Answer:
book resting on a shelf has no potential energy relative to the shelf since it has a height of zero meters relative to the shelf. However, if the book is elevated to some height above the shelf, then it has potential energy proportional to the height at which it resides above the shelf.
An object can have both kinetic and potential energy at the same time. For example, an object which is falling, but has not yet reached the ground has kinetic energy because it is moving downwards, and potential energy because it is able to move downwards even further than it already has. The sum of an object's potential and kinetic energies is called the object's mechanical energy.
As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.
Another important concept is work.
A skater embeds a firecracker in a large snowball and pushes it across a frozen pond at 7.6 m/s. The firecracker explodes, breaking the snowball into two equal-mass chunks. One chunk moves off at 9.3 m/s at 19° to the original direction of motion. Discover the speed and direction of the second chunk.
Answer:
[tex]v_2 = 6.406 / cos25.29[/tex]
Explanation:
From the question we are told that
Speed of snow ball 7.6m/s
Speed of chunk 9.3m/s at [tex]19 \textdegree[/tex]
Generally the equation for the conservation of momentum is mathematically given as
Let the snow ball be b
and the chunk b/2
According to conservation of momentum we have
[tex]b_u = (b /2) v_1 cos\theta_1 + (b / 2) v_2cos\theta_2[/tex]
[tex]v_2 cos\theta_2 = 2 * 7.6 -9.3 * cos_19[/tex]
[tex]v_2 cos\theta_2 = 15.2 -8.793 = 6.406[/tex]
Therefore
[tex](b/2)v_1 sin\theta_1 = (b/2) v_2sin\theta_2[/tex]
[tex]v_2 sin\theta_2 = 9.3 * sin_19[/tex]
[tex]v_2 sin\theta_2 = 3.027[/tex]
Mathematically From above equations
[tex]tan\theta_2 = 0.4726[/tex]
[tex]\theta_2 = 25.29 \textdegree[/tex]
Therefore the speed and direction of second chunk
[tex]v_2 = 6.406 / cos25.29[/tex]
[tex]v2 = 7.0855m/s[/tex]
Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7 r, respectively What is the ratio of their orbital speeds
Answer:
The ratio of their orbital speeds are 5:4.
Explanation:
Given that,
Mass of A = 5 m
Mass of B = 7 m
Radius of A = 4 r
Radius of B = 7 r
The orbital speed of satellite A,
[tex]v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}[/tex]......(I)
The orbital speed of satellite B,
[tex]v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}[/tex]......(I)
We need to calculate the ratio of their orbital speeds
Using equation (I) and (II)
[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}[/tex]
Put the value into the formula
[tex]\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}[/tex]
[tex]\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}[/tex]
Hence, The ratio of their orbital speeds are 5:4.
I NEED HELP WITH THE LAST QUESTION I’L MARK U AS BRANLIEST
The question is: how would you connect the cells?
Answer:
gap junctions,tight junctions,and desmosomes
two pictures show friends playing with a string telephone. in which picture can they hear each other
Answer:
In picture one because string telephones are best heard when there is more tension on the string
Explanation:
What is the force of gravitational attraction between an object with a mass of 100 kg and another object that has a mass of 300 kg and are at a distance of 2m apart
Answer:
5 x 10⁻⁷N
Explanation:
Given parameters:
Mass of object 1 = 100kg
Mass of object 2 = 300kg
Distance = 2m
Unknown:
Force of gravitational attraction between the objects = ?
Solution:
Newton's law of universal gravitation states that the gravitational force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
From Newton's law of universal gravitation we derive an expression:
Fg = [tex]\frac{Gm_{1} m_{2} }{r^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
m is the mass
r is the distance between the bodies
Now insert the parameters and solve;
Fg = 6.67 x 10⁻¹¹ x [tex]\frac{100 x 300}{2^{2} }[/tex] = 5 x 10⁻⁷N
How far does a zebra travel if it gallops at an average speed of
25km/hr for 2 hours?
Answer:
50km i think but i am not very sure
A mechanism which uses mechanical energy to produce electrical energy is known as an) -
O electric motor
O transformer
O electromagnet
O electrical generator
CAN I HAVE SOME HELP PLEASE
give listing the law used the intensity of current carried by each lamp?
