The speed and direction of the combined vehicles just after impact is 7.2 m/s west.
What is final speed of the cars after the collision?
The final speed of the cars after the collision is obtained by applying the principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v ( m₁ + m₂ )
where;
m₁ is the mass of the first carm₂ is the mass of the second caru₁ is the initial velocity of the first car = 55 km/h = 15.28 m/su₂ is the initial velocity of the second car = 65 km/h = 18.1 m/sv is the final velocity of the two cars(1600 x 15.28) - ( 3300 x 18.1) = v ( 1600 + 3300)
-35,282 = 4,900v
v = -35,282 / 4,900
v = -7.2 m/s
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the ability to do activities for more than a few minutes is
The ability to do activities for more than a few minutes is called aerobic ability.
Endurance is the ability to maintain an activity for extended periods of time and usually refers to aerobic ability. For example, children can play actively for hours. We need this ability to perform repetitive activities of daily living, such as stirring food while cooking, using a hair dryer to dry our hair, or walking up steps.
Recreational and job-related activity also often require a high level of endurance. Local muscle endurance is best defined as the ability to resist muscular fatigue and demonstrate how a given type of contraction can be sustained, typically measured in terms of the number of repetitions.
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What is the net work required to slow down a 5,020kg car from 41.0m/s to 21.0m/s?
a. Wnet = 6.22 · 106J
b. Wnet = 5.02 · 104J
c. Wnet = -3.11· 106J
d. Wnet =-9.85 · 105J
The net work that has been done is obtained as 5.02 * 10^4J. Option B
What is the net work that is required?We have to note that the work that is done is defined as the product of the force that has been applied and the acceleration of the object. In this case ,we can see that there is work that has been done as the velolcity of the object is changed.
Hence;
Work done = 1/2mΔv^2
m = mass of the object
v = velocity of the object
Thus;
Work done = 0.5 * 5020 * (21 - 41)^2
Work done = 5.02 * 10^4J
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consider the energy consumed by a 300 w light bulb in an hour. how many flights of stairs would you have to climb to equal the work of the lightbulb?
The number of flights of stairs you would have to climb to equal the work of the lightbulb in this time is approximately 589 stairs.
Given that,
Mass = 88 kg
Height of stairs = 2.1 meters.
Power = 300 Watt
Time = 1 hour
To calculate how many flights of stairs you would need to travel to accomplish the same amount of work as the lightbulb in this period of time:
First, we would figure out how much effort is required to ascend a set of steps that is 2.1 metres high.
Work done, mathematically, is given as,
W = m g h = 88 × 10 × 2.1 = 1848 J
Energy is given by, E = power × time
E = 300 × 3600 = 1080000 J
We can now determine how many flights of stairs you would need to climb to accomplish the same amount of work as the lightbulb in this period:
Flights = Energy/work done = 1080000/1848 = 588.41 ≈ 589 stairs
The given question is incomplete. The complete question is 'Assume your mass if 88 kg and the acceleration of gravity is 10 m/s². How much work do you do against gravity when you climb a flight of stairs that is 2.1 meters high?'
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a ball is thrown upward from the ground with an initial speed of 17.2 m/s; at the same instant, another ball is dropped from a building 20 m high. after how long will the balls be at the same height above the ground?
The balls will be at the same height above the ground at 1.16s.
Equations of motion in physics are equations that explain how a physical system behaves in terms of how its motion changes over time. The behaviour of a physical system is described in more detail by the equations of motion as a collection of mathematical functions expressed in terms of dynamic variables.
Before they collide, the falling ball travels a distance of (20m-h), where h is the height above the earth. We submit
yf= yi + vit - 0.5 gt²
to the falling ball to obtain,
-(20-h)= - 0.5 gt²
Applying yf= yi + vit - 0.5 gt² to the rising ball gives,
h= 17.2t - 0.5gt²
Combining both equations,
t= [tex]\frac{20}{17.2}[/tex] = 1.16 s
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A 2-kilogram box on a horizontal frictionless surface is acted upon by a 9-newton horizontal force to the left and a 1-newton horizontal force to the right. The acceleration of the box is
(1) 5 m/s^2 to the right
(2) 5 m/s^2 to the left
(3) 4 m/s^2 to the right
(4) 4 m/s^2 to the left
The acceleration of the box is 4 m/s² to the left.
option 4 is the correct answer.
What is the acceleration of the box?
The acceleration of the box is calculated by applying Newton's second law of motion as follows;
∑F = ma
where;
∑F is the net forcem is the mass of the boxa is the acceleration of the boxF₁ - F₂ = ma
1 N - 9 N = ( 2 kg ) a
-8 N = ( 2 kg ) a
divide both sides by 2
- 8 / ( 2 kg ) = a
- 4 m/s² = a
Thus, the acceleration of the box will be directed to the left which has greater force.
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what is the minimum number of degrees that the star below could be rotated so that it would look exactly the same as it did at its starting point?
The minimum number of degrees is 108°.
Based on the accompanying figure,
The angle of rotation from point A to point B is the smallest angle at which the star can overlap itself.
Let a° be the angle formed by A and B.
By connecting the star's vertices, we can form a pentagon.
The formula to calculate a polygon's internal angle is
The internal angle measurement is
=> 180 x (n-2)/n
=> 180 x (5-2)/5
=> 180 x 3/5
=> 36
ABC measured as:
= 108°
Since the inner angle is divided into three equal parts by a side star,
36° is the internal angle measurement.
the AOB,
The sum of a triangle's inner angles is 180 degrees, or m(OAB) + m(AOB) + m(ABO).
36° + a + 36° = 180°
a + 72° = 180°
a = 180 - 72
a = 108°
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The full question is :
What is the minimum angle of rotation (in degrees) that will carry the star onto itself?
I need answers, do your thing, I don’t need breakdowns or explanations, just answers.
a) The gravitational force on the earth is 81 times that on the moon
b) The slope of the graph is the universal gravitational constant
c) The force is 5 * 10^19 N
d) The acceleration due to gravity is the same.
What is gravity?We know that the term gravity has to do with the force that acts on any object that we can be able to find on the surface of the earth. We know that the larger the mass of the object, the greater the magnitude of the gravitational force on the object.
We can see that in the question the distance between the earth and the moon was halved hence the force would now be one quarter of its initial value hence;
New force = 1/4 * 2 * 10^20
= 5 * 10^19 N
The acceleration due to gravity does not change hence it is the same both at the surface and near the surface.
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Now imagine the same bouncing basketball as it is moving down. Does it have a positive or a negative acceleration? Why?
Imagine the same bouncing basketball as it is moving down, it is having a negative acceleration, as it is moving down.
This is because acceleration is defined as the rate of change of velocity, and negative acceleration occurs when an object's velocity is decreasing. The basketball is moving down, it is losing height, and its velocity is decreasing, thus it has a negative acceleration. To be more precise, the negative acceleration of the basketball is due to the force of gravity acting upon it, which is pulling it down toward the earth. The force of gravity is a constant force, acting downward, and it causes all objects to accelerate downward at a rate of 9.8 m/s^2 (on the surface of the earth).
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a baseball is thrown vertically into the air with a velocity v, and reaches a maximum height h. at what height was the baseball moving with one-half its original velocity? assume air resistance is negligible.
At the height of 3h/4 meter, the velocity of the baseball will be half of its original velocity v, if it reaches a maximum height h.
Maximum height reached by the baseball, = h
The velocity of the ball at maximum height v₁ = 0
Initial velocity, = v
by the 3rd equation of motion, v₁² = v² - 2gh
0 = v² - 2gh
v² = 2gh
Now let the height at which the velocity will be half of the original velocity = h₁
Then by the 3rd equation of motion,
(v/2)² = v² - 2gh₁
2gh₁ = v² - v²/4 = 3v²/4
h₁ = 3v²/8g
We know v² = 2gh, So
h₁ = 3(2gh)/8g = 3h/4 m.
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cart 1 of mass m moves with a velocity of v toward cart 2 which has a mass of 9m and is at rest. cart 1 collides elastically with cart 2. what is the velocity of both carts after the collision?
After the collision, cart 1 will have a velocity of v/10, and cart 2 will have a velocity of 9v/10.
The velocity of both carts after the collision can be calculated by using the conservation of momentum principle. In an elastic collision, both momentum and kinetic energy are conserved.
Before the collision, the total momentum of the system is equal to the momentum of cart 1, which is equal to m*v where the mass of cart 1 is multiplied by its velocity.
After the collision, the total momentum of the system is equal to the combined momentum of the two carts.
Consider the velocity of carts 1 & 2 after the collision as v1 & v2.
The total momentum after the collision is:
mv1 + 9mv2 = m*v (conservation of momentum)
By solving v1 & v2, we get:
v1 = (mv) / (m + 9m) = v / 10
v2 = (9mv) / (m + 9m) = 9v / 10
So after the collision, cart 1 will have a velocity of v/10, and cart 2 will have a velocity of 9v/10.
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What makes a thermometer sensitive
A thermometer sensitive is a bulb with a thin glass wall. so, option (a) is correct.
What is thermometer?
A thermometer is a device that measures temperature. The three most used ways to measure temperature are in degrees Celsius, degrees Fahrenheit, and degrees kelvin. The metric system incorporates the Celsius scale.
What is thermometer sensitive?
The degree of change in a thermometer's thermometric property for a unit change in temperature is referred to as its sensitivity. It speaks of the smallest temperature difference that may be seen or quantified. The difference between the maximum and least temperature is referred to as a thermometer's range.
Therefore, a thermometer sensitive is a bulb with a thin glass wall. so, option (a) is correct.
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what is the magnitude of the electric force on one of the masses? express your answer with the appropriate units.
Two 1.6 kg masses are 1.3 m apart (center to center) on a frictionless table. Each has + 9.6 µC of charge.
What is the magnitude of the electric force on one of the masses? Express your answer to two significant figures and include the appropriate units.
ANSWER:
F -0.49N
About electric forceThe electric force is the force experienced by a charged object that is in an electric field. The formulation of the electric force is sometimes confused with Coulomb's law, even though the electric force is more general in nature than this law, which only applies to two point charges./ the force possessed by electrically charged objects
Electric force formulaElectric force, as is generally the case, is denoted by the letter F or usually given a small index under E (electric) or L (electricity).
F=qE
with:
q: is the payload of the object
E : is the electric field
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the resultant of a 40-n force at right angles to a 30-n force is
The magnitude of resultant force of the two forces (40 N and 30 N) at right angles is 50 N.
The Pythagorean theorem is used to calculate the consequence of two forces acting at right angles to one another. According to the Pythagorean theorem, the square of the hypotenuse, the longest side opposite the right angle, in a right triangle, equals the sum of the squares of the other two sides. (the two shorter sides).
In this instance, the two forces combine to form the triangle's two shorter sides, with the hypotenuse as the third force.
Therefore, the amount of the resulting force can be determined as follows:
⇒ √(40² + 30²) = √(1600 + 900) = √2500 = 50 N
Hence, the magnitude of the resultant force of the two forces (40 N and 30 N) at right angles is 50 N.
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cold tcs food should be stored at an internal temperature of
Cold TCS food should be stored at an internal temperature of
41° F or below.
What is TCS food?TCS full form is Time/Temperature Control for Safety. Foods that typically spoil more quickly than other foods are referred to as TCS. These foods may either have a high initial microorganism load naturally or possess qualities that are ideal for the growth of pathogens.
