Given:
Two objects collide and stick together.
To find:
Effect on the kinetic energy due to the collision.
Explanation:
When perfectly inelastic bodies moving along the same line collide, they stick together. An inelastic collision is a collision in which there is a loss in kinetic energy.
Let m1 and m2 be the masses and velocity v1 and v2 be the velocities of objects along the same line before the collision.
Let V be the velocity of objects after they collide and stick together.
By the law of conservation of linear momentum,
[tex]m_1v_1+m_2v_2=m_1V+m_2V[/tex]Rearranging the above equation, we get:
[tex]V=\frac{m_1v_1+m_2v_2}{m_1+m_2}\text{ ......\lparen1\rparen}[/tex]The total kinetic energy of objects before the collision is:
[tex]\text{Kinetic Energy \lparen before collision\rparen}=m_1v_1^2+m_2v_2^2\text{ ......\lparen2\rparen}[/tex]The kinetic energy after collision is:
[tex]\text{Kinetic energy \lparen after collision\rparen}=\frac{1}{2}(m_1+m_2)V^2\text{ ......\lparen3\rparen}[/tex]Using equations (1), (2), and (3), the loss in kinetic energy is given as:
[tex]\begin{gathered} \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2}\lbrack m_1v_1^2+m_2v_2^2-\frac{m_1v_1+m_2v_2}{m_1+m_2}\rbrack \\ \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2}\lbrack\frac{m_1m_2\left(v_1^2+v_2^2-2v_1v_2\right)}{m_1+m_2}] \\ \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{m_1m_2\left(v_1-v_2\right)^2}{2(m_1+m_2)} \end{gathered}[/tex]From the above result, we see that the energy loss in the kinetic energy is positive.
Thus, the kinetic energy of decreases, when two objects collide and stick together.
Final answer:
The kinetic energy of decreases, when two objects collide and stick together.
A .5kg book is thrown at a speed of 3m/s. What type of energy does it have and how much?
Given:
The mass of the book is m = 0.5 kg
The speed of the book is v = 3 m/s.
To find the type of energy and how much.
Explanation:
As the book is in motion, the energy is kinetic energy.
The formula to calculate kinetic energy is
[tex]K.E.\text{ =}\frac{1}{2}mv^2[/tex]On substituting the values, the kinetic energy will be
[tex]\begin{gathered} K.E.=\frac{1}{2}\times0.5\times(3)^2 \\ =2.25\text{ J} \end{gathered}[/tex]Thus, the kinetic energy will be 2.25 J
Some electric power companies use water to store energy.Water is pumped from a low reservoir to a high reservoir.To store the energy produced in 1.0 hour by a 180-MWelectric power plant, how many cubic meters of water willhave to be pumped from the lower to the upper reservoir?Assume the upper reservoir is an average of 380 m abovethe lower one. Water has a mass of for every 1.00 x 10^3 kg for every 1.0m^3.
Answer:
1.74 x 10⁵ m³
Explanation:
The power is equal to:
[tex]P=\frac{E}{t}\rightarrow E=Pt[/tex]Where P is the power, E is the energy and t is the time. So, we can calculate the energy replacing
P = 180 MW = 180 x 10⁶ W
t = 1 hour = 3600 s
So, the energy is
[tex]\begin{gathered} E=(180\times10^6\text{ W\rparen\lparen3600 s\rparen} \\ E=6.48\times10^{11}\text{ J} \end{gathered}[/tex]On the other hand, the energy stored will be potential, so
[tex]E=mgh[/tex]Where m is the mass, g is the gravity and h is the height. Solving for m and replacing E = 6.48 x 10^11 J, g = 9.8 m/s², and h = 380 m, we get:
[tex]\begin{gathered} m=\frac{E}{gh} \\ \\ m=\frac{6.48\times10^{11}\text{ J}}{9.8\text{ m/s}^2\text{ \lparen380 m\rparen}} \\ \\ m=1.74\times10^8\text{ kg} \end{gathered}[/tex]Finally, we can calculate the volume of the water in cubic meters using the given ratio that Water has a mass of 1.00 x 10^3 kg for every 1.0m^3, so
[tex]1.74\times10^8\text{ kg }\times\frac{1\text{ m}^3}{1\times10^3\text{ kg}}=1.74\times10^5\text{ m}^3[/tex]Therefore, the answer is 1.74 x 10⁵ m³
A train car with a mass of 1 kg is stuck to another train car with a mass of 2 kg. An Explosion pushes the two carts away from each other. After the recoil, the 1 kg train car is now moving at -2 m/s and we need to find the Velocity of the 2 kg train car.
We know that the momentum is conserved on the system this will mean that:
[tex]p_i=p_f[/tex]where:
[tex]p=mv[/tex]At the beginning the cars are stuck together and we can assume they are at rest; we know that after the explosion the velocity of the 1 kg car is -2 m/s; then we have:
[tex]\begin{gathered} (1+2)(0)=(1)(-2)+2v \\ 0=-2+2v \\ 2v=2 \\ v=\frac{2}{2} \\ v=1 \end{gathered}[/tex]Therefore, the velocity of the 2 kg car is 1 m/s.
A football kicker is attempting a field goal from 44m out. The ball is kicked and just clears the lower bar with a time of flight of 2.9s. If the angle of the kick was 45°, what was the initial speed of the ball, assuming no air resistance?Group of answer choices
Given,
Distance travelled, d=44 m.
The time of flight t=2.9 s
The angle is
[tex]\theta=45^o[/tex]Let the initial velocity be v
The motion is projectile.
Thus the horizontal distance travelled is
[tex]\begin{gathered} d=(v\cos \theta)t \\ \Rightarrow44=v\times\cos 45^o\times2.9 \\ \Rightarrow v=21.5\text{ m/s} \end{gathered}[/tex]Thus the answer is 21.5 m/s
A cat is thrown straight up at 24.9 m/s What is the cat's velocity 4.3 seconds later?
Answer:
-17.3 m/s
Explanation:
a = -9.81 mm/s^2 vo = 24.9 m/s
vf = vo - at
= 24.9 - 9.81 ( 4.3) = -17.3 m/s ( cat is already coming back down)
Calculate the x-component and the y-component (in m) of the vector with magnitude 26.0 m and direction 45.0°.x = m y = m
The components can be found as:
[tex]\begin{gathered} V_x=V\cdot\cos (45)=26\cdot\cos (45) \\ V_x=13\sqrt[]{2} \\ V_x\approx18.38m \\ ----------------------- \\ V_y=V\cdot\sin (45)=26\cdot\sin (45) \\ V_y=13\sqrt[]{2} \\ V_y\approx18.38m \end{gathered}[/tex]Inquiry is a process promoted by a Blank
Inquiry is a process prompted by curiosity, observation or question.
Explanation:Inquiry is driven by curiosity, wonder, interest and/or a need to answer a question. Being able to ask rich questions enables students to construct their knowledge and develop an understanding of concepts and experiences.
If Rodger wants to determine whether a baseball thrown at 100 m/s or a bowling ball thrown at 100 m/s will have more momentum, what is one measurement device he could use to figure this out, without actually throwing either ball?A. a rulerB. a stopwatchC. a balance scaleD. an accelerometer I chose C, I am not completely sure that I’m right
The momentum of an object is given by the product of its mass and its velocity:
[tex]p=mv[/tex]If both objects have the same velocity (in this case, 100 m/s), the object with the greater mass will have the greater momentum.
In order to measure the mass of an object, we can use a balance scale.
Therefore the correct option is C.
What is force equal to?
Answer 1:Mass plus acceleration
Answer 2:Mass minus acceleration
Answer 3:Mass multiplied by acceleration
2
Carly mentions to her friends that her new scented oils have been helping heal her headaches. With no scientific evidence that this "aromatherapy"
actually works, her friends should caution her and tell her to be careful. What do Carly's friends think she may be diving into?
O A. astrology
OB. phrenology
OC. pseudoscience
O D. astronom
Carly's friends think she may be diving into pseudoscience .
When presented as pseudoscience, a claim, finding, or explanatory framework lacks the objectivity needed for scientific inquiry. Research that is based on dubious ideas, a defective experimental design, or unreliable data can also lead to pseudoscience.The word "pseudoscience" can be used to describe a single assertion or claim that is allegedly supported by facts or science but falls apart when put to the test of reason. A complicated system, like astrology, that claims to explain how astronomical occurrences cause and effect world events is another example of a pseudoscience. Many pseudosciences, like astrology, are reasonably safe.To know more about pseudoscience visit : https://brainly.com/question/12257058
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I need to find wfg,wnet,the acceleration of the car and how long will it take Tyler to get to the gas station
The free body diagram representing this scenario is shown below
From the information given.
mass of car = 1200kg
Recall,
weight = mg
where
m = mass
g = acceleration due to gravity = 9.8 m/s^2
Thus,
weight of car = 1200 x 9.8 = 11760 N
Normal reaction, N has same magnitude with the weight but the direction is opposite. thus,
N = 11760 N
Frictional force = N x coefficient of friction
From the information given,
coefficient of friction = 0.67
Thus,
Frictional force = 11760 x 0.67 = 7879.2 N
Net force of the car = applied force - frictional force
applied force = 1.3 x 10^4 N = 13000 N
Thus,
Net force = 13000 - 7879.2 = 5120.8N
Recall,
net force = mass x acceleration
acceleration = net force/mass = 5120.8/1200 = 4.27 m/s^2
We want to calculate the time it will take Tyler to get to the gas station. We would apply one of Newton's equation of motion which is expressed as
s = ut + 1/2at^2
where
s = distance covered
t = time
a = acceleration
u = initial velocity
From the information given,
s = 15
a = 4.27
u = 0 because the car was at rest(Since it was broken down)
By substituting these values into the equation, we have
15 = 0 x t + 1/2 x 4.27 x t^2
15 = 0 + 2.135t^2
15 = 2.135t^2
Dividing both sides of the equation by 2.135, we have
15/2.135 = 2.135t^2
t^2 = 7.026
t = square root of 7.026
t = 2.65
The time required is 2.65 s
When you drop a 0.37 kg apple, Earth exerts
a force on it that accelerates it at 9.8 m/s² to-
ward the earth's surface. According to New-
ton's third law, the apple must exert an equal
but opposite force on Earth.
If the mass of the earth 5.98 x 10²4 kg, what
is the magnitude of the earth's acceleration
toward the apple?
Answer in units of m/s².
The earth is accelerating toward the apple at a rate of 6.2 × 1025 m/s2.
How is this acceleration determined?The apple weighs m = 0.37 kg.
The apple's speed when it approaches the earth's surface is 9.8 meters per second.
Earth's mass, M, is 5.98 × 1024 kg.
Using Newton's Second Law of Motion, we may now:
The strength of the force exerted by Earth on the apple is,
F = ma
⇒ F = 0.37 × 9.8
⇒ F = 3.626 N
We are informed that the apple must exert an equal but opposite force on Earth in accordance with Newton's third law of motion.
Therefore, the force exerted by the apple on Earth will be of the following magnitude:
F = 3.626 N
Let "A" be the acceleration of the earth relative to the apple in m/s2.
Thus,
The following will be used to determine how quickly the earth is moving toward the apple:
F = MA
⇒ 3.626 = [5.98 × 10²⁴] × A
⇒ A = 3.626 / [5.98 × 10²⁴]
⇒ A = 0.606 × 10⁻²⁴
⇒ A = 6.06 × 10⁻²⁵ m/s²
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A small sphere with charge q = 5.00 µC and mass 0.500 g is traveling horizontally toward the east at a height of 60.0 cm above the ground. The sphere has a speed of 2.00 m/s as it enters a region of uniform horizontal electric field with magnitude E directed to the east.
What is E if the sphere has a speed of 5.00 m/sm/s just before it strikes the ground?
If the sphere has a speed of 5.00 m/s just before it strikes the ground the value of E is given as 468.4N
How to solve for the value of Ewe have the following values to solve the problem with.
charge q = 5μc,
speed v = 2 m/s
height h = 60 cm
speed of sphere = 5 m/s
The Kinematic equation is given as
[tex]Y = V_{oy} T+\frac{1}{2} a_{y}t^{2}[/tex]
This is the time that it would take the sphere to hit the ground.
we have to insert the values
60 x 10⁻² = 0 + 1/2 (9.8)t²
t = 0.3499 sec
The sphere hits the ground at 0.3499 seconds
Vy = Voy + ayt
= 0 + 9.8 x 0.3499
= 3.429 m/s
Vn = 2 + qE/m (0,.3499)
= 2 + ((5 x 10⁻⁶) E /0.5 x 10⁻³)0.3499
= 2 + 2.499x10⁻³E
V = √Vx² + Vy²
5² = (2 + 3.499 x 10⁻³E)² + (3.429)²
13.241959 = (2 + 3.499 x 10⁻³E)²
3.638 = 2 + 3.499 x 10⁻³E
take like terms
1.6389 = 3.499 x 10⁻³E
to get E
1.6389 / 3.499 x 10⁻³
E = 468.4N
The value of E given the speed is given as 468.4 Newton
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The principal of a local high school gave all students breakfast for one week. Teachers commented that students were more alert that week in the morning.
Students commented they were more alert and could concentrate in the morning after eating breakfast. What can be concluded based on the observations?
All principals should give free breakfast every morning.
If students eat breakfast they will be able to concentrate in school in the afternoon.
Teachers should serve breakfast to their students.
Students should eat breakfast to be more alert at school during the morning.
Answer:
D Students should eat breakfast to be more alert at school during the morning.
Explanation:
Because based on the information given they said the principal gave students breakfast and teacher's said the students were more alert and the only common sense answer is D because they are more active in class in the morning not afternoon.
caracteristicas del sonido
In general, intensity, tone, timbre, and duration are the four subjective qualities used to describe musical sounds. Each of these characteristics depends on one or more physically measurable parameters.
What is sound and what are its characteristics?In physics, sound is any phenomenon that involves the transmission of mechanical waves, whether audible or not, through a fluid or solid medium while causing a body to vibrate.
The sensation of sound is one that is brought on by the vibrations of objects and occurs in the ear. An object vibrating generates airborne waves that are transmitted. People communicate with one another through sound, doing so by expressing their thoughts, emotions, and desires.
The sounds are divided into three categories based on their frequency: agudos (high frequency), medios (medium frequency), and graves (baja frecuencia).
There is no change in phase in the reflection when the airborne sound waves (also known as pressure waves) come into contact with a hard surface. To put it another way, when the high pressure portion of a sound wave strikes a wall, it is reflected as a high pressure rather than a low pressure, which would be a phase change.
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As the swings ride hits top speed, what applies centripetal force to keep the riders traveling in a circle?
When the swing moves from the lowest point up to either peak, the main force acting is momentum; and when the swing falls from either peak to its lowest point, the main force acting is gravity. If you do not pump your legs fast enough, air resistance along with gravity will keep you down.
What provides centripetal force?The centripetal force is the resultant force because the swings move in a circle. The centripetal force—the force that keeps the rider in a circle and the horizontal component of the tention force—is produced by the total of the forces acting along the x-axis.Gravity is the centripetal force that drives astronomical orbits according to Newtonian mechanics. A common illustration of centripetal force is when a body moves uniformly fast in a circular motion.A net force that keeps an object moving in a circular motion is known as a centripetal force.To learn more about : Centripetal force.
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Jennifer hits a stationary 0.20-kg ball, and it leaves her racket at 40 m/s. Time-lapse photography shows that the ball was in contact with the racket for 40 ms.What average force did the ball exert on the racket?Express your answer using two significant figures.What is the ratio of this force to the weight of the ball?Express your answer using two significant figures.
Given:
The mass of the stationary ball is: m = 0.20 kg.
The velocity of the ball when it leaves the rocket is: Vf = 40 m/s.
The time for which the ball was in contact with the rocket is: t = 40 ms = 0.04 s.
To find:
The average force the ball exerts on the rocket.
The ratio of the force on a rocket to the weight of the ball.
Explanation
As the ball is initially stationary, its initial velocity Vi is zero. After the ball was hit, it moves with the velocity Vf = 40 m/s for time t = 0.04 s.
Thus, using Newton's second law, we get:
[tex]\begin{gathered} F=ma \\ \\ F=m\times\frac{V_f-V_i}{t} \end{gathered}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} F=0.20\text{ kg}\times\frac{40\text{ m/s}-0\text{ m/s}}{0.04\text{ s}} \\ \\ F=200\text{ kg.}^\text{m/s}^2 \\ \\ F=200\text{ N} \end{gathered}[/tex]The weight of the ball on the earth is the product of its mass and acceleration due to gravity g. The value of the acceleration due to earth is: g = 9.8 m/s^2.
The weight W of the ball on earth is given as:
[tex]W=mg[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} W=0.20\text{ kg}\times9.8\text{ m/s}^2 \\ \\ W=1.96\text{ N} \end{gathered}[/tex]The ratio of the force F exerted by the ball on the rocket and the weight of the ball is:
[tex]\begin{gathered} \frac{F}{W}=\frac{200\text{ N}}{1.96\text{ N}} \\ \\ \frac{F}{W}=\frac{102.04}{1} \end{gathered}[/tex]Thus, F : W = 102.04 : 1.
Final answer:
The force exerted by the ball on the rocket is 200 N. The ratio of this force and the weight of the ball is 102.04 : 1.
evaluate when 8.1p + 7.5r when p=6 and r=7
You have the following expression:
8.1p + 7.5r
if p=6 and r=7, you obtain for the previous expression:
8.1(6) + 7.5(7) = 48.6 + 52.5 = 101.1
Hence, the result is 101.1
how the water falls connected to gravity?
Answer:
At the top of the waterfall, the water is higher in the gravitational field of the Earth and has gravitational potential energy. When it falls, the potential energy turns into kinetic energy. So energy is not created, it was there at the beginning, stored as potential energy.
I hope this helps :)
How fast must a 10 kg object be traveling to have the SAME momentum as a 2kg object at 8 m/s?
Recall that the momentum is given by
[tex]p=m\cdot v[/tex]Where m is the mass and v is the velocity.
[tex]p=2\cdot8=16\;\;\frac{kg\cdot m}{s}[/tex]So, the momentum of the object is 16 kg m/s.
Now, we need to equate this momentum to find the velocity of the other object.
[tex]\begin{gathered} p=m\cdot v \\ 16=10\cdot v \\ v=\frac{16}{10} \\ v=1.6\;\frac{m}{s} \end{gathered}[/tex]Therefore, the object must be traveling at 1.6 m/s to have the same momentum as the other object.
What is the maximum speed of a wave on a cable with length of 50 m, mass of 200 kg, and a tension force of 1000 N apply to it?V= √ Ft/e
Given:
the length of the cable is
[tex]l=5\text{0 m}[/tex]the mass of the cable is
[tex]m=200\text{ kg}[/tex]the tension force on the cable is
[tex]T=1000\text{ N}[/tex]Required: velocity of the wave.
Explanation:
to find the velocity of the wave on a cable we use the formula that is given by
[tex]v=\sqrt[2]{\frac{T}{\mu}}[/tex]Where
[tex]T[/tex]is the tension force and
[tex]\mu[/tex]is mass per unit length.
first, we calculate the mass per unit length.
[tex]\begin{gathered} \mu=\frac{200\text{ kg}}{50\text{ m}} \\ \mu=4\text{ kg/m} \end{gathered}[/tex]now put this value in the above relation and solve for velocity, we get
[tex]\begin{gathered} v=\sqrt[2]{\frac{T}{\mu}} \\ v=\sqrt[2]{\frac{1000\text{ N}}{4\text{ kg/m}}} \\ v=15.81\text{ m/s} \end{gathered}[/tex]Thus, the speed of the wave is 15.81 m/s.
velocity differs from speed in that velocity indicates a particle's __________ of motion.
velocity differs from speed in that velocity indicates a particle's Direction of motion.
For the majority of us, speed and velocity can be a little puzzling. Speed and velocity are different in that speed just gives us an indication of how quickly an object is going, whereas velocity not only tells us an object's speed but also the direction in which it is travelling. In contrast to velocity, which is defined as a function of displacement, speed can be defined as a function of the distance travelled. A body's instantaneous velocity is its speed right now. Average velocity is defined as total displacement divided by total time and is provided by the formula v = x/t, where x is the body's total displacement and t is the time. Average speed is never greater than average velocity, and vice versa.