Answer:
By teh way is isn't it question of law or science and the picture is of what electric light or not I have read it so I was asking isn't is question of science
camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive
The question is incomplete. Here is the complete question.
The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined for this image, so the cars you see are in fact the same car, but photographed at differene times.
Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?
Answer: v = 6.5 m/s
Explanation: The question asks for velocity of the car. Velocity is given by:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:
Δx = 7(5.3)
Δx = 37.1 m
The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:
1.3 frames = 1 s
7 frames = Δt
Δt = 5.4 s
Then, the car was driving:
[tex]v=\frac{37.1}{5.4}[/tex]
v = 6.87 m/s
The car drove at, approximately, a velocity of 6.87 m/s
The velocity of the car will be 6.5 m/s.The rate of change of displacement is defined as speed.
What is velocity?The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
t is the time for camera deliver= 1.3 frames per second
l is the length = 5.3 meters
The instantaneous velocity is given as;
[tex]\rm v = \frac{\triangle x }{\triangle t } \\\\ \rm \triangle x = 7 \times 5.3 \\\\ \rm \triangle x = 37.1 m[/tex]
The time engaged is find as;
1.3 frames = 1 s
[tex]\rm \triangle t= 7 \ frames \\\\ \rm \triangle t=5.4 sec[/tex]
Hence the velocity of the car driving;
[tex]\rm v= \frac{37.1}{5.4} \\\\ \rm v= 6.87 m/sec[/tex]
Hence the velocity of the car will be 6.5 m/s.
To learn more abouty the velocity refe to the link;
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I NEED HELP WITH NUMBER 7 I WILL GIVE U BRAINLIEST
Answer:
the blue one
Explanation:
The rock has potential energy due to the gravitational potential with the Earth. As the rock is falling, this gravitational energy becomes less. What happens to that energy? Question 11 options: The energy disappears. The energy is transformed into kinetic energy. The energy is absorbed by the ground and the air The energy is transformed into chemical potential energy.
Answer:
Explanation:
while it is still falling it is turned into kinetic energy. While on the floor it turns onto potential energy. Remember energy cannot be created or the destroyed so the other ones are automatically out.
while snowboarding mr. brooks (100 kg) takes a chair lift up to 800 m to the top of the mountain. if he snowboards down the mountain without stopping of falling, how fast will he be going when he get back to the bottom of the mountain
Given that,
Mass of Mr. Brooks = 100 kg
A chair lift up to 800 m to the top of the mountain.
To find,
How fast will he be going when he get back to the bottom of the mountain.
Solution,
Let v is the speed when he get back to the bottom of the mountain. Using the conservation of energy to find it.
[tex]\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 800} \\\\v=125.21\ m/s[/tex]
So, the required speed is 125.21 m/s.
A merry-go-round initially at rest at an amusement park begins to rotate at time t=0. The angle through which it rotates is described by θ(t)=πk(t+ke−t/k), where k is a positive constant, t is in seconds, and θ is in radians. The angular velocity of the merry-go-round at t=T is
Answer:
Angular velocity of merry-go-round is πk - 1 at t= T
Explanation:
From the question it is given that
[tex]\theta(t) = \pi k(t+k_e-\frac{t}{k} )[/tex] ..........................(1)
since mathematically, angular velocity is defined as
[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] ........................(2)
on substituing the value of θ(t) from equation 1 in equation (2) we get
[tex]\omega(t) = \frac{d\theta(t)}{dt}[/tex] = [tex]\frac{d\pi k (t + k_e - \frac{t}{k} )}{dt}[/tex] ............................(3)
on differentiating equation (3) with respect to time we get
ω(t) = πk(1 -[tex]\frac{1}{k}[/tex]) = πk - 1 angular velocity of merry-go-round
Therefore, angular velocity of merry-go-round is πk - 1 at t= T
The angular velocity of the merry-go-round at t=T is πk - 1. The pace at which angular displacement changes is defined as angular velocity.
What is angular velocity?The rate of change of angular displacement is defined as angular velocity.
It is stated as follows:
ω = θ t
The equation for the angular displacement is given as;
[tex]\theta (t) = \pi k (t+k_e-\frac{t}{k} )[/tex]
The angular velocity is found by the differentiation of the angular displacement.
[tex]\rm \omega(t) = \frac{d(\theta)}{dt} \\\\ \rm \omega(t) = \frac{d (\pi k (t+k_e-\frac{t}{k} )}{dt} \\\\ \rm \omega(t) = \pi k-1[/tex]
Hence the angular velocity of the merry-go-round at t=T is πk - 1.
To learn more about the angular speed refer to the link;
https://brainly.com/question/9684874
"A 60 g tennis ball is dropped from a 2 m height, strikes a horizontal sidewalk and rebounds to a height of 1 m. Find the average force with which the ball hits the ground if the collision lasts 0.02s."
Answer:
F = Changing momentum / time
Explanation:
change in momentum = final momentum - initial momentum
but before finding the force we have to find the initial and final vertical velocity of the ball in both cases (1st bounce and 2nd bounce) so first let's find the vertical velocities of the ball in the first bouncehere the ball dropped - it means (initial velocity is zero 0ms-1)but there is a final vertical velocity...let's find the final vertical velocity v² = u² + 2as (here a = g = 10ms^-2)v = √0²+2(10)2v = 6.32ms^-1After that let's find the vertical velocities of the ball in the second bounceso here our final velocity is zero (because vertical velocity in maximum heigh )is zeothen let's find our initial vertical velocityv² = u² + 2as0 = u² + 2(-10)1 (here -10 cause gravitational acceleration act positive only in the downward direction )u = √20u = 4.47ms^-1ok, now we found all the velocities then let's find the forceF = Changing momentum / timechanging momentum = impulse= mv-m(-u)= 0.06 (6.32+4.47)=0.6474 Nsso we found the chang in momentumthe let's find the forceF = Changing momentum / timeF = 0.6474 / 0.02F = 32.37NAn AC voltage source, with a peak output of 200 V, is connected to a 50-Ohm resistor. What is the effective (or rms) current in the circuit
Answer:
2.8 A
Explanation:
We know that;
Vrms = Vo/√2
Where;
Io = peak value of voltage = 200V
Hence;
Irms = 0.707 × 200
Irms = 141.4 V
Irms = Vrms/R = 141.4/50 = 2.8 A
A 100 kg cart is moving at 3 m/s. Calculate the cart’s kinetic energy.
Answer:
450
Explanation:
Given,
Mass= 100kg
Velocity= 3 m/s
Kinetic Energy= ?
Kinetic Energy= 1/2 mv^2
= 1/2× 100× 3^2
= 1/2× 900
= 450.
HOPE IT HELPED :)
Will get Brainlest 5 star and heart looking for someone who knows 7 grd science flvs work dm and friend
Which of these best describes Earth's crust?
A inner layer consisting of two parts
B middle layer, density increases with depth
C Top portion called asthenosphere, thickest layer
D Thinnest under the oceans and thickest under continents
A 620 kg moose is standing in the middle of a train track. A 10,000 kg train moving at 10 m/s is unable to stop and the moose ends up riding the cowcatcher down the track. What type of interaction is this, and what is the new combined velocity
Answer:
This is an example of Inelastic colission
Explanation:
Step one:
given:
mass of moose m1 = 620 kg
mass of train m2= 10,000kg
Initial velocity of moose u1= 0 m/s
Initial velocity of train v1 = 10m/s
combined velocity of the system is given as v
Applying the conservation of momentum equation we have
m1u1+ m2u1= (m1+m2)V
substitutting we have
620*0+10000*10= (620+10000)V
100000= 10620V
divide both sides by 10620
V = 100000/10620
V=9.41m/s
The velocity of the moose after impact is 9.41m/s
What is the resistance of five 10Ohm resistors in parellel?
Answer:
The equivalent resistance is 2 ohms.