The TCS food definition makes it clear that any time and temperature control has a significant impact on the foods that fall under this category. Food handlers may find it extremely beneficial to know how TCS foods behave in order to better formulate, prepare, and serve their food products to consumers. In order to avoid causing foodborne illnesses and food waste, this task is a crucial duty of both food managers and food handlers.
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The maximum horizontal distance a boy can throw a ball is 50 m. Assume he can throw
with the same initial speed at all angles. How high does he throw the ball when he throws
it straight upward?
(Hint: At what angle should the boy throw the ball to get furthest distance?]
The height the ball reaches when the boy throws the ball up in the air is 12.6m.
Given the maximum horizontal distance a boy can throw a ball (s) = 50m
The initial speed of the ball = um/s
The maximum height reached = ym
The motion in vertical direction is given as:
y = ut1 + 1/2at1^2 where v is the speed in vertical direction and a is the acceleration and t1 is the time taken to travel upward distance.
s = 0 + gt^2/2 then t = √2s/g = √2 x 50/9.8 = 3.18s
We know that as the velocity is same at any point in distance travelled distance = speed x time = 50m
So t1 = t/2 as it covers half time when reached maximum height.
The velocity (v) = gt = 9.8 x 3.18 = 31.164m/s
The maximum height reached (y) = vt1 + 1/2gt1^2 where t1 - t/2
y = vt/2 + 1/2g(t/2)^2
y = 50/2 + 1/2(9.8)(3.18/2)^2
y = 12.6m
Hence the maximum height the ball can be thrown is 12.6m.
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what is the maximum elevation above the roof the ball will reach? b) how long will it take to reach that maximum elevation? c) how long will it take to come back to the roof level? d) if the building is 15 m high, how long will it take for the ball to hit the ground after release?
Equations of motion for constant acceleration to find maximum elevation, time and velocity are as follows :
y = yi + vi*t + (1/2)*a*t^2
vf = vi + a*t
V^2= U^2 + 2 a s
Givens:
vf = 0 m/s (the ball stops at the time in question)
yi = 0 m (or assume whatever initial height you wish, and add to the final answer)
a = -9.8 m/s (we can add more significant digits, but this is close enough)
b) The maximum elevation above the roof the ball will reach can be found by, y= yi + vi*t + (1/2)*a*t^2
c)Time to reach maximum height can be obtained from v = u + at
at = v - u
therefore, time required is t = v - u / a
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The speed of an object undergoing constant acceleration increases from 8.0 meters per second to 16.0 meter per second in 10 seconds. How far does the object travel during the 10 seconds?
As acceleration increases from 8.0 m/s^2 to 16.0 m/s^2 in 10 seconds, the object travel during this 10 seconds is 84 meters.
To calculate the distance traveled by an object undergoing constant acceleration, we need to use the equation of motion for uniformly accelerated motion, which is given by:
d = v_0t + 0.5 at^2
where d is the distance traveled, v_0 is the initial velocity (8.0 m/s in this case), t is the time interval (10 s in this case), and a is the acceleration. To find the acceleration, we can use the following equation:
v = v_0 + at
Putting the final velocity (16.0 m/s) and the initial velocity (8.0 m/s) and the time interval (10 s), we get:
16 = 8 + a x 10
Solving for acceleration:
a = 0.8 m/s^2
Now that we have the acceleration, we can use the first equation to find the distance traveled:
d = v_0t + 0.5at^2
d = 8 x 10 + 0.5 x 0.8 x 10^2
d = 80 + 4
d = 84 m
So, the object travels 84 meters during the 10 seconds.
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Suppose you were to make a three-dimensional model of magnetic force using magnetic field lines and a magnet. Which two characteristics might your model have?
A. The field lines exist only at the top and the bottom of the magnet.
B. The field lines near the magnet are far apart.
C. Arrows on the field lines point from the magnets North Pole to its South Pole.
D. The field lines do not cross paths as they go from one pole of the magnet to the other.
Answer:
C
Explanation:
it shows how forces are converted from a certain pole to the other
is that point to the left of the first two particles, to their right, or be- tween them? (b) should the third particle be positively or negatively charged? (c) is the equilibrium stable or unstable?
