Answers
Answer: the angle between the resultant vector and the vector of the 18 pound is 28°
Explanation:
given that data in the question; as its interpreted in the diagram below;
from the cosine rule, we know that;
a² = b² + c² - 2bc
so
(13)² = (6)² + (18)² - (2 × 6 × 18 ) cos∅
169 = 36 + 324 - 216cos∅
169 = 360 - 216cos∅
216cos∅ = 360 - 169
216cos∅ = 191
cos∅ = 0.8842
∅ = cos⁻¹ ( 0.8842 )
∅ = 27.8° ≈ 28° {nearest whole number}
Therefore the angle between the resultant vector and the vector of the 18 pound is 28°
Related Questions
How can you measure the strength of any electric field?
Answers
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.
Explain the difference between a high tide and a low tide.
Answers
Answer: Tidal range
Explanation:
Tides are considering the rise and fall of sea levels and there are two types of it which are called high tide and low tide. The difference between high tide and low tide is called the tidal range.
The tidal range is not constant and it is considering height difference. It can change and it is depending on the locations of the Sun or the Moon.
High tide is the highest level of the place where the water rises because when the water rises to its highest level, then the water is reaching its high tide.When it comes to low tide, then it is the opposite of high tide. Water is reaching its lowest level.
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
Answers
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
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
Answers
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]
A constant force of magnitude 23 N acts on an object for 3.1 s. What is the magnitude of the impulse
Answers
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
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
Answers
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.
10. During 4th period we put Klaudia in a box because she was talking too much. We still heard her voice through the box so we decided to push her outside. The force of friction of the ground on the box was 68 N. If Mr.Whitmore can apply a force of 25 N and every other 7th grade student can apply a force of 6 N. How many students would Mr. Whitmore need to make the box start moving and go outside. (Think quickly, the faster we move the box out, the quicker she stops talking)
Answers
Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside
Explanation:
Given that;
friction force of ground box = 68 N
student of 7th grade = n
Whitmore can apply a force of 25 N
every other 7th grade student can apply a force of 6 N.
now
friction force = forced applied by whitmore + total force ny 7th grade student
we substitute
68 = 25 + 6n
6n = 68 - 25
6n = 43
n = 43/6
n = 7.17
Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside
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
Answers
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
A pinball machine launches a .045 kg ball at a speed of 9.2 m/s. Determine the potential energy of the spring just before it launched the pinball
Answers
Given :
A pinball machine launches a .045 kg ball at a speed of 9.2 m/s.
To Find :
The potential energy of the spring just before it launched the pinball.
Solution :
We know, their is no external force applied on system.
It means that kinetic energy will remains constant.
Initial Energy = Final Energy
[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\0 + P.E_i= \dfrac{mv^2}{2}+ 0\\\\P.E_i = \dfrac{0.045 \times 9.2^2}{2}\ J\\\\P.E_i = 1.9044\ J[/tex]
Therefore, the potential energy of the spring just before it launched the pinball os 1.9044 J.
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?
Answers
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].
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
Answers
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]
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
Answers
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]
draw simple vector diagram and resolve 60N at an angle of 30° from the horizontal.. plz help guys
Answers
Explanation:
The attatched figure shows the vector diagram for a force that has magnitude of 60 N and it is acting at an angle of 30° from the horizontal.
When it is resolved, the horizontal and vertical components are given by :
[tex]F_x=F\cos\theta\\\\=60\times \cos30\\\\=51.96\ N[/tex]
And
[tex]F_y=F\sin\theta\\\\=60\times \sin30\\\\=30\ N[/tex]
Hence, this is the required solution.
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?
Answers
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
You have to choose one of these Kpop members for the purge, or you can group them into three. Who do you choose, what are their skills, why are they one of the best choices?
- Namjoon
- Wonho
- Felix
- Seonghwa
- Bangchan
- Zico
Answers
Answer:
Wonho, Seonghwa, and Bangchan
Explanation:
Answer:
namjoon,seonghwa,bangchan
Explanation:
Why do you need air resistance in a parachute ?what happens without it?
And
How does air resistance affect acceleration and terminal velocity
Answers
air resistance slows acceleration down( air resistance increases for an object that is accelerating)
an object that encounters air resistance will eventually reach a terminal velocity (look at newton’s second law)
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
Answers
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
In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.
Answers
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
a guitar string is 0.620 m long, and oscillates at 234 Hz. if a player uses his finger to shorten the string to 0.480 m, what is the new frequency?
Answers
Answer:
the new frequency is : 302.25Hz
using the formula F2 = [tex]\frac{F1 L1}{L2}[/tex]
Explanation:
meaning of frequencyfrequency of a string is the number of vibrations of a plucked string per second. it is measured in Hertz.
the frequency of a string is inversely proportional to twice the length of the string. which means the longer the string, the smaller the freqency and the higher the string the higher the frequency.
f ∝ 1/2L.
f = k/2L
where f = frequency
L = length of string
k = constant
k = 2fL is a constant
given data
L1 = 0.62m
f1 = 234Hz
L2 = 0.48m
2f1L1 = 2f2L2
f1L1 = f2L2
f2 = f1L1/L2
f2 = [tex]\frac{234 x 0.62}{0.48}[/tex] = 302.25Hz
in conclusion, the new frequency is 302.25Hz
learn more about frequency of a vibrating string: brainly.in/question/1149252
SPJ2
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
Answers
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
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.
Answers
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:
difference between incident ray and refracted ray
Answers
Answer:
** incident ray.
Incident ray - the ray of light falling on the surface AB is called the incident ray
reflected ray.
** Reflected ray - the incident ray bouncing back in the same medium after striking the reflecting surface is called reflected ray.
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?
Answers
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
Is anyone good at science I need help with 2 tests
Answers
Answer:
i am!
Explanation:
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 .
Answers
Answer:
the answer is B
Explanation:
A mechanic uses a lift to raise an 11,760 N car 0.50m off the ground. How much work does the lift do on the car?
Answers
Answer:
5880 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 11760 × 0.5
We have the final answer as
5880 JHope this helps you
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!!
Answers
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 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).
Answers
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
A 50.0-kg child stands at the rim of a merry-go-round of radius 2.00 m, rotating in such a way that it makes one revolution in 2.09 s. What minimum coefficient of static friction between her feet and the floor of the merry-go-round is required to keep her in the circular path
Answers
Answer:
2.01
Explanation:
First, we need to find the centripetal acceleration.
We're given that the merry go round rotates 1 revolution in 2.09 seconds. Converting to rpm, we know that it rotates 30 revolution per minute
Now this speed gotten in rpm will be converted to m/s, to ease the calculation
30 rpm = πdN/60 m/s
30 rpm = (3.142 * 4 * 30)/60
30 rpm = 377.04/60
30 rpm = 6.284 m/s
a(c) = v²/r
a(c) = 6.284²/2
a(c) = 39.49 / 2
a(c) = 19.74 m/s²
F = ma
F = 50 * 19.74
F = 987 N
Also, Normal Force, F(n) =
F(n) = mg
F(n) = 50 * 9.81
F(n) = 490.5
We then use this to find the coefficient of static friction, μ
μ = F/F(n)
μ = 987 / 490.5
μ = 2.01
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
Answers
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