Answer:
5. 32I
Explanation:
The moment of inertia of a solid sphere about its central axis is given by
I = [tex]\frac{2}{5} MR^2[/tex] ------------------(i)
Where;
M = mass of the sphere
R = radius of the sphere.
From the question;
Case 1: The aluminum sphere has a radius R and moment of inertia I.
This means that we can substitute these values of R and I into equation (i) and get;
I = [tex]\frac{2}{5} MR^2[/tex] --------------(ii)
M is the mass of the aluminum sphere and is given by;
M = pV
Where;
p = density of aluminum
V = Volume of the sphere = [tex]\frac{4}{3} \pi R^3[/tex]
=> M = p([tex]\frac{4}{3} \pi R^3[/tex]) --------------------(*)
Case 2: An aluminum sphere with a radius of 2R instead.
Let the moment of inertia in this case be I' and mass be M'
Substituting R = 2R, M = M' and I = I' into equation (i) gives
I' = [tex]\frac{2}{5} M'(2R)^2[/tex] ------------------(iii)
Where;
M' = pV'
p = density of aluminum
V' = volume of the sphere = [tex]\frac{4}{3} \pi (2R)^3[/tex]
=> M' = p([tex]\frac{4}{3} \pi (2R)^3[/tex])
Rewriting gives;
M' = p([tex]\frac{4}{3} \pi (2)^3(R)^3[/tex])
M' = p([tex]\frac{4}{3} \pi8(R)^3[/tex])
M' = 8p([tex]\frac{4}{3} \pi R^3[/tex])
From equation (*), this can be written as
M' = 8M
Now substitute all necessary values into equation (ii)
I' = [tex]\frac{2}{5} M'(2R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(2R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(2)^2(R)^2[/tex]
I' = [tex]\frac{2}{5} (8M)(4)(R)^2[/tex]
I' = [tex]\frac{2}{5} (32M)(R)^2[/tex]
I' = [tex]32[\frac{2}{5}MR^2][/tex]
Comparing with equation (ii)
I' = [tex]32[I][/tex]
Therefore, the moment of inertia about a central axis of a solid
aluminum sphere of radius 2R is 32I
High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.
Answer:
The speed of the golf ball just after the impact is 73.04 m/s.
Explanation:
Given that,
The mass of golf club, m₁ = 183 g = 0.183 kg
The mass of golf ball, m₂ = 46.6 g = 0.0466 kg
The initial speed of golf club, u₁ = 58.6 m/s
The initial speed of a golf ball, u₂ = 0
The final speeds of club, v₁ = 40 m/s
We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s[/tex]
So, the speed of the golf ball just after the impact is 73.04 m/s.
a brick of mass 0.8 kg is accidentally dropped from a high scaffolding. it reaches the ground with a kinetic energy of 240 J. How high is scaffolding ?(Take acceleration due to gravity g be 10 m s-¹)
Answer:
30 m
General Formulas and Concepts:
Energy
Gravitational Potential Energy: [tex]\displaystyle U_g = mgh[/tex]
m is mass (in kg)g is gravityh is height (in m)Kinetic Energy: [tex]\displaystyle KE = \frac{1}{2}mv^2[/tex]
m is mass (in kg)v is velocity (in m/s²)Law of Conservation of Energy
Explanation:
Step 1: Define
Identify variables
[Given] m = 0.8 kg
[Given] g = 10 m/s²
[Given] U = 240 J
[Solve] h
Step 2: Solve for h
[LCE] Substitute in variables [Gravitational Potential Energy]: (0.8 kg)(10 m/s²)h = 240 JMultiply: (8 kg · m/s²)h = 240 JIsolate h [Cancel out units]: h = 30 mbowling ball weighing 710 N attached to the ceiling by a rope of length 3.87 mThe ball is pulled to one side and released it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.51 m/s
What is the acceleration of the bowling ball, in magnitude and direction, at this instant?
What is the tension in the rope at this instant?
Answer:
a = 5.256 m / s² with the vertical direction upwards and T = 1090.8 N
Explanation:
to find the acceleration we must use Newton's second law, in the lowest part of the motion
T - W = m a
where the acceleration is centripetal
a = v² / r
let's calculate
a = 4.51² / 3.87
a = 5.256 m / s²
As the acceleration is centripetal it is directed towards the center of the circle, which in the lower part coincides with the vertical direction upwards.
Let's find the tension of the rope with the first equation
T = W + m a
W = m g
let's calculate
T = 710 + 710 5.256 / 9.8
T = 1090.8 N
A car with a mass of 1 892kg, traveling at 12m/s, was heading towards a wall. It applied 70N of force on its brakes for 6 seconds until it slows down.
Find:
1. The momentum of the car.
2. The resulting impulse of the car.
Answer:
MOMENTUM=205 IMPULSE = 5
Explanation:
The momentum of car is 22704 kg m/s.
The resulting impulse of car is 420 N-s.
What is momentum?Momentum is product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.
p= mv
p= 1892*12
p= 22704 kg m/s
What is impulse?Impulse is the motion produced by a starting force.
[tex]\Delta p = F\Delta t[/tex]
[tex]\Delta p = 70*6\\\Delta p = 420 Ns[/tex]
To learn more about Momentum here
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you are traveling in a convertible with the top down. the car is moving at a constant velocty of 25 m/s due east along falt ground. you throw a tennis ball straight upward at a speed of 15 m/swhen the ball just leaves your hand, what is the speed of the ball as measured by an observer on the side of the road
Answer:
Teh velocity of ball with respect to the ground is 29.2 m/s.
Explanation:
Velocity of car, vc = 25 due m/s east
velocity of ball with respect to car, v(b,c) = 15 m/s due Z axis
Write the velocities in the vector form
[tex]\overrightarrow{v}_{c}=25 \widehat{i}\\\\\overrightarrow{v}_{(b,c)}=15 \widehat{k}[/tex]
The velocity of ball with respect to ground is
[tex]\overrightarrow{v}_{(b,c)}= \overrightarrow{v}_{(b,g)}-\overrightarrow{v}_{(c,g)}\\\\15 \widehat{k} = \overrightarrow{v}_{(b,g)} - 25 \widehat{i}\\\\\overrightarrow{v}_{(b,g)} = 25 \widehat{i} + 15 \widehat{k}\\\\v_{(b,g)}=\sqrt{25^2 + 15^2}=29.2 m/s[/tex]
While taking a psychology exam, Lori got up and shut the classroom door due to outside noise that affected her concentration. Which theory explains Lori’s behavior?
Group of answer choices
Drive-reduction theory
Expectancy theory
Instinct theory
Arousal theory
Answer:
drive reduction theory
Explanation:
I would say that because of all the cars beeping and making A Lot of cachos on the street so that will definitely affect her taking her exam
Theory explains Lori’s behavior is b) Drive-reduction theory
What is Drive-reduction theory?
It is based on the idea that the primary motivation behind all human behavior is to reduce drives . A drive is a state of discomfort which is triggered by a person's physiology or biological need .
so , she removed discomfort that can happen by the noise outside the classroom in order to maintain student's concentration
correct answer is b) Drive-reduction theory
learn more about Drive-reduction theory
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Question 1 of 25
Which equation is an example of a synthesis reaction?
A. HNO3 + KOH → KCI + H20
B. 2Li+ CaCl2 - 2LiCl + Ca
O C. S+ 02 - S02
7
O D. CH4 + 202 - 2H2O + CO2
Answer:
C. S + 02 → S02
Explanation:
A synthesis or combination reaction is that reaction involving two elements as reactants to form a single compound as product.
In the reaction given below;
S + 02 → S02
Sulphur and oxygen are elemental substances that combine to synthesize sulfur IV oxide (SO2), and hence it is an example of synthesis reaction.
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of charge from one sphere to the other? Answer in Joules.
Answer:
Use 2>60n to get 19>089
Explanation:
The idea of a field was needed for which reason?
A. There was no way to calculate forces.
B. Forces were exerted without physical contact.
C. There was no way to know the direction of forces.
D. Forces affect only certain things.
Answer:
i believe b
Explanation:
for example, gravity works without directly contacting objects so a field helps visualize how gravity objects without physical interaction
Answer:
Explanation:
giâir
5. The disks, or pads, that exist between bones as gliding joints are made of which substance?
Answer:
cartilage
Explanation:
Answer:
Cartilage
Explanation:
Correct in AE
what was the significance of jumping a.keep the snake b.keep feet cleans c.avoid the hot water d.avoid the Bumbo stick
Answer:
D I think I’m not for sure
Explanation:
Planets closer to a star will have what type of average temperature
Answer:
Mercury - 800°F (430°C) during the day, -290°F (-180°C) at night. Venus - 880°F (471°C) Earth - 61°F (16°C) Mars - minus 20°F (-28°C)30-Jan-2018
cho hệ cơ học như hình vẽ hai đầu dây buộc hai vật có khối lượng tương ứng là m1=2kg và m2>m1 lấy g=10m/s sau 1s kể từ lúc bắt đầu chuyển dộng hệ vật đi được 50 cm tính m2 và sức căng của dây
xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

help me with this please
Answer:
Where is the Question?
Explanation:
I am glad to h help you out.
Answer:
Sorry is there any attachments to put with this question, If please publish question properly please so that brainly answerers will know what to answer.
A block whose weight is 45.8 N rests on a horizontal table. A horizontal force of 36.6 N is applied to the block. The coefficients of static and kinetic friction are 0.697 and 0.371, respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? If the block does not move, give 0 m/s2 as the acceleration?
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
(4.56 x 10^-13)-(1.17 x 10^-13)
show your full steps
Answer:
x = 13552.6 m
Explanation:
This is a projectile throwing exercise
Let's start by looking for the time it takes for the pump to reach the ground y = 0
y = y₀ + v_{oy} t - ½ g t²
as the plane flies horizontally the vertical speed is zero
0 = y₀ + 0 - ½ g t²
t = [tex]\sqrt{2y_o/g}[/tex]
t = [tex]\sqrt{2 \ 10000/9.8}[/tex]
t = 45.175 s
with this time we can find the distance it travels horizontally
x = v₀ₓ t
x = 300 45,175
x = 13552.6 m
the bomb must be dropped at this distance before hitting the target
New alleles arising from mutations in a population will
A 25N[L] force acts on a 1kg object that is already travelling at 10m/s[R]. The force acts for t=2s. What is its final speed? *
A) 25m/s
B) 25m/s/s
C) 40m/s[R]
D) 40m/s[L]
show your full work please
Answer:
C) 40m/s [R]
Explanation:
Given the following data;
Force = 25 N [L]
Mass = 1 kg
Initial velocity = 0 m/s
Velocity = 10 m/s [R]
Time = 2 seconds
To find the final velocity;
First of all, we would determine the acceleration of the object.
Force = mass * acceleration
25 = 1 * acceleration
Acceleration = 25/1
Acceleration = 25 m/s²
Next, we would determine the final velocity by using the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the formula, we have;
[tex] V = 0 + 25*2[/tex]
[tex] V = 0 + 50 [/tex]
V = 50 m/s
Final velocity = 50 - 10
Final velocity = 40 m/s [R]
A bullet is shot straight up into the air from ground level. It reaches a maximum
height at h = 739 m.
Complete question:
A bullet is shot straight up into the air from ground level. It reaches a maximum height at h = 739 m. Calculate the initial velocity of the bullet.
Answer:
the initial velocity of the bullet is 120.35 m/s
Explanation:
Given;
maximum height reached by the bullet, h = 739 m
let the initial velocity of the bullet = u
At maximum height the final velocity of the bullet, v = 0
Apply the following kinematic equation to determine the initial velocity of the bullet.
v² = u² - 2gh
0 = u² - 2gh
u² = 2gh
u = √2gh
u = √(2 x 9.8 x 739)
u = 120.35 m/s
Therefore, the initial velocity of the bullet is 120.35 m/s
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555.
(a) What is the magnitude of the frictional force?
(b) If the player comes to rest after 1.22 s, what is his initial speed?
Answer:
[tex]v=6.65m/sec[/tex]
Explanation:
From the Question we are told that:
Mass [tex]m=97.6[/tex]
Coefficient of kinetic friction [tex]\mu k=0.555[/tex]
Generally the equation for Frictional force is mathematically given by
[tex]F=\mu mg[/tex]
[tex]F=0.555*97.6*9.8[/tex]
[tex]F=531.388N[/tex]
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
[tex]a_f=-\mu g[/tex]
[tex]a_f=-0.555 *9.81[/tex]
[tex]a_f=-54455m/sec^2[/tex]
Therefore
[tex]v=u-at[/tex]
[tex]v=0+5.45*1.22[/tex]
[tex]v=6.65m/sec[/tex]
A speedometer in a car gives the car’s speed at that given moment, or the?
A. General speed
B. Instantaneous speed
C. Average speed
D. Constant speed
It’s not C or D!
Answer:
a because it is at a given moment
Explanation:
did u
A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.240 T/s .
Requried:
If the loop has a resistance of 0.700Ω. Find the current induced in the loop.
Answer:
The induced current is 26.7 mA
Explanation:
Given;
area of the loop, A = 0.078 m²
initial magnetic field, B₁ = 3.8 T
change in the magnetic field strength, dB/dt = 0.24 T/s
The induced emf is calculated as;
[tex]emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V[/tex]
The resistance of the loop = 0.7 Ω
The induced current is calculated as;
[tex]V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA[/tex]
Leakage of liquid petroleum gas can be detected from a distance . name the process
Answer:
ultrasonic
Explanation:
they detect the acoustic emission created when a pressured gas expands in a low pressure area through a small orifice(the leak).
A 2.80 kg mass is dropped from a
height of 4.50 m. Find its potential
energy (PE) when it is 3.00 m
above the ground.
Answer:
PE = 82.32J
Explanation:
PE = m*g*h
PE = 2.80kg*9.8m/s²*3m
PE = 82.32J
Under which conditions would a nucleus be least likely to undergo radioactive decay? A. When the strong nuclear forces are less than the electrostatic forces of repulsion
B. When the strong nuclear forces are greater than the electrostatic forces of repulsion
C. When the strong nuclear forces are less than the electrostatic forces of attraction
D. When the strong nuclear forces are greater than the electrostatic forces of attraction
Answer:
b is the correct answer i believe
Explanation:
the strong nuclear force encourages particles to come together so that they can nuetralize, a process the opposite of the electrostatic force of repulsion. This lets the atom stay intact because if the repulsion was more than the nuclear forces strength the atom would lose electrons and slowly decay
You sit on ice and shove a heavy box with your feet with a given force. What will you and the box share? *
A) Same acceleration
B) equal and opposite acceleration
C) the equal and opposite force
D) same force
explain please
Answer:
the correct answer is C
Explanation:
In this exercise we will analyze the situation.
When the person is on the ice, the friction coefficient is very small, if the box is in a place where there is no ice, the coefficient is different, so the friction force on each body is different.
Therefore the acceleration that each body acquires is different.
If we apply the conservation of momentum, each body moves in the opposite direction, but with different speeds.
If we use Newton's third law, the force applied to each body has the same magnitude and opposite direction, which is why the force is of the action and reaction type
Consequently the correct answer is C
1. Given an object that follows this time-dependent velocity function: ~v(t) = 2 m/s 2 tˆi − 3 m/s 3 t 2ˆj, and assuming the object begins at the origin at t=0s, where will the object be at t=2.0s?
2. Suppose an object of mass 2.0kg begins at rest and is acted upon by a force F(t) = 5.0N e^−0.1t . What will the object’s speed be after ten seconds?
Answer:
Explanation:
V=ds/dt
Where s =distance traveled
Given
V= 2ti - 3t²j
ds/dt=2ti - 3t²j
ds=(2ti - 3t²j)dt
Integrating
S= 2t²i/2 - 3t³j/3 + C
S=t²i - t³j + C
Since the object starts from Rest when t=0 and s(distance)=0
0=0²i - 0³j + C
C=0
Therefore
S=t²i - t³j
At t=2sec
S=(2)²i - (2)³j
S=4i - 8j
Magnitude of S(distance) = √4²+(-8)²
S= 4√5 meters
S=8.94meters.
2.mass =2kg
F(t)=5e^-0.1t
From Newton 2nd Law
F(t) = mdv/dt
5e^-0.1t= 2dv/dt
2dv = (5e^-0.1t)dt
Integrating
2V(t) = 5e^-0.1t/(-0.1) + C
2v(t) = - 50e^-0.1t + C
Since it started from rest at t=0. That is v(0)=0
2(0) = -50e^-0.1(0) + C
0 = -50 + C
C= 50
v(t) = -50e^-0.1t + 50
At t=10sec
v(10) = -50e^-0.1(10) + 50
V= -18.39 + 50
V= 31.61ms-¹.
While using your calc to evaluate -50e^-0.1(10)
Don't forget the -(minus) sign at the top of the exponential.
If you neglect it... You'll have a different answer.
Hope this helps.
Have a great day!!!
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.
Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?
Answer:
a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b) [tex]a=30.7[/tex]
c) [tex]a=35.91[/tex]
Explanation:
From the question we are told that:
Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
Initial Length [tex]L_1=0.600m[/tex]
Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]
Final Length [tex]L_2=0.900m[/tex]
a)
Generally the rotation with the greater speed is
[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b)
Generally the equation for centripetal acceleration at 8.16 is mathematically given by
[tex]a=\omega_1^2*L_1[/tex]
[tex]a=8.16 rev/s*0.6[/tex]
[tex]a=30.7[/tex]
c)
At 6.35 rev/s
[tex]a=6.35 rev/s*0.9[/tex]
[tex]a=35.91[/tex]
Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second.
Answer:
[tex]F = 14.2 N[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.15kg[/tex]
Radius [tex]r=0.6[/tex]
Angular Velocity [tex]\omega=2rev/s[/tex]
[tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]
Generally the equation for Force applied is mathematically given by
[tex]F =mrw2[/tex]
[tex]F=0.15*0.6* (4*x3.14^)2[/tex]
[tex]F = 14.2 N[/tex]