A third charged particle (does not matter whether it is a positive or negative charged particle) can be placed in between -3q and -q at x along .
Is equilibrium stable or unstable?
Stable equilibrium exists when the object is in its lowest energy condition; metastable equilibrium exists when additional energy (ΔG) must be introduced before the object can reach true stability; unstable equilibrium exists when no additional energy is needed before reaching metastability or stability.An equilibrium is asymptotically stable when f'(x)<0\ ; that is, the slope of f is negative. It is unstable when f'(x)>0\ . The left two equilibria in the figure are hyperbolic (f'(x) \neq 0), the others are non-hyperbolic because the slope (eigenvalue) is zero.A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. For example, a marble at the bottom of a bowl will experience a restoring force when displaced from its equilibrium position.To learn more about equilibrium refers to:
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What things do we need to know to determine how much force was used on an object? Select 2 correct answer(s) Question 5 options: How far it went How much it weighs The object's mass Speed How fast it's moving The object's acceleration
Answer:
The object's mass
The object's acceleration
Explanation:
To determine how much force was used on an object, we need to know the object's mass and acceleration. The force acting on an object can be calculated using the equation force = mass x acceleration (F = m*a) So, the more massive the object is, the more force is required to accelerate it, and the greater the acceleration is, the greater the force applied to the object.
Five balls are placed one after the other along a straight line as shown in the figure. Initially, all the balls are at rest. Then the second ball has been projected with speed v0 towards the third ball. Mark the correct statements. (Assume all collisions to be head-on and elastic)
a. total number of collision in the process is 5
b. the velocity of separation between the first and fifth ball after the last possible collision is v0
c. finally, three balls remain stationary
d. all of the above
All of the given options are correct as the velocity of each ball is transferred via collision
Given,
5 balls are placed one after the other
the velocity of the second ball is V0
if the second ball is given the velocity, it will rebound back and hit the first one. Also, all the other balls will undergo collision due to the velocity given to the second ball.
therefore, 5 collisions will take place
as the second ball collides with the 3rd one, its momentum is transferred to the 4th and will carry on to 5th ball and thus separation velocity of 5th and 1st ball will be the same.
Again, as the collision ends up with only 1st and 2nd ball being separated by velocity V0 , then other 3 balls remain stationary
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for the tower below, the resultant of the tension in the two cables acts downward. what is the magnitude of this resultant (in units of kn)? the tension force in cable ac is 8 kn.
The effect of the stress between the two cables acts downward on the tower below. the consequent 11.96 kN's magnitude. 8 kN is the tension force in cable ac.
Tac = 8kN [40/sqrt(402+602) I - 60/sqrt(402+602)]
Tac is equal to 4.438 i-6.656 j kN.
If the tension force in AB is T, then Tab = T is equal to T [- 50 / sqrt(502 + 602) I - 60 / sqrt(502 + 602)].
The resulting Tnet = Tac + Tab = Tab = - 0.64 T I - 0.768 T j k N (4.438 - 0.64T) Since the consequent of I - (6.656 + 0.768 T) j is along j, the component along I will be zero.
magnitude of resultant = 6.656 + 0.768 T = 6.656 + 0.768(6.91) = 11.96 kN 4.438 - 0.64 T = 0 T = 6.91 kN
Simply said, "distance or amount" is the definition of size. In terms of motion, it shows the absolute or relative size, direction, or movement of an item.
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what are the major examples of functional groups? i. which are polar? which are non-polar? which are charged? how can you tell? be able to identify each one as either polar, non-polar, charged based on its structure
A polar molecule is usually formed when the one end of the molecule is said to possess more positive charges and whereas the opposite end of the molecule has negative charges, creating an electrical pole.
What is polar molecules?
The unequal sharing of electrons between the atoms and the unsymmetrical shape of the molecule means that a water molecule has two poles -
a positive charge on the hydrogen pole (side) and
a negative charge on the oxygen pole (side).
We say that the water molecule is electrically polar.
What is non -polar molecule?A nonpolar molecule is one whose charge distribution is spherically symmetric when averaged over time; since the charges oscillate, a temporary dipole moment exists at any given instant in a so-called nonpolar molecule.
These temporary dipole moments fluctuate rapidly in magnitude and direction,…
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a merry-go-round rotates at the rate of 0.27 rev/s with an 85 kg man standing at a point 2.2 m from the axis of rotation. what is the new angular speed when the man walks to a point 0 m from the center? consider the merry-go-round is a solid 67 kg cylinder of radius of 2.2 m. answer in units of rad/s.
The new angular speed of the merry go round after the man came to the point 0m on the merry go round will be 1.37 rev/s.
The merry go round is rotating at the speed of 0.27 rev/s with a man of mass 85 kg on it at a pint of 2.2 m form he center of the axis of rotation.
We have to find the new speed of the merry go round when the person walk to the center of the circle.
The mass and radius of the merry go round are given to be 67 kg and 2.2m.
Using the conservation of momentum as the system is not under any external force,
Initial momentum = final momentum
I₁W₁ = I₂W₂
I₁ and W₁ are initial moment of inertia and angular speed, I₁ and W₁ are final moment of inertia and angular speed.
Putting values,
(85+67)(2.2)₂0.27 = (67)W₂
1.34 rev/s = W₂
So, the final angular speed is 1.34 rev/s.
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The velocities of a rollercoaster car at points A and B are shown in the picture. The car has a mass of 25 kg. Apply the Work-Energy theorem to find the net work done by the car between point A and B.
The work done by the car is equal to the change in kinetic energy.
The work-energy theoremChange in kinetic energy = (1/2)*mass*(velocity A^2 - velocity B^2) Net work done = (1/2)*25*(6^2 - 9^2) Net work done = -225 JThe work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In the case of a rollercoaster car, the work done can be determined by calculating the change in the car's kinetic energy as it moves from point A to point B.The car's kinetic energy at point A can be calculated by using the equation KE=1/2mv^2, where m is the mass of the car and v is its velocity. We can substitute in the mass (25 kg) and the velocity (20 m/s) at point A to get a kinetic energy of 250 J.We can calculate the car's kinetic energy at point B in the same way. Substituting in the mass (25 kg) and the velocity (30 m/s) at point B yields a kinetic energy of 375 J.The change in the car's kinetic energy is equal to the work done by the car between point A and B. This can be calculated by subtracting the kinetic energy at point A from the kinetic energy at point B. Doing this yields a net work of 125 J. This means that the car does 125 J of work between point A and B.To learn more about The work-energy theorem refer to:
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Answer:A
Explanation:
a propeller plane and a jet travel 3000 miles. the velocity of the plane is 1/3 the velocity of the jet. it takes the prop plane 10 hours longer to complete the trip. what is the velocity of the jet?
The velocity of the jet is calculated to be 900 miles/hr when the relation between velocities of both propeller and the jet is given.
The distance travelled by a propeller plane and a jet = 3000 miles
Velocity of the plane Vp = 1/3 Velocity of the jet Vj
Vp = 1/3 Vj
Vj = 3 Vp ----(1)
Time taken by the propeller plane to complete the trip = 10 hours
Velocity of the propeller plane = Distance of the propeller plane/Time
⇒ 3000/10 = 300 miles/hr
The velocity of the jet = 3 times the velocity of the propeller plane
⇒ 3 × 300 miles/hr = 900 miles/hr
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how many watts of power are generated by an object with mass 6 kg that accelerates from rest to 9 m/s in 9 seconds?
The power generated by the object with a mass 6 kg and accelerate from rest to 9 m/s in 9 seconds is 27 watts.
How do you calculate power in watts?Power is energy transformed by the device per unit time.
The symbol for power is P and the unit of power is watt.
The formula used to calculate the power is as follows:
Power = [tex]\frac{E}{t}[/tex], which E is kinetic energy and t is time
To calculate the kinetic energy (E), can use the following formula :
[tex]E = \frac{1}{2} mv^{2}[/tex] or [tex]E = \frac{1}{2} m(v2^{2} - v1^{2})[/tex]
By combining both formulas above, here is the new formula that can be applied :
Power = [tex]\frac{\frac{1}{2} m(v2^{2} - v1^{2})}{t}[/tex] = [tex]\frac{\frac{1}{2} 6(9^{2} - 0^{2})}{9}[/tex] = [tex]\frac{\frac{1}{2}x6x81}{9}[/tex] = 27 watts
Thus, as the object has a mass 6 kg and accelerate from rest to 9 m/s in 9 seconds will generate power 27 watts.
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Shilpa found some small creatures moving in the fresh water near her house. What are those small creatures? Do these animals show complete or incomplete metamorphosis in their life cycle
Without more information about the specific characteristics and appearance of the small creatures that Shilpa found, it is difficult to determine exactly what they are.
However, there are many different types of small aquatic animals that can be found in fresh water, including insects, crustaceans, and mollusks. Some examples of small aquatic creatures that are commonly found in fresh water include:
Aquatic insects: Mayflies, stoneflies, caddisflies, and dragonflies are examples of aquatic insects that can be found in freshwater. These insects typically have a complete metamorphosis, where they go through four distinct stages in their life cycle: egg, larva, pupa, and adult.
Crustaceans: Freshwater crustaceans include crayfish, freshwater shrimp, and freshwater crabs. These animals have a complete metamorphosis, where they go through a series of stages in their life cycle, including a planktonic stage, megalopa stage, and juvenile stage before reaching adulthood.
Mollusks: Freshwater snails, freshwater mussels, and freshwater clams are examples of mollusks that can be found in freshwater. These animals typically have a complete metamorphosis, where they go through different stages in their life cycle, including a planktonic stage, a glochidium stage, and a juvenile stage before reaching adulthood.
It is important to note that without more information about the specific characteristics of the creatures that Shilpa found, it is impossible to determine their exact species and if they show complete or incomplete metamorphosis in their life cycle.
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how much work is done on a 3kg block that is pushed with a 12 newtons force for a distance of 9 meters
The work done on a block of certain mass when pushed with specific force is -162 J.
Mass of the block = 3 kg
Force with which the block is pushed = 12 N
Distance pushed = 9 m
The net force is nothing but the force minus the frictional force on the block.
Mathematically, Fnet = F - mg = 12 - (3×10) = -18 N
The work done is given by the formula, W = F × d
where, F is the force
d is the distance
W is the work done
Plugging in the values in equation, we have,
W = F × d = -18 × 9 = -162 J
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two small conducting point charges, separated by 0.5 m, carry a total charge of 180 c. they repel one another with a force of 120 n. for the universal constant k use the value 8.99 109 n m2/c2. find the charge on the larger of the two point charges:
The charge on the larger of the two-point charges is 90 coulombs
To determine the charge on the larger point charge, we can use Coulomb's law, which states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Using the known values of the force (120 N), distance (0.5 m), and the universal constant (8.99 x 10^9 Nm^2/C^2), we can use the equation F = k(q1q2)/r^2 to solve for one of the charges.
By rearranging the equation and substituting the known values, we get q1 = (Fr^2)/k. Then by substituting the given values, we get q1 = (120 N * (0.5 m)^2) / (8.99 x 10^9 Nm^2/C^2) = 90 coulombs.
Since we know the total charge is 180 coulombs and q1 is 90 coulombs, then the charge on the second point charge would also be 90 coulombs.
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