(a) The speed of the mass m2 just before it hits the floor for frictionless table is 1.98 m/s.
(b) The speed of the mass m2 just before it hits the floor when there is friction is 1.53 m/s.
Apply Newton's second law of motion for object m₁[tex]-f_k + T = m_1a\\\\ T = m_1 a + f_k \ ---\ (1)\\\\ [/tex]
Apply Newton's second law of motion for object m₂[tex]m_2g - T = m_2a \ ---- (2)\\\\ [/tex]
Solve (1) and (2) together
[tex]m_2 g - (m_1 a+ f_k) = m_2 a\\\\ m_2 g - f_k = m_1 a + m_2 a\\\\ m_2 g - f_k = a (m_1 + m_2)\\\\ a = \frac{m_2 g- f_k}{m_1 + m_2} [/tex]
When the table is frictionless, the acceleration of the masses is calculated as follows;
[tex]a = \frac{m_2 g}{m_1 + m_2} \\\\ a = \frac{1 \times 9.8}{2 + 1} \\\\ a = 3.27 \ m/s^2[/tex]
When the coefficient of friction between m1 and the table is 0.2, the acceleration of the masses is calculated as follows;
[tex]a = \frac{m_2 g - f_k}{m_ 1+ m_2} \\\\ a = \frac{m_2 g - \mu_k m_1g}{m_1 + m_2} \\\\ a = \frac{g(m_2 - \mu_km_1)}{m_1 + m_2} \\\\ a = \frac{9.8(1 - \ 0.2\times 2)}{2+1 } \\\\ a = 1.96 \ m/s^2[/tex]
The height of the mass m2 above the ground = 60 cm = 0.6
Speed of the mass for frictionless tableThe speed of the mass m2 just before it hits the floor for frictionless table is calculated as follows;
[tex]v^2 = u^2 + 2ah\\\\ v^2 = 0 + 2ah\\\\ v^2 = 2ah\\\\ v= \sqrt{2ah} \\\\ v = \sqrt{2 \times 3.27 \times 0.6} \\\\ v = 1.98 \ m/s[/tex]
The speed of the mass when there is frictionThe speed of the mass m2 just before it hits the floor when there is friction is calculated as follows;
[tex]v = \sqrt{2ah} \\\\ v = \sqrt{2 \times 1.96 \times 0.6} \\\\ v = 1.53 \ m/s[/tex]
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Which kind of EM Radiation carries the most energy, microwaves
or IR?
Answer:
gamma-rays
Explanation:
a car travels for 3 hours to a distance city 200km due East (1) what was its speed in (a)m/s (b) km/h (2) what was its velocity in (a) m/s (b) km/h
Hope you could understand.
If you have any query, feel free to ask.
1 2 3 TIME REMA 56:1 Dierdra began writing a summary of the relationship between the first and second laws of thermodynamics. Which best completes Dierdra's summary? Thermal energy in a system is conserved and can increase the system's internallenergy and/or be used to do work. O Thermal energy moves from cooler objects to warmer objects O Thermal energy moves from warmer objects to cooler objects. 0 Thermal energy is destroyed when it leaves the system. 0 Thermal energy is created when it enters the system
Answer: Option (b) is the correct answer.
Explanation:
The energy present within the molecules of an object is known as thermal energy.
When an object is warmed then its molecules gain more kinetic energy and results in more number of collisions. Then they release extra energy in the form of heat into the atmosphere.
And when a cool object is placed adjacent to a warmer object then heat transfers from warm object to cool object.
Thus, we can conclude that thermal energy moves from warmer objects to cooler objects.
PLEASE HELP ANYONE?!!!!!
Answer:
for question 3 the answer is A.
Explanation:
Pulse is not a part of the body but rather a measurement, same as with strength.
Find the acceleration of a car that goes from 16 m/s to 48 m/s in 8 seconds. Help asappp and put units plssss
Answer:
48 m/s - 16 m/s = 32 m/s
(32 m/s )/ (8 s) = 4 m/s²
If Ms. Moss is going 70m/s how long will it take her to go 4,200 m?
Answer:
from
velocity= displacement/time
then
time=displacement/velocity
time= 4200/70
time=60s
time will take 60seconds
A sprinter accelerates from the starting line in the 100 meter dash at a rate of 7.0ft/s and reaches a velocity of 28.2 ft/s. What amount of time did it take to reach the final velocity?
The time taken for the sprinter to reach the final velocity is 5.68 s.
What is velocity?
This can be defined as the ratio of displacement of a body to time. The S.I unit of velocity is m/s.
The time taken for the sprinter to reach its final velocity can be calculated using the formula below.
Formula
S = (v+u)t/2.................... Equation 1Where:
S = Total distance covered by the sprinterv = Final velocityu = Initial velocityt = time taken to reach its final velocitymake t the subject of the equation
t = 2S/(v+u)................. Equation 2From the question,
Given:
S = 100 mu = 7 ft/sv = 28.2 ft/sSubstitute these values into equation 2
t = 2(100)/(28.2+7)t = 200/35.2t = 5.68 sHence, The time taken for the sprinter to reach the final velocity is 5.68s
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A car traveling with an initial velocity of 31 m/s slows down at a constant rate of 5.4 m/s2 for 3 seconds. What is its velocity at the end
of this time?
If a car traveling with an initial velocity of 31 meters/seconds slows down at a constant rate of 5.4 meters/seconds for 3 seconds, the velocity at the end would be 47.2 m/s.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
As given in the problem , a car traveling with an initial velocity of 31 meters/seconds slows down at a constant rate of 5.4 meters / seconds² for 3 seconds,
By using the second equation of motion,
S = ut + 1/2*a*t²
=31×3 + 0.5×5.4×3²
= 93 + 24.3
=117.3
Now by using the third equation of motion,
v² - 31² = 2×5.4×117.3
v = 47.2 m/s
Thus, the velocity of the car at the end of the three seconds would be 47.2 m/s.
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Examine The
graph and determine which girl
swam the farthest during the workout
Answer:
Mary
Explanation:
because if you look at her line, her distance is exceedin the others faster and higher. She has swam 2,200 m in 30 minutes while the other girls only swam 1,800 m and 1,200 m.
hence, Mary swam the most distance after 30 min.
have a nice night! :)
How is height related to work and force?
What would the weight of a 10kg rock be on Saturn? w=mg
90 N
9 N
568 N
120,536 N
Answer:
Not one of your choices .....closest is 90 N
Explanation:
Saturn g = 10.44 m/s^2
10 kg * 10.44 m/s^2 = 104.4 N
If you play a violin what kind of energy are you using
Answer:
mechanical energy.
Explanation:
When the body of the violin and the strings vibrate and make sound
Explain a charged by neutral body negatively by induction
Katarina’s therapist thinks that Katarina is experiencing an identity crisis. How is Katarina MOST likely demonstrating characteristics of this crisis?
A.
She has a new group of friends each week.
B.
She is proud of the woman she is becoming.
C.
She is constantly worried about school exams.
D.
She has purchased clothes that make her feel confident.
Answer:
A. She has a new group of friends each week.
Explanation:
This seems the most logical of all the options. Looking for new friends constantly may mean that she is trying to find somewhere where she can fit in. Therefore, she is not even aware of where that is herself.
I just got it right on my test as well.
If we apply a constant force of 100 N on a 10 kg object that is located on a frictionless surface, what is the acceleration of the object?
Answer:
10 m/s²
Explanation:
We know F = ma.
=> 100N = 10kg x a
=> a = 100 / 10 = 10 m/s²
BRAINLIEST.
BAM!
If we apply a constant force of 100 N on a 10 kg object that is located on a frictionless surface, the acceleration of the object would be 10 meters/second².
What is acceleration?The rate of change of the velocity with respect to time is known as the acceleration of the object.
As given in the problem If we apply a constant force of 100 N on a 10 kilograms object that is located on a frictionless surface, then we have to find out the acceleration
acceleration on the frictionless surface = force /mass
= 100 Newtons/10 kilograms
=10 meters/second²
Thus, the acceleration of the object is 10 meters/second².
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Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x [tex]10^{11}[/tex] N and 66 degrees
ELECTRIC FORCE (F)
F = [tex]\frac{KQq}{d^{2} }[/tex]
Where K = 9 x [tex]10^{9}[/tex] N[tex]m^{2}[/tex]/[tex]C^{2}[/tex]
The distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex] can be calculated by using Pythagoras theorem.
d = [tex]\sqrt{33^{2} + 33^{2} }[/tex]
d = 46.7 cm = 0.467 m
For force [tex]F_{1}[/tex], substitute all the parameters into the formula above
[tex]F_{1}[/tex] = (9 x [tex]10^{9}[/tex] x 3 x 1)/[tex]0.467^{2}[/tex]
[tex]F_{1}[/tex] = 2.7 x [tex]10^{10}[/tex]/0.218
[tex]F_{1}[/tex] = 1.24 x [tex]10^{11}[/tex] N
For force [tex]F_{4}[/tex], substitute all the parameters into the formula above
[tex]F_{4}[/tex] = (9 x [tex]10^{9}[/tex] x 3 x 4)/[tex]0.33^{2}[/tex]
[tex]F_{4}[/tex] = 1.08 x [tex]10^{11}[/tex]/0.1089
[tex]F_{4}[/tex] = 9.92 x [tex]10^{11}[/tex] N
For force [tex]F_{2}[/tex], substitute all the parameters into the formula above
[tex]F_{2}[/tex] = (9 x [tex]10^{9}[/tex] x 3 x 2)/[tex]0.33^{2}[/tex]
[tex]F_{2}[/tex] = 5.4 x [tex]10^{10}[/tex]/0.1089
[tex]F_{2}[/tex] = 4.96 x [tex]10^{11}[/tex] N
Summation of forces on Y component will be
[tex]F_{y}[/tex] = [tex]F_{4}[/tex] - [tex]F_{1}[/tex] Sin 45
[tex]F_{y}[/tex] = 9.92 x [tex]10^{11}[/tex] - 1.24 x [tex]10^{11}[/tex] Sin 45
[tex]F_{y}[/tex] = 9.04 x [tex]10^{11}[/tex] N
Summation of forces on X component will be
[tex]F_{x}[/tex] = [tex]F_{2}[/tex] - [tex]F_{1}[/tex] Cos 45
[tex]F_{x}[/tex] = 4.96 x [tex]10^{11}[/tex] - 1.24 x [tex]10^{11}[/tex] Sin 45
[tex]F_{x}[/tex] = 4.08 x [tex]10^{11}[/tex] N
Net Force = [tex]\sqrt{F_{x} ^{2} + F_{y} ^{2} } }[/tex]
Net force = [tex]\sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2} }[/tex]
Net force = 9.9 x [tex]10^{11}[/tex] N
The direction will be
Tan ∅ = [tex]F_{y}[/tex]/[tex]F_{x}[/tex]
Tan ∅ = 9.04 x [tex]10^{11}[/tex] / 4.08 x [tex]10^{11}[/tex]
Tan ∅ = 2.216
∅ = [tex]Tan^{-1}[/tex](2.216)
∅ = 65.7 degrees
Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x [tex]10^{11}[/tex] N and 66 degrees approximately.
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The addition of vectors and Coulomb's law allows us to find the result for the force on the charge q₃ is:
The modulus is F = 5.71 N The direction is: tes = -65º
Given parameters,
The value of the charges, q₁= +q, q₂= - 2 q, q₃= -3 q, q₄= -4q The value of the charge q=+ 2.4 μC = 2.4 10⁻⁶ C The side of the square is d = 33 cm = 0.33 mTo find.
The force on q3Electric Force.
The electric force is given by Coulomb's law, which states that the force is proportional to the charges and inversely proportional to the square of the distance.
F =[tex]k \frac{q_1q_2}{r_{12}^2}[/tex]
where F is the force, k Coulomb's constant, q the charges and r the distance between the charges.
Sum Vectors.
Force is a vector magnitude, so vector algebra must be used for vector addition, an easy and efficient way is to use an analytical method for addition:
We decompose each vector We make the sum of the components We construct the resulting vector.
In the attachment we have a diagram of the forces, the sum on each axis is:
x axis
Fₓ= F₂- F₁ₓ
y axis
[tex]F_y = -F_4 + F_{1y}[/tex]
Let's use trigonometry to find the component of force.
cos 45 = [tex]\frac{F_1_x}{F_1}[/tex]
cis 45 = [tex]\frac{F_1_y}{F_1}[/tex]
F₁ₓ= F₁ cos 45
[tex]F_1_y[/tex] = F₁ sin 45
Let's look for the distance between the charges, for charges 2 and 3 charges 4 and 3 the distance of is equal to the side of the square
r=r₂₃ = r₄₃ = d
The distance between charges 1 and 3 is the diagonal of the square that we can find with the Pythagorean theorem.
R₁₃² = d² + d²
R₁₃² = 2d²
We substitute in Coulomb's law.
F₂₃ = [tex]k \frac{q_2q_3}{d^2}[/tex]
F₄₃ = [tex]k \frac{q_4q_3}{d^2}[/tex]
F₁₃ = [tex]k \frac{q_1q_3}{2d^2 }[/tex]
Let us substitute in the components of the force on each axis.
x axis
Fₓ= F₂- F₁ₓ
[tex]F_x = k \frac{q_2q_3}{d^2} - k \frac{q_1q_3}{2d^2} cos 45 \\F_x = k \frac{q_3}{d^2} ( q_2 - \frac{q_1 cos 45}{2})[/tex]
y axis
[tex]F_y = - F_4 + F_1_y \\F_y = k \frac{q_4q_3}{d^2} - k \frac{q_1q_3}{2d^2} sin 45 \\F_y = k \frac{q_3}{d^2} ( -q_4 + \frac{q_1sin 45}{2})[/tex]
We substitute in value of each charge.
x axis
[tex]F_x = k \frac{3q}{d} \ q( 2 - \frac{1 \ cos 45}{2} ) \\F_x = 3k (\frac{q}{d})^2 1.6464[/tex]
y axis
[tex]F_y = k \frac{3q}{d^2} ( -4 + \frac{1 \ sin45}{2} ) \\F_y = 3k (\frac{q}{d})^2 ( -3.64645)[/tex]
The resulting vector is
F = [tex]F_x \hat i + f_y \hat j[/tex]
F = [tex]3k(\frac{q}{d} )^2 ( 1.6464 \hat i - 3.64645 \hat j )[/tex]3k q² /d² ( 1.6464 i^ - 3.64645 j^ )
We calculate.
F = 3 9 10⁹ ( 2.4 10-6/0.33)² ( 1.6464 i^ - 3.64645 j^ )
F = 1.428( 1.6464 i⁻ 3.64645 j^
We build the resulting vector, for the module we use the Pythagorean theorem.
[tex]F = \sqrt{F_x^2 + F_y^2}\\F = 1.428 \ \sqrt{1.6464^2+ 3.64645^2}[/tex]
F = 5.71N
Let's use trigonometry for the angle.
[tex]tan \theta = \frac{F_y}{F_x}[/tex]
[tex]\theta = tan^{-1} (\frac{-3.64645}{1.6464} )[/tex]
θ = -65º
This angle is half clockwise from the positive side of the x-axis.
in conclusion using vector addition and Coulomb's law we can find the result for the force on the charge q3 is:
The direction is: tes = -65ºThe modulus is F = 5.71 N
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PLEASE SAVE ME FROM THIS HW!!
A model rocket climbs 200m in 4 seconds. If was moving 10 m/s to begin with, what is its final velocity
Explanation:
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Consider two simple pendula. Pendulum 1 has a length of L and a bob mass of m. Pendulum 2 has a length of 4L and a bob mass of 4m. If the period of pendulum 1 is 4 seconds, then the period of pendulum 2 is...
2 s
16 s
1 s
8 s
Answer:
8s
Explanation:
period of pendulum = 2π*√(l/g)
l = length of pendulum
period for 1st pendulum = 4 sec
thats 2π*√(l/g) = 4
for 2nd pendulum
length = 4 l
substitute 4l in formula
2π*√4l/g = 2*4= 8s
Given the data from the question, the period of the pendulum 2 is 8 s
Data obtained from the questionLength of pendulum 1 (L₁ ) = LPeriod of pendulum 1 (T₁ ) = 4 sLength of pendulum 2 (L₂) = 4LPeriod of pendulum 2 (T₂) = ?How to determine the period of pendulum 2The period of pendulum 2 can be obtained as illustrated below:
T²₁ / L₁ = T²₂ / L₂
4² / L = T²₂ / 4L
Cancel out L
4² = T²₂ / 4
Cross multiply
T²₂ = 4² × 4
T²₂ = 64
Take the square root of both sides
T₂ = √64
T₂ = 8 s
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A body starts from rest and moves with uniform acceleration of 6m/s?. What distance does it cover in the third second?
We are given –
n = 3Initial velocity, u = 0 m/s Acceleration, a = 6 m/s²Distance covered in nth seconds –
↠Sn = u + 1/2 × a × (2n-1)
↠Sn = 0 + 1/2 × 6 × (2× 3 -1)
↠Sn = 0 + 3 × (6-1)
↠Sn = 0 + 3 × 5
↠Sn = 15 m
Henceforth, distance covered in the 3rd sec will be 15 m.PLSSSSS.......HELPPPPPP.
Answer:
volume of mass and size of mass
Explanation:
I remeber this from school trust me
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• A steel wire 4.7 m long stretches 0.11 cm when it is subjected to a tension of 360 N. What is the spring constant of the wire?
Answer:
Explanation:
length = 4.7 m
extension x = (0.11 /100) m + 4.7 m
= 4.7011 m
force = 360 N
spring constant k = ?
formula : F = kx
k = F/ x
k = 360 ÷ 4.7011
k= 76.578 Nm^-1
The spring constant of the wire of a steel wire 4.7 m long stretches 0.11 cm when it is subjected to a tension of 360 N is 76.57 N/m.
What is the spring constant?Simple, symmetrical motion i.e., SHM is a genuinely fascinating kind of agitation. It is continuously used in the objects' oscillatory motion. Most springs have SHM. The "spring constants" that are intrinsic to springs determine how stiff they are. A well-known law, Hooke's law, describes the SHM and provides a formula for the applied force using the spring constant.
Simple harmonic motions and Hooke's law are related to the definition of the spring constant. Therefore, we must first examine Hooke's rule before attempting to define the spring constant and comprehend how it operates.
Given:
The tension force, T = 360 N,
The length of the wire, L = 4.7 meters,
The elongation, l = 0.11 cm,
Calculate the spring constant by the following formula,
k = T / l
( l = x(0.11 /100) m + 4.7 m= 4.7011 m)
Here, k is the spring constant.
Substitute the values,
k = 360 / 4.7011
k = 76.57 N / m
Therefore, the spring constant of the wire of a steel wire 4.7 m long stretches 0.11 cm when it is subjected to a tension of 360 N is 76.57 N/m.
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An especially violent lightning bolt has an average current of 1.16 x 103 A lasting 0.139 s. How much charge is delivered to the ground by the lightning bolt
Charge is current times time :
Q = I × t
Q = 1.16 × 10³ × 0.139
Q = 161.24 C
A 12-V car battery supported a current of 150 mA to bulb calculate the power absorbed by the bulb over an interval of 20 mintues?
Answer:
+1.8W2,160JExplanation:
First, we need to solve for the power absorbed by the bulb.
Formula [ p = vi ]
p = 12 * 150 * 10^-3
p = +1.8W
Second, we can solve for the power aborbed in a interval of 20 minutes.
Formula [ e = vit ]
e = 12 * 150 * 10^-3 * 20 * 60
e = 2,160J
Just a note that "20 * 60" represents the time given in seconds.
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An object of height 5cm is placed 20cm in front of pin hole camera from a cubic box of side 6cm . Determine the height of the image formed
does the mass of a parachute affect terminal velocity?
Answer:
The greater weight increases the terminal velocity by acting as an extra force against gravity and air resistance.
This is the WIDTH of the wave HELP ME!!!
A Crest
B Trough
C Wavelength
D Amplitude
1) 2 objects are floating in space only 50 m apart. They are pulling on each other with the force of 2.668 X10-9 N. If one object has a mass of 100 kg, what is the mass of the 2nd object? (1000 kg)
Hope you could understand.
If you have any query, feel free to ask.
Part F
Rub one balloon with wool, fur, or hair and the other balloon with a piece of plastic wrap. Again,
suspend the two balloons so that they're about six inches apart. Do they attract or repel each
other?
Answer:
Two balloons both rubbed with plastic wrap, however, have the same result. The two equal (positive) charges push away from each other. When one balloon is rubbed with wool and one balloon with plastic wrap, the result is opposite charges. In this case, opposites truly attract.
Explanation:
10. What is the resistance of a lamp that is connected to a 110 voltage source that draws 2 amps of
current?
Answer:
According to ohms law
V= I x R
R= V ÷ I
R= 110 ÷ 2
R= 55 ohms
Explanation:
Questions: 1. Give the names of the following 1.1 NaF
Answer:
Sodium fluoride
Explanation:
Sodium Fluoride