Explanation:
let the first resistance = R₁ = 10 ohm
let the second resistance = R₂ = 10 ohm
let the third resistance = R₃ = 10 ohm
let the fourth resistance = R₄ = 10 ohm
let the fifth resistance = R₃ = 10 ohm
The equivalent resistance is calculated as;
[tex]\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \\\\\frac{1}{R_T} = \frac{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)}{R_1R_2R_3R_4R_5} \\\\R_T = \frac{R_1R_2R_3R_4R_5}{(R_2R_3R_4R_5)+(R_1R_3R_4R_5)+(R_1R_2R_4R_5) +(R_1R_2R_3R_5)+(R_1R_2R_3R_4)} \\\\R_T = \frac{(10^5)}{(10^4)+(10^4)+(10^4)+(10^4)+(10^4)} \\\\R_T = \frac{10^5}{5(10^4)} \\\\R_T = \frac{10}{5} \\\\R_T = 2 \ ohms[/tex]
Therefore, the equivalent resistance is 2 ohms.
What is the net force acting on the box?
A. 285 N
B. 185 N
C. 85 N
D. 65 N
Answer:
65N
Explanation:
Answer:
65 N
Explanation:
i just did the assignment on edge 2021
A 19 kg child is riding a 5.6 kg bike with a
velocity of 5.0 m/s to the northwest.
a) What is the total momentum of the child
and the bike together?
Answer in units of kg. m/s.
What is the momentum of the child?
Answer in units of kg. m/s.
c) What is the momentum of the bike?
Answer in units of kg. m/s
Answer:
[tex]\mathrm{a.\:}123\:\mathrm{kg\cdot m/s},\\\mathrm{b.\:}95\:\mathrm{kg\cdot m/s},\\\mathrm{c.\:}28\:\mathrm{kg\cdot m/s}[/tex]
Explanation:
The momentum of an object is given by:
[tex]p=mv[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
a)
The mass of the child and bike together is [tex]19+5.6=24.6[/tex] kilograms. Since they're moving at a velocity of 5.0 m/s, their momentum is:
[tex]p=24.6\cdot 5=\fbox{$123\:\mathrm{kg\cdot m/s}$}[/tex].
b)
The mass of the child is given as 19 kg. Since the child is on the bike moving at 5.0 m/s, it's implied the child is as well. Therefore, the momentum of the child is:
[tex]p=19\cdot 5=\fbox{$95\:\mathrm{kg\cdot m/s}$}[/tex].
c) The mass of the bike is given as 5.6 kg and it is moving at 5.0 m/s. Therefore, the momentum of the bike is:
[tex]p=5.6\cdot 5=\fbox{$28\:\mathrm{kg\cdot m/s}$}[/tex]
An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free-fall acceleration on the moon?
Answer:
the free-fall acceleration on the moon is 1.68 m/s^2
Explanation:
recall the formula for the gravitational potential energy (under acceleration of gravity "g"):
PE = m * g * h
replacing with our values for the problem:
46 J = 91 * g * 0.3
solve for the "g" on the Moon:
g = 46 / (91 * 0.3)
g = 1.68 m/s^2
a)
F E
60. The temperature of source and sink of cannot engine are 400K
and 300K respectively. What is its efficiency?
h) 75%
c) 33.3% d) 25%
Answer: 25%
Explanation:
From the question, we are informed that the temperature of source and sink of cannot engine are 400K
and 300K respectively.
The efficiency will be calculated as:
= 1 - t/T
= 1 - 300/400
= 1 - 3/4
= 1/4
= 25%
Which occupies more volume: 3m^3, or 100 ft^3?
Answer:
3 m³ will occupy more volume.
Explanation:
We need to tell which occupies more volume : 3m³ or 100 ft³.
To compare, we first make the units same.
We know that,
1 m = 3.28 feet
1 m³ = 35.31 foot³
3 m³ = 105.94 foot³
Now, we can compare 105.94 foot³ and 100 ft³. It can be clear that 3 m³ will occupies more volume.
A worker for a moving company is pushing on a refrigerate with a force of 1500 Newtons but the refrigerator just won’t move. How much work did the worker do on the refrigerator?
Answer:
0J
Explanation:
work done= force x distance in direction of force.
since distance=0, work done=0
A certain car can accelerate from 0 to 60 mph in 7.9 s. What is the car's average acceleration in mph/s?
Answer:
a = 7.6\ mph/s
Explanation:
Motion With Constant Acceleration
It's a type of motion in which the velocity of an object changes uniformly in time.
The equation that describes the change of velocities is:
[tex]v_f=v_o+at[/tex]
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
Solving the equation [for a:
[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]
The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:
[tex]\displaystyle a=\frac{60 \ mph-0}{7.9}[/tex]
a = 7.6\ mph/s
A pulley system consists of two fixed pulleys and two movable pulleys that lift a load that has a weight of 300 N. If the effort force used to lift the load is 100 N, what is the mechanical advantage of the system
Answer:
Mechanical advantage = 3
Explanation:
Given:
Weight = 300 N
Force load = 100 N
Find:
Mechanical advantage:
Computation:
Mechanical advantage = Weight/Force Load
Mechanical advantage = 300/100
Mechanical advantage = 3
Mechanical advantage is the ratio of output force to the input force.It is also used to find the efficiency of the system.The value of mechanical advantage is 3.
What is mechanical advantage?Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of forces in levers and pulley.
It is an effective way of amplify the force in simple machines like lever The theoretical mechanical advantage is defined as ratio of the force responsible for he useful work in system to the applied force.
The theoretical value may be grater than the actual value of mechanical advantage because in the theoretical mechanical advantage friction is assumed to be absent.
The given data in the problem is;
weight that lifted (output)=300
Effort force(input) =100
Mathematically it is given as;
[tex]\rm M A= \frac{F_O}{F_I} \\\\\rm M A= \frac{300}{100}\\\\\rm M A=3[/tex]
Hence the value of mechanical advantage is 3.
Learn more about the mechanical advantage refer to the link;
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A girl and a boy pull in opposite directions on strings attached to each end of a spring balance. Each child exerts a force of 20N. What will the reading on the spring balance be? 20 N 10 N 40 N 0 N
Answer:
yo they deleted my answer. The answer is 0N
Explanation:
so when two forces pull on an object from opposite sides with the same force (in this case its 20N), then the object is in equilibrium at 0N.
So its clear that there is one person on the the opposite side.
SOOO generally: (left or down) would be considered negative in an equation. And the other person (right or up) would be considered positive. So if both forces are the same numbers on opposite sides then the answer is 0 (if you add both of them).
0 is the number of equilibrium.
OK, so the equation would be -20N + 20N and then badda bing badda boom viola, the answer: 0N
thanks for coming to my TED talk. I hope they don't delete this answer.
I really need help. I want a detailed solution
Answer:
v_{f} = -0.693 m / s
Explanation:
The acceleration of the runner can be obtained from Newton's second law
F = m a
a = F / m
the bold are vectors, therefore the acceleration throughout the journey varies as the force has variations.
For the part of finding the velocity of the body we can use the relationship between the momentum and the variation of the momentum
I = Δp
∫ F Δt = m [tex]v_{f}[/tex] - m v₀
int F dt = m (v_{f}-v₀)
1) To find the change in velocity we must find the area under the curve of the graph, this can be done analytically if we know the functional of the curve or approximate it by intervals
a) between 0 <t <0.20 s
v) between 0.20 <t <0.30 s
a reasonable curve shape can be a Gaussian.
2) If we do not have the form of the cure, we can perform a graphical integration to find the area under the curve, we can do this by dividing the curve into small rectangles, finding the area of each one and adding them.
3) Another even more approximate way is to create an average force in each interval and find the area of this force, the average force is the average value of the force in the interval, let's use this method in the exercise
a) first interval 0 <t <0.20 Average force [tex]F_{mean}[/tex] = 300 N
area = F_{mean} Δt
area = I = 300 0.20
I = 60 N s
the speed change is
I = m Δv
Δv = I / m
Δv = 60/65
Δv = 0.923 m / s
If we assume that the runner starts from rest, his final velocity is v = 0.923 m / s in the direction of the force.
b) second interval 0.2 <t <0.30s average force F_mean = 150 N
area = I = 150 (0.30 - 0.20)
I = 15 N s
the speed change is
Δv = 15/65
Δv = 0.23 m / s
Note that in this case the initial speed is not zero and since the two impulses are in the opposite direction the speed decreases
[tex]v_{f}[/tex]= -0.923 + 0.23
v_{f} = -0.693 m / s
You throw a ball into the air. Which two forces cause the ball to gradually stop moving upward and then fall back to Earth?
A.
Balanced forces
B.
Friction
C.
Normal force
D.
Gravitational force
will mark brainliest
Answer: Gravitational Forces
Explanation: