10 log [ \frac{I_f }{I_o } ][/tex]
=> [tex]\beta_f = 10 log [ \frac{ 0.0003126 }{1*10^{-12}}[/tex]
=> [tex]\beta_f = 84.95 \ dB[/tex]
An l-c cirucit with a 70 mh inductor and a ,54 F capacitor oscillates. The maximum charge on the capacitor is 11.5 C. What are the oscillation frequency and the maximum current in this circuit
This question is incomplete, the complete question is;
An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit ;
Options
a) 1.07 kHz, 63.4 mA
b) 4.38 kHz, 101.3 mA
c) 6.74 kHz, 55.7 mA
d) 2.31 kHz, 93.5 mA
e) 0.82 kHz, 59.1 mA
Answer:
the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.
so Option e) 0.82 kHz, 59.1 mA is the correct answer
Explanation:
Given that;
inductor L = 70 mH = 70 × 10⁻³ H
Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f
Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c
oscillation frequency in L-C circuit;
f = 1/2π√(LC)
we substitute our values;
f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶ )
f = 0.0818 × 10⁴ Hz
f = 0.082 × 10³ Hz ≈ 0.82 kHz
Maximum circuit in L-C circuit is given by
I_max = Qmax/√(LC)
we substitute
I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶ )
= 59.1 × 10³ A ≈ 59.1 mA
Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.
so Option e) 0.82 kHz, 59.1 mA is the correct answer
First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is
The question is incomplete. The complete question is :
First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:
The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number
What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.
Solution :
a). Given :
[tex]$Q_1 = 42 \ J$[/tex] , [tex]$Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J $[/tex]
Net heat transfer
[tex]$Q= Q_1+Q_2$[/tex]
= 42 + (-22)
= 20 J
Total work
[tex]$W= W_1+W_2$[/tex]
= 6 + (-6)
= 0 J
∵ ΔU = Q - W
= 20 - 0
= 20 J
This is the net change in the internal energy of the system.
b). ΔU = Q + W
= (-140) + (-165)
= -305 J
c). ΔU = Q + W
= (-440) + (645)
= 205 J
d). ΔU = Q + W
= (-450) + (690)
= 240 J
The charge on an electron is
Answer:
there is a negative charge on electron.
Explanation:
Question 1 of 10
A wave meets a large barrier that has a small opening in it. The part of the
wave that meets the opening bends as it passes through. Which statement
best describes what has happened to the wave?
OA. The opening in the barrier absorbed all of the wave's energy.
OB. The wave was reflected as it passed through the opening in the
barrier.
OC. The opening in the barrier caused the wave to change speed and
refract.
OD. The wave diffracted as it passed through the opening in the
barrier.
Answer:
OD. The wave diffracted as it passed through the opening in the barrier.
Explanation:
A progressive wave (i.e waves in motion) has the capacity to bend around an obstacle on its path. This is one of the general properties of waves called diffraction. Others are: reflection, refraction, interference. Note that only transverse waves undergo polarization.
Diffraction of waves is the ability of waves to bend around an obstacle on its path during progression.
Thus, the bending of the part of waves as it passes through the barrier implies that the wave diffracted as it passed through the opening in the barrier.
Answer:
The wave diffrected as it passed through the opening
Explanation:
In Space, an astronaut releases a wrench from his hand. The wrench has a mass of 4 grams and is traveling with a velocity of -15m/s. The Astronaut’s mass is 70kg. What is his Velocity?
Answer:
[tex]v_=-8.5\times 10^{-4}\ m/s[/tex]
Explanation:
Given that,
Mass of a wrench, m₁ = 4 g = 0.004 kg
Speed of wrench, v₁ = -15 m/s
Mass of the Astronaut, m₂ = 70 kg
We need to find Astronaut's velocity. Let it is v₂. Using the conservation of linear momentum to find it.
[tex]m_1v_1=m_2v_2\\\\v_2=\dfrac{m_1v_1}{m_2}\\\\v_2=\dfrac{0.004\times (-15)}{70}\\\\=-8.5\times 10^{-4}\ m/s[/tex]
So, the speed of Astronaut is [tex]8.5\times 10^{-4}\ m/s[/tex].
Is anyone good at science I need help with 2 tests
Answer:
i am!
Explanation:
A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The mass is released, separating it 5 cm from its equilibrium position. Find the period of the motion
Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
The signal from the oscillating electrode is fed into an amplifier, which reports the measured voltage as an rms value, 1.0 nV . What is the potential difference between the two extremes
Answer:
The value is [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]
Explanation:
From the question we are told that
The measure voltage is [tex]E_{rms} = 1.0 \ n V = 1.0 *10^{-9} \ V[/tex]
Generally the peak voltage is mathematically represented as
[tex]E_{max} = \sqrt{2} * E_{rms}[/tex]
=> [tex]E_{max} = \sqrt{2} * 1.0 *10^{-9}[/tex]
=> [tex]E_{max} = 1.4142 *10^{-9} \ volts[/tex]
Generally the potential difference between the two extremes is mathematically represented as
[tex]V = 2 * E_{max}[/tex]
=> [tex]V = 2 * 1.4142 *10^{-9}[/tex]
=> [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]
The orbital radius of Venus
is 10.8 x 1010 m. Its rotation
period is 224.7 days. What is its
orbital speed?
A 20 km s-1 C 35 km s-1
B 25 km s-1 D 62 km 5-1
Answer:
v = 35 km/s
Explanation:
Given that,
Orbital radius of Venus, [tex]r=10.8\times 10^{10}\ m=10.8\times 10^{10}\times 10^{-3}=10.8\times 10^{7}\ km[/tex]
The rotation period is, [tex]T = 224.7\ \text{days}=19414080\ s[/tex]
We need to find the orbital speed of the Venus. The formula for the orbital speed is given by :
[tex]v=\dfrac{2\pi r}{T}\\\\v=\dfrac{2\pi \times 10.8\times 10^{7}}{19414080}\\\\=34.95\ km/s[/tex]
or
v = 35 km/s
So, the orbital speed of Venus is 35 km/s.
Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car.
a. Joel does 75% more work than Jerry.
b. Joel does 25% more work than Jerry.
c. Jerry does 50% more work than Joel.
d. Joel does 50% more work than Jerry
e. None of the above .
Answer:
the answer is B
Explanation: