Answer:
hope you will understand
Explanation:
If an object has a mass of 100 kg and is accelerating at a rate of 3.3 m/s², how much force is the object receiving?
Answer:
133 kg
Explanation:
the statement was the last time I get home and get some sleep and then we can
Answer:
need my points back
Explanation:
help me with b and d please
A 65 kg diver stands still on a tower, 100 m above the water
(a) Calculate the gravitational potential energy of the diver, relative to the height of the water. Assume the gravitational field constant is 9.8 N/kg
(b) The diver reaches the surface of the water at a speed of 14 m/s. Calculate the diver's kinetic energy
(c) Compare your answers to (a) and (b). Explain your comparison using concepts learned in this course.
Assume there is no air resistance during the dive (d) Calculate the speed of the diver 5.0 m above the water.
Following are the solution to the given points:
Given:
[tex]\to mass(m)= 65\ kg\\\\\to h=10.0 \ m\\\\\to g=9.8 \ \frac{m}{s^2}[/tex]
To find:
gravitational potential energy(GPE)=?
kinetic energy(KE)=?
concept=?
v=?
Solution:
For point a:
[tex]\to GPE= mgh[/tex]
[tex]= 65 \times 10 \times 9.8\\\\ = 6370 \ J\\\\[/tex]
For point b:
[tex]\to KE=\frac{1}{2} m v^2\\\\[/tex]
[tex]=\frac{1}{2} \times 65 \times 14^2 \\\\=\frac{1}{2} \times 65 \times 14 \times 14 \\\\= 65 \times 14 \times 7 \\\\=6370\ J\\[/tex]
For point c:
As we've seen, there really is no force acting, and the mechanic KE: GPE is used. Because legitimate is conserved. Home energy loss means GPE must be equal to KE gain.
For point d:
A diver has risen 5 feet above the surface. As a result, according to the energy conservation law:
[tex]\to U_i + K_i= U_f + K_f\\\\ \to mg \times h + 0 = mg \times \frac{h}{2} + \frac{1}{2} m v^2\\\\ \to 9.8 \times 10 = 9.8 \times 5 + \frac{v^2}{2}\\\\\to 9.8 \times 10 - 9.8 \times 5 = \frac{v^2}{2}\\\\\to \frac{v^2}{2} =9.8( 10 - 5) \\\\\to v^2 =9.8( 10 - 5) \times 2 \\\\\to v^2 =9.8\times 5 \times 2 \\\\\to v^2 =98 \\\\\to v =98\ \frac{m}{s^2} \\\\[/tex]
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the game in the suburbs and rural areas of the Philippines is/are called_____??
Answer:
Games such as Patintero, Tumbang Preso, Piko, Sipa, Turumpo, and many others, are still played daily in neighborhoods.
Explanation:
Hope this helps... Maybe
A rock is held steady over a cliff and dropped. 1 seconds later, another rock is thrown straight down at a speed of 11.3 m/s, and hits the first rock. How far have the rocks dropped before they collide? How long is the first rock in the air before it gets hit by the second rock?
What is the displacement of the rocks when they collide? __________ (Hint: be careful of sign -- the rocks drop.)
The first rock is in the air for __________ before it is hit.
Answer:
4.27s
Explanation:
If "t" represents the time traveled from the time rock 2 is dropped until the collision, then the time traveled for rock 1 = t + 1. And, since rock #1 is dropped making its initial velocity = 0, then:
The distance rock 1 travels is
x = (0)(t + 1) + 1/2(-9.8)(t + 1)2 = -4.9(t2 + 2t + 1) = -4.9t2 - 9.8t - 4.9
The distance rock 2 travels
x = -11.3t + 1/2(-9.8)t2 = -11.3t - 4.9t2
For the distances must be equal when the rocks collide:
-4.9t2 - 9.8t - 4.9 = -11.3t - 4.9t2
-9.8t - 4.9 = -11.3t
-4.9 = -1.5t
t = 3.267 s
Now, the distance they traveled can be found by plugging the 3.267 s back into either equation:
x = -11.3(3.267) - 4.9(3.267)2 = -89.2 m or 89.2 m below where they began
The time the first rock was in the air is t + 1 = 3.267 + 1 = 4.267 s = 4.27 s
Let
First rocks time be x
Second rocks time be x+1
initial velocity=u=11.3m/s
Distance of both rocks be s1 and s2
[tex]\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2[/tex]
Now
[tex]\\ \sf\longmapsto s1=11.3x+5x^2[/tex]
[tex]\\ \sf\longmapsto s2=11.3(x+1)+5(x+1)^2[/tex]
As both collide then
[tex]\\ \sf\longmapsto s1=s2[/tex]
[tex]\\ \sf\longmapsto 11.3x+5x^2=11.3x+11.3+5(x+1)^2[/tex]
[tex]\\ \sf\longmapsto 5x^2=11.3+5x^2+10x+1[/tex]
[tex]\\ \sf\longmapsto 10x+12.3=0[/tex]
[tex]\\ \sf\longmapsto x=1.23s[/tex]
Displacement
[tex]\\ \sf\longmapsto 11.3(1.23)=13.8m[/tex]
Done
Which statement describes Kepler’s third law of orbital motion?
Answer:
The square of orbital period is proportional to the cube of the semi-major axis.
Explanation:
I just took the quick check
A 5.7 kg block slides along a horizontal surface toward a horizontally mounted spring with a spring constant of 150 N/m. At one instant, the block is 5.0 m from the end of the spring and has a speed of 12.0 m/s. If the surface is frictionless, what is the maximum compression of the spring
Hi there!
We can use the conservation of energy:
Ei = Ef
The initial energy involves kinetic energy while the final energy involves maximum elastic potential energy, so:
1/2mv² = 1/2kx²
We can cancel out the 1/2s for ease of solving:
mv² = kx²
Plug in given values to solve for x:
√(mv²/k) = x
√(5.7)(12²)/(150)) = x
x = 2.339 m
Which of these describes Kepler’s third law of orbital motion?(1 point)
T3 ∝a2
T∝a2
T2 ∝ a3
T2∝a
Answer:
1) which planet's orbit was Kepler first studying when he began to developed his laws of orbital motion? Answer: Mars
2) which is the best description of Earth's orbit? Answer: almost circular
3) which orbital shape did most scientists in the early 1600s think the planets of our solar system have? Answer: circular
4) which of these describes Kepler's third law of orbital motion? Answer T^2 OX a^3
5) which statement describes Kepler's third law of orbital motion? Answer The square of orbital period is proportional to the cube of the semi-major axis.
Explanation:
I just took the connexus quick check and got a 100%
T2 ∝ a3 is the relation that describes Kepler’s third law of orbital motion, therefore the correct answer is option C.
What are Kepler's laws of planetary motion?There are three laws of Kepler as follows,
The orbit of the planet is elliptical and the sun is present at one of the two foci of the elliptical route followed by the rotating planet.
A line segment linking a planet and the Sun covers equal regions throughout equal intervals of the time period.
The semi-major axis length of an elliptical planet's orbit is equal to the square of the planet's orbital period.
T2 ∝ a3 is the relation that describes Kepler’s third law of orbital motion,
Thus, the correct answer is option C.
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3
Which finding would have disproved Virchow's hypothesis?
O Animalcules are not organisms.
O Animals are made of non-cell-based matter.
Plants are made of non-cell-based matter.
O Cells can generate from nonliving matter.
Answer:
the correct answer is cells can generate from nonliving matter
what is the force the when does when Gravity pushes you
Answer:
Image result for what is the force the when does when Gravity pushes you
The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.
Explanation:
Image result for what is the force the when does when Gravity pushes you
The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.
Answer:
It is called gravitational force.
Explanation:
[tex]{ \boxed{ \rm{F = \frac{GM_{a} m _{a}}{ {r}^{2} } }}}[/tex]
F is the gravitational forceMa is the mass of a planetary bodyma is mass lf body being attractedr is the separation distanceG is the universal gravitational constant→ Assume you are in space and you have a mass of ma, and you are r metres from the Earth of mass Ma. The force F will be experienced and you will tend to be pulled towards the Earth when you are in its gravitational field.
What are the similarities between coral reefs and costal ecosystems
Answer:
A coral reef is an underwater ecosystem characterized by reef-building corals. Reefs are formed of colonies of coral polyps held together by calcium carbonate. ... Unlike sea anemones, corals secrete hard carbonate exoskeletons that support and protect the coral.
Coastal oceans are commonly defined as the areas from the shoreline to the outer edge of the continental margin. They connect the continents to the open ocean and serve as a link for transporting organic and inorganic, natural and anthropogenic material from land to sea.
Coastal oceans in tropical areas have similar characteristics to their temperate counterparts. ... Coral reefs are biologically diverse and rich habitats that occur in relatively nutrient-poor environments along coastlines. Areas that have too many nutrients often harbor too much algal growth to sustain coral growth.,
Explanation:
Two skaters face one another and stand motionless, then push against each other on the ice. If one's mass is 55 kg and he moves to the left after the push at a velocity of 4.3 m/s, how fast was the other skater traveling after the push if her mass was 48 kg.
4.9 m/s to the right
Explanation:
We are going to use the conservation law of linear momentum to find her velocity. Initially, they are motionless so their initial total momentum is zero. Let's assume that anything moving to the left has a negative velocity and anything moving to the right has a positive velocity. We can write the conservation law of momentum as
[tex]0 = M_mv_m + M_wv_w[/tex]
where
[tex]M_m[/tex] = mass of the man = 55 kg
[tex]v_m[/tex] = velocity of the man = -4.3 m/s
[tex]M_m[/tex] = mass of the woman = 48 kg
[tex]v_w[/tex] = velocity of the woman
[tex]\Rightarrow 0 = (55\:\text{kg})(-4.3\:\text{m/s}) + (48\:\text{kg})v_w[/tex]
Solving for [tex]v_w,[/tex]
[tex]v_w = \left(\dfrac{55\:\text{kg}}{48\:\text{kg}}\right)(4.3\:\text{m/s}) = 4.9\:\text{m/s}[/tex]
going to the right.
An advertisement shows a 1239 kg car slowly pulling at constant speed a large passenger airplane to demonstrate the power of
its newly designed 85.7 hp (horsepower) engine. During the pull, the car passes two landmarks spaced 26.0 m apart in 14.3 s.
The passenger plane being towed is a Boeing 707 with a total weight of 62.5 tons.
Calculate the force that opposes the motion. Assume that the efficiency of the car is such that 15.4% of the engine power is
available to propel the car forward.
How much work is performed by the car on the airplane during this time? How much work is performed by the airplane on the car during this time?
Answer:
Power = Work / Time
available power = .154 * 85.7 = 13.2 hp
1 hp = 746 watts (joule/sec)
13.2 hp = 13.2 * 746 = 9850 J/sec
Work = P * t = 9850 J/sec * 14.3 s = 141,000 joules
This is the work performed on the airplane and the car
Work = F * s (force * distance)
F = 141000 J / 26 m = 5420 Newtons opposes the motion
A ball rolls with a constant acceleration of 2.5 m/s^2 for 5.0 s with an initial velocity of 5.0 m/s. Calculate the
ball's distance from the starting point at 1.0 s intervals. Make a position-time graph for the object's motion. In
your response, show what you are given, the equation that you used, any algebra/work required, a table of
data, and your graph.
The kinematics allows finding the answers for the positions of the body as a function of time and making its graph are:
The positions according to time are in the table In the attachment we can see a graphic representation
Given parameters
The acceleration a = 2.5 m / s² Time t = 5.0 s The initial velocity vo = 5.0 m / sTo find
The distance each Dt = 1.0 s
Kinematics studies the motion of bodies, gives relationships between the position, velocity and acceleration of bodies, let's use the relationship
x = v₀ t + ½ a t²
Where x is the position, v or the initial velocity, at the acceleration and t is the time.
Let's substitute
x = 5 t + ½ 2.5 t²
x = 5 t + 1.25 t²
Let's calculate the position and write it in the table
t = 0 s x = 0 m
t = 1 s x = 5 1 + 1.25 1² = 6.25 m
t = 2 s x = 5 2 + 1.25 2² = 15 m
t = 3s x = 5 3 + 1.25 3² = 26.25 m
t = 4 s x = 5 4 + 1.25 4² = 40 m
t = 5 s x = 5 5 + 1.25 5² = 56.25 m
In the following table the time and positions of the body are given, these given are used for the attached graph
time (s) position (m)
0 0
1 6.25
2 15
3 26.25
4 40
5 56.25
In conclusion, using kinematics we can find the answers for the positions of the body as a function of time and make its graph are:
The positions according to time are in the tableIn the attachment we can see a graphic representation
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Elvis and Noah want to move the piano in their living room. Elvis pushes from behind with a force of
400 N, and Noah pulls from the front using a rope with a force of 150 N. The piano has a mass of 225 kg,
and the friction force of the floor and the piano is 100 N.
Calculate the acceleration of the piano.
Round your answer to the nearest tenth.
Answer:
2.0 m/s
Explanation:
Remember that F = maF=maF, equals, m, a, where FFF is the net force, mmm is the mass of the object, and aaa is its acceleration. To solve this problem, we will need to first find the net force acting on the piano and then use F = maF=maF, equals, m, a to find the piano’s acceleration.
Hint #22 / 4
The net force is the sum of all of the forces on the piano. In this circumstance, Elvis's and Noah's forces are in the same direction, and the friction force is in the opposite direction. So, our net force equation would be:
=400 N+150 N−100 N
=450 N
Now, we can use F = maF=maF, equals, m, a and solve for the acceleration:
225 kg
450 N
=2.0
The acceleration of the piano is 2.0 m/s
The acceleration of the piano is 2m/s².
We know the formula for force is
F =ma
where f is force, m is mass and a is acceleration.
We need to calculate the total force used in moving the piano
F= Total force- Frictional force
F= 400N+150N-100N= 450N.
We have F=450N and m= 225kg
F=ma
450N= 225ˣ a
a= 450/225
a= 2m/s²
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The driver of a 1,500 kg car is traveling at 25 m/s. A deer runs into the road and the driver slams on the brakes. If it takes the car 5 seconds to stop, the acceleration of the car is ____ m/s2.
a.
0.25 m/s2
b.
-0.25 m/s2
c.
5 m/s2
d.
-5 m/s2
Answer:
c.
5 m/s2
Explanation:
Δv = at
a = Δv/t
a = (0 - 25)/5
a = 5 m/s²
Over the 10 second period, how far does Juan run?
Answer:
C. 10
Explanation:
On the graph you can see that at the 10 second mark it is at the 10 meters mark as well.
draw the diagram of third class lever showing fulcrum, effort and load
Answer:
In a third class lever, the effort is located between the load and the fulcrum. If the fulcrum is closer to the load, then less effort is needed to move the load. If the fulcrum is closer to the effort, then the load will move a greater distance. ... These levers are useful for making precise movements.
Determine the MASS of the object in kg. (use g=10 m/s/s
Explanation:
Sorry but the question is incomplete
A ball is thrown at a speed of 40m/s. If the ball has a mass of 0.33kg, what is the ball’s momentum?
Answer:63745.6573
Explanation:
7. A 77 kg girl and a 62 kg boy face each other on friction free roller skates. The girl pushes the boy, who moves away at a speed of 2.5 m/s. What is the girl's speed?
Answer:
[tex]\sf\longmapsto \: 2.0m/s[/tex]
Explanation:
Initial momentum = 0
[tex]\sf\longmapsto \: 77 \: velocity + 62 \times 2.5 = 0[/tex]
[tex]\sf\longmapsto \: 77 \: velocity = - 155[/tex]
[tex]\sf\longmapsto \: velocity = \frac{ - 155}{77} [/tex]
[tex]\sf\longmapsto \: \: velocity = - 2.012[/tex]
[tex]\sf\longmapsto \: 2.0 \: m/ s[/tex]
Find out which of the following sentences is a false statement. Group of answer choices The potential energy of a pair of positively charged bodies is positive. The potential energy of a pair of oppositely charged bodies is negative. The total work required to assemble a collection of discrete charges is the electrostatic potential energy of the system. The potential energy of a pair of negatively charged bodies is negative. The potential energy of a pair of oppositely charged bodies is positive.
The statements which are false from the choices above are:
The potential energy of a pair of negatively charged bodies is negativeThe potential energy of a pair of oppositely charged bodies is positiveIn scenarios of like charges, the potential energy is always positive, that is because we need to put energy in the system to bring like charges closer together.
Conversely, for unlike charges, the potential energy of the pair of unlike charges is negative, as a result of the energy expended by the system in pushing the charges away from each other.
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A motor with an operating resistance of 30.0 ohms is connected to a voltage source 4.0 amps of current flow in the circuit. What is the voltage of the source
Answer:
120 V.
Explanation:
voltage V=IxR
= 4x30
= 120 V.
when the frequency of a wave decreases the what increases
Answer:
The wavelength
Explanation:
As the frequency decreases, the wavelength gets longer.
A spring scale is used to measure the weight of various objects. When a 100 g empty glass bottle is placed on the scale, the spring stretches down 1.8 cm. How much water should be poured into the glass bottle to stretch the spring down by 2.7 cm?
A: 44 mL
B: 50 mL
C: 33 mL
D: 150 mL
50 mL of water have to poured to the glass bottle of 100 g to increase the stretching of the spring from 1.8 cm to 2.7 cm.
What is spring stretching?A spring can hang when it loads a weight. The stretching of spring is the change in length of the spring end from its initial length. Spring balance works with the principle of Hooks law.
The stretching is made by the elasticity of the spring which make it in a wave like motion.
Given that the glass bottle with 100 g weight when loaded on the spring scale, it will stretch down to 1.8 cm. Then the weight required to make a stretch of 2.7 cm is calculated as follows:
weight required = (100 g × 1.8 cm) /2.7 cm
= 150 g.
A total of 150 g is needed to stretch into 2.7 cm. Thus we have to add 50 g of water to the glass weighing 100 g . 50 g is equivalent to 50 ml, since density of water is 1 g/L.
Hence, 50 ml of water should be poured to the glass.
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Un hamster fait tourner sa cage de 27cm de diamètre a raison de 14tours par minute .quel est le module de l accélération centripète d un morceau de nourriture collé sur la paroi extérieure de la cage ?
Answer:
Explanation:
a_c = ω²R
14 rev/min (2π rad/rev)(1 min/60 s) = 1.466 rad/s
a_c = 1.466²(0.27/2)
a_c = 0.29 m/s²
Le hamster est lent
EMERGENCY! PLS HELP
An isotope of Cesium-137 has a half-life of 30 years. If 6 grams of Cesium-137 decays over a period of 90 years, how many grams of Cesium would remain?
Answer:
0.75 grams
Explanation:
Today we have 6 grams
In 30 years, 3 grams remain
In 60 years 1.5 grams remain
in 90 years 0.75 grams remain
mathematically It would look like this
m = 6(0.5⁽ⁿ/³⁰⁾) where n is number of years
How does the suns energy contibute to the movement of water in the water cycle
Explanation:
The sun is what makes the water cycle work. The sun provides what almost everything on Earth needs to go—energy, or heat. Heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds... clouds that move over the globe and drop rain and snow.
Answer:
"The sun is what makes the water cycle work. ... Heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds... clouds that move over the globe and drop rain and snow. This process is a large part of the water cycle."
Explanation:
Which kind of precipitation is formed, in part, by wind?
A. Rain
B. Snow
C. Sleet
D. Hail
Answer:
Rain
Explanation:
What is Juan’s velocity between 0-5 seconds?
Juan’s velocity between 0-5 seconds would be 2 m/s, the slope of the position time graph represents the velocity of the object and between the 0-5 seconds slope of the line is 2.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
velocity= total displacement/Total time
As given in the problem we have to find the velocity of Juan between 0-5 seconds.
The displacement covered between 0-5 seconds is 10 meters.
the time taken to change position is 5 seconds,
the velocity of the Juan = 10/5
= 2 meters/seconds
Thus, its velocity of Juan comes out to be 2 meters/second.
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con lắc lò xo có m=0,6KG K=30N/m cho nó giao động điều hòa với biên độ 6cm tính thế năng động năng của con lắc khi x=4CM
Answer:
hdsjdjsksksjjdjd
Explanation:
ndnskskdjdkskdkxjdis bdjdkdnd