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For a body to be in SHM, the restoring force in the ___________ direction as the displacement from equilibrium.A. OppositeB. PerpendicularC. SameD. Not enough info
A. Opposite
ExplanationSimple harmonic motion (SHM)is a very important type of periodic oscillation where the acceleration is proportional to the displacement from equilibrium, in the direction of the equilibrium position.
the oscillation of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's law
The restoring force is the force that brings the object back to its equilibrium position (red); the restoring force acts in the direction opposite to the displacement.
therefore, the answer is
A. Opposite
I hope this helps you
A 3500 kg van hits a 2500 kg car with a force of 1480 N [E].a) What force does the van experience?b) Calculate the acceleration of both vehicles after the collision.
Given
m1 = 3500 kg (van)
m2 = 2500 kg (car)
F = 1480 N
Procedure
a) In a collision between object 1 and object 2, the force exerted on object 1 (F1) is equal in magnitude and opposite in direction to the force exerted on object 2 (F2). In equation form:
[tex]F1=-F2[/tex]The above statement is simply an application of Newton's third law of motion to the collision between objects 1 and 2.
b) Accelaration
[tex]\begin{gathered} F=ma \\ a=\frac{F}{m} \\ a=\frac{1480N}{3500+2500} \\ a=0.246\text{ m/s} \end{gathered}[/tex]The acceleration would be 0.246 m/s
While standing on a 3-foot ladder, a grapefruit is tossed straight up with an initial velocity of 45 fu/sec. The initial position of the grapefruit is 7 feet above the ground when it is released.Its height at time t is given by y = h(t) = -16t^2 + 45t + 7.A) How high does it go before returning to the ground? Round time to 2 decimal places to compute height. ___feet.b) How long does it take the grapefruit to hit the ground? Round time to 3 decimal places. ___ seconds
ANSWER:
A) 38.64 feet.
B) 2.960 seconds
STEP-BY-STEP EXPLANATION:
We have that the function that models the situation of the statement is the following:
[tex]h(t)=-16t^2+45t+7[/tex]A)
We can calculate this height, which would be the maximum height it can reach before it begins to decay, as follows:
[tex]\begin{gathered} t_v=-\frac{b}{2a} \\ \\ \text{ in this case b = 45, a = -16, we replacing:} \\ \\ t_v=-\frac{45}{-16\cdot2}=\frac{45}{32} \\ \\ t_v=1.406\text{ sec} \\ \\ \text{ Now, we replace in the function this time like this: } \\ \\ h(t)=-16\left(1.406\right)^2+45\left(1.406\right)+7 \\ \\ h(t)=38.64\text{ ft} \end{gathered}[/tex]Therefore, the height is equal to 38.64 feet.
B)
To determine the time when the ground, we must make the height 0 and solve for t, just like this:
[tex]\begin{gathered} 0=-16t^2+45t+7 \\ \\ -16t^2+45t+7=0 \\ \\ \text{ We use the general formula for quadratic equations} \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ a=-16,b=45,c=7 \\ \\ \text{ We replacing} \\ \\ t=\frac{-45\pm\sqrt{45^2-4(-16)(7)}}{2(-16)} \\ \\ t=\frac{-45\pm\sqrt{2473}}{-32} \\ \\ t_1=\frac{-45+\sqrt{2473}}{-32}=-0.148 \\ \\ t_2=\frac{-45-\sqrt{2473}}{-32}=\:2.960 \end{gathered}[/tex]Therefore, the time is equal to 2.960 seconds
A satellite is 100 miles above the earth, it has a mass of 100kg. If the mass of the satellite were doubled the force of gravity of the satellite on the earth would ?
Given:
Distance of satellite from Earth = 100 miles
Mass of satellite = 100 kg
If the mass of the satellite were doubled, let's find what would happen to the force of gravity of the satellite on earth.
Apply the gravitational force formula:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]Where:
F is the force of gravity
m is the mass.
From the formula above, we can see the force of gravity (F), is directly proportional to the mass.
Since they are directly proportional to each other, if the mass is doubled, the force of gravity will also be doubled.
Therefore, if the mass of the satellite is doubled, the force of gravity of the satellite on the earth will also be doubled.
ANSWER:
Force of gravity will be doubled when the mass of the satellite is doubled.
A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.49 radians with the horizontal. The frictional coefficients between the crate and plane are (s = 0.66, k = 0.21 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?
The acceleration of the crate as it slides up the incline is 2.06 m/s².
What is the acceleration of the crate?
The acceleration of the crate as it slides up the inclined plane is calculated as follows;
Apply the principle of Newton's second law of motion to determine the acceleration of the crate.
Ff = ma
kmg = ma
kg = a
where;
k is the coefficient of kinetic frictiong is acceleration due to gravitya is the acceleration of the cratea = (0.21 x 9.8 m/s²)
a = 2.06 m/s²
Thus, the acceleration of the crate as it slides up the incline depends on the coefficient of kinetic friction since the crate is in motion.
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To a mass m1 = 100 g of water at θ1 = 10°C we add a mass m2 = 60 g of water at θ2 = 55°C. Calculate the final temperature of the mixture? Data: Specific heat capacity of water in the liquid state: 4.18 kJ.kg -1 .K
Given:
The mass of water is m1 = 100 g = 0.1 kg
The temperature of the water is
[tex]\begin{gathered} \theta1=10^{\circ}\text{ C} \\ =10+273 \\ =283\text{ K} \end{gathered}[/tex]The mass of water is m2 = 60g = 0.06 kg
The temperature of the water is
[tex]\begin{gathered} \theta2=\text{ 55 }^{\circ}\text{C} \\ =55+273 \\ =328\text{ K} \end{gathered}[/tex]To find the final temperature of the mixture.
Explanation:
The final temperature of the mixture can be calculated by the formula
[tex]\begin{gathered} T=\frac{(m1\theta1+m2\theta2)}{(m1+m2)} \\ \text{ } \end{gathered}[/tex]On substituting the values, the final temperature will be
[tex]\begin{gathered} T=\frac{(0.1\times283)+(0.06\times328)}{(0.1+0.06)} \\ =293.625\text{ K} \\ =20.625\text{ }^{\circ}C \end{gathered}[/tex]Thus, the final temperature of the mixture is 20.625 degrees Celsius.
What is the ratio of the force exerted on the floor by the ground to the persons body weight
We will have the following:
We are given:
m = 67kg
F = 2300N
Now, we have that the average are of a human foot is approximately 76 cm^2.
So, we will deterine the ratio of force excerted on the floor in terms of pressure, that is:
[tex]P=\frac{2300N}{0.0076m^2}\Rightarrow P=302631.5789Pa[/tex][tex]\Rightarrow P\approx302.632kPa[/tex]So, the ratio of the force exerted on the floor by the ground is approximately 302.632 Kilopascals.
A 3.00-kg object is initially moving northward at 15.0 m/s. Then a force of 15.0 N, toward the east, acts on it for 2.70 s.
A) At the end of the 2.70 s, what is the magnitude of the object’s final velocity?
B) What is the direction of the final velocity? Enter the angle in degrees where positive indicates north of east and negative indicates south of east.
C) What is the change in momentum during the 2.70 s? Take +y to be north and +x to be east. Enter a positive answer if the change in momentum is toward the east and a negative answer if the change in momentum is toward the west.
The magnitude of the object’s final velocity, the direction of the final velocity and the change in momentum during the 2.70 s
V=0.1 m/s[tex]\theta[/tex]=48.0^0 + from east-4.5kgm/s toward westThis is further explained below.
What is the magnitude of the object’s final velocity?F_x=m a_x=15N
f is force
m is mass
a is acceleration
[tex]&a_x=\frac{15}{3} \\\\ m s^2=5 m_{s^{-2}} \\[/tex]
v_x=0+5 * 2.7
v_x=13.5
v_y=15
[tex]V=\sqrt{v_x^2+v_{y^2}}\\[/tex]
V=0.1 m/s
b)
[tex]\tan \theta=\frac{V_y}{V_x} \\\\ =\frac{18}{13.5} \\\\[/tex]
=1.11
[tex]\theta[/tex]=48.0^0 + from east
c)
Initial momentum,
p_{i x}=0, p_{i y}=m v_y=45 kgmg
final momentum,
P_{f x}=13.5 *3
=40.5kgm
P_{f y}=45Kgm/s
In conclusion, D P_x=P_{1 \text { In }}-P_{i x}=-4.5kgm/s toward west
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