if a 5-kg bowling ball is projected upward witha velocity of 2.0 m/s, then what is the recoil velocity of the Earth(mass=6.0×10^24kg)

1. Meter Rule  

Meter rule is used to measure the small length or distance between any two points.

2. Measuring Tape

It is also use to measure length and distance in meters and centimeters.

3. Vernier calipers.

It can be use to measure small diameter or radius of objects up-to 10th part of a millimeter.

4. Micrometer screw gauge.

It is also use to measure diameter or radius unto 100th part of a millimeter.

5.Beam balance

It is use to measure mass upto 10th part of a gram.

6. Physical balance

it is use to measure mass upto 100th part of a gram.

7.Electronic balance

It is use to measure mass upto 100th part of a gram.

8. Stop watch.

It is use to measure time upto 10th and 100th part of a second.

9.Measuring cylinder

It is use to measure volume of a small object.

10. Hydrometer

It is use to measure density of liquids.

11. Barometer.

It is use to measure the atmospheric pressure.

12. Thermometer

It is use to measure temperature.

13. Ammeter

It is used to measure amount of electric current in circuit.

14.Photometer.

It is use to measure intensity of light.

15. Protector,.

It can be use to measure angle.

16. Lactometer.

It can be use to measure specific gravity of milk.

The energy which is due to position is potential energy. So when the snow is lying on the mountain. It possess potential energy but when suddenly, it starts to move down the mountain, the potential energy is converted into the kinetic energy. Yet some force is exerting on the snow to stop the smooth flow of snow through mountains.

This example of frictional force may be due to presence of rough surface or stones. Generally, there are four types of friction as static, rolling, sliding and fluid friction. Though in this case when snow is lying it possess static friction, when flows then it possesses sliding and fluid friction both.

We know that either team is winning which means that the cumulative forces (sum of each girl force) on both teams are equal.

This means,

[tex]250 + 250 + 250 + 250 = 260 + 260 + 260 + x[/tex]

being x the force of the fourth girl on the east team.

Solving to x, you get

[tex]x = (250 + 250 + 250 + 250) - (260+260+260)

x = 1000 - 780

x = 220[/tex]

The fourth girl on the east is pulling with a force of 220N.

The force applied by fourth girl in east team is 220 N.

If neither side is wining, it means that force applied on both sides are equal.

Total force by team west = 250 N * 4 = 1000 N

Total force of 3 girls on east side = 260 N * 3 = 780 N

As we know the east team apply total force of 1000 N.

Force applied by the fourth girl in east team = 1000 N - 780 N = 220 N

gravitational potential energy is given as

[tex]PE = mgH[/tex]

here given that

m = 75 kg

g = 9.8

H = 100 m

now plug in all value above

[tex]PE = 75 * 9.8* 100[/tex]

[tex]PE = 73500 J[/tex]

now for kinetic energy we can use

[tex]K = \frac{1}{2}mv^2[/tex]

here we have

m = 75 kg

v = 60 m/s

now we have

[tex]K = \frac{1}{2}*75*60^2[/tex]

[tex]K = 135000 J[/tex]

now for mechanical energy we have

[tex]ME = KE + PE[/tex]

so now we will add above to energy to find mechanical energy

[tex]ME = 73500 + 135000 [/tex]

[tex]ME = 208500 J[/tex]

Blow dart is projected from a height of 1.2 m

Dart will hit the target at height 0.6 m

so the vertical displacement of the dart will be 1.2 - 0.6 = 0.6 m

here we can say that in vertical direction kinematics is applied

[tex]y = v_i* t + \frac{1}{2}at^2[/tex]

[tex]0.6 = 0 + \frac{1}{2}*9.8*t^2[/tex]

[tex]t = 0.35 s[/tex]

now we can say that dart will hit the target after t = 0.35 s

so in the same time it will cover the distance 12 m in horizontal direction

so here in order to find the initial speed we can say

[tex]d = v * t[/tex]

[tex]12 = v* 0.35[/tex]

[tex]v = \frac{12}{0.35}[/tex]

[tex]v = 34.3 m/s[/tex]

so dart is projected with speed 34.3 m/s in horizontal direction

The blow dart is launched with an initial velocity of around 34.3 m/s. This is determined by calculating the time of flight using the height difference and gravitational acceleration, then determining the velocity using that flight time and the horizontal distance to the target.

The subject of this question is physics, as it requires an understanding of kinematic equations for projectile motion to solve. First, let's figure out how long the dart is in the air. We're going to use the formula h = 0.5 * g * t^2, where h is the height difference (1.2m - 0.6m = 0.6m), and g is the acceleration due to gravity (approximately 9.81 m/s^2). Solving for t, we have sqrt((2*h) / g) = sqrt((2*0.6) / 9.81) = approximately 0.35 seconds.

Now that we know how long the dart was in the air, we can find out its initial horizontal velocity. Since horizontal motion is the constant velocity in the absence of air resistance, the initial velocity is simply the total horizontal distance divided by the time. Therefore, the initial velocity, v, is d / t = 12m / 0.35s = approximately 34.3 m/s. So the dart was launched with an initial velocity of roughly 34.3 m/s.

https://brainly.com/question/34917979

#SPJ3

The maximum magnitude of their resultant vector is when the two vectors are parallel and in the same direction, so they lie on the same axis. In this case, the magnitude of their resultant vector is simply the sum of the two magnitudes:

[tex]R=2.5 km+6.5 km=9.0 km[/tex]

The minimum magnitude of their resultant vector is when the two vectors are parallel but in opposite direction. In this case, the magnitude of their resultant vectors is just the difference between the two magnitudes:

[tex]R=6.5 km-2.5 km=4.0 km[/tex]

I should assume the answer would be C, to determine the affects of sunlight on plant growth. In the above paragraph, everything else in the experiment was the same, and the only variable was the amount of sunlight reaching each plant. Hope this helped!

This procedure will demonstrate the effect of sunlight on plant growth.  If that wasn't the purpose that James had in mind, then he picked the wrong experiment.

These are all Examples of Models

                Option-A " The intermolecular forces holding the caffeine molecules together are weaker than the ionic bonds in CuCl₂ ".

                         There are two types of interactions among the atoms and molecules. One are known as intramolecular forces while the other are known as intermolecular forces.

Examples of Intramolecular forces are ionic bonds and covalent bonds e.t.c. while examples of intermolecular forces are hydrogen bond interactions, dipole-dipole interactions e.t.c.

Remember that intramolecular forces ar far more greater in strength than the intermolecular forces. Hence, in given statement the interactions between Caffeine molecules are intermolecular forces while, that between Cu and Cl ions in CuCl₂ are intramolecular forces.

Final answer:

The lower melting point of caffeine compared to copper (II) chloride is due to the weaker intermolecular forces in caffeine as opposed to the stronger ionic bonds in CuCl2.

Explanation:

The most important reason caffeine (C8H10N4O2) has a lower melting point than copper (II) chloride (CuCl2) is that the intermolecular forces holding the caffeine molecules together are weaker than the ionic bonds in CuCl2. Option A correctly identifies that caffeine, being a covalent molecular compound, is held together by intermolecular forces which are generally less strong than the ionic bonds found in ionic compounds like CuCl2.

This is due to the fact that ionic bonds result from the electrostatic attraction between ions of opposite charge, which is a significantly stronger force than that of the intermolecular forces present in molecular compounds like caffeine.

When a paper is crumpled and then released the air drag on the paper is less while when we put the flat paper and released it from rest we can see that there is large surface area due to which it will have more air drag and hence it will reach later

So we can say that air friction or air drag depends on the surface area of object

More will be the surface area more is the drag force of air and hence less is the effective acceleration due to gravity

So when we drop the two objects in vacuum it will not have any friction force

and hence they both will have same gravity downwards

So in that case the two objects will reach the ground in same time

so its answer will be

Both papers will reach the ground at the same time.

Answer:

the answer is they both hit at the same time

Explanation:

As we know that sphere roll without slipping so there is no loss of energy in this case

so here we can say that total energy is conserved

Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

[tex]\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2+ mgh'[/tex]

as we know that ball start from rest

[tex]v_i = 0[/tex]

height of the ball initially is given as

[tex]h = Lsin\theta[/tex]

[tex]h = 1200sin53 = 960 cm[/tex]

also we know that

[tex]I = \frac{2}{5}mR^2[/tex]

also for pure rolling

[tex]v = r\omega[/tex]

also we know that

[tex]480 = m*9.8 [/tex]

[tex]m = 49 kg[/tex]

now plug in all data in above equation

[tex]480*9.60 + 0 = \frac{1}{2}*49*(0.40*\omega)^2 + \frac{1}{2}*\frac{2}{5}*49*(0.40)^2\omega^2 + 0[/tex]

[tex]4608 = 3.92\omega^2 + 1.568\omega^2[/tex]

[tex]\omega^2 = 839.65[/tex]

[tex]\omega = 29 rad/s[/tex]

So speed at the bottom of the inclined plane will be 29 rad/s

Seagull is at height 11 m

it drops a shell from rest

so here we know that

acceleration of shell is due to free fall = -9.81 m/s^2

displacement = - 11 m

initial speed = 0

now we will use kinematics equation here in order to find the final speed

[tex]v_f^2 = vi^2 + 2as[/tex]

now plug in all values in it

[tex]v_f^2 = 0 + 2*(-9.81)*(-11)[/tex]

[tex]v_f^2 = 215.82[/tex]

[tex]v_f = 14.7 m/s[/tex]

so here the speed by which shell will strike with the rocks is 14.7 m/s

Final answer:

The velocity of a shell dropped from rest at a height of 11m when it hits the ground is approximately 14.69 m/s. This is calculated using the kinematic equation for uniformly accelerated motion, considering gravity as the acceleration.

Explanation:

The question refers to the velocity of a shell dropped by a seagull from a specific height until it hits the ground. To solve this, we can use the equations of motion under gravity. Specifically, we can use the following equation, which comes from the kinematic equations for uniformly accelerated motion:

v² = u² + 2as

v represents the final velocity

u is the initial velocity (which is 0 in this case, since the shell is dropped from rest)

a is the acceleration due to gravity (approximately 9.81 m/s² on Earth)

s is the distance the object has fallen, in this case, 11 meters

Plugging the numbers into the equation gives us:

v² = 0⁻ + 2*9.81*11

v² = 215.82

v = √(215.82)

v = 14.69 m/s (to two decimal places)

Therefore, the velocity of the shell when it hits the rocks will be approximately 14.69 m/s.

Answer:

55 miles per hour

Explanation:

Given

Distance d = 55 miles

t_o =1.00 pm

t_f = 2.00 pm

Solution

Duration of the trip

[tex]\Delta t = t_f - t_o\\\\\Delta t = 2.00 pm - 1.00 pm\\\\\Delta t = 1 hour[/tex]

Average velocity

[tex]v = \frac{d}{\Delta t} \\\\v = \frac{55}{1} \\\\v = 55 mph[/tex]

7.  First write down all the known variables while separating the values for each direction:

x-direction:

vix = 20m/s

vfx = 20m/s

x = 39.2m

y-direction:

viy = 0m/s

ay = -9.8m/s^2

y = ?

Based on the knowns, the first step is to calculate the time of flight from the x-direction as it will be the same as value for the y-direction.  Find the correct kinematic equation to do so:

x = (1/2)(vix+vfx)t

(39.2) = (1/2)(20+20)t

1.96s = t

Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the y-direction:

y = viy*t + (1/2)ay*t^2

y = (0)(1.96) + (1/2)(-9.8)(1.96)^2

y = -18.82m  (Value is negative because gravity constant was negative.  It is the height reference that from the top of the building down, which is why it is negative.  The sign can be ignored for this question.)

8.  First write down all the known variables while separating the values for each direction:

x-direction:

x = 12m

vfx = 0m/s

vix = ?

y-direction:

y1 = 1.2m

y2 = 0.6m

viy = 0m/s;

ay = -9.8m/s^2

First find time in the y-direction as it would be the same value for the x-direction.

(y2 - y1) = viy*t + (1/2)ay*t^2

(-0.6) = (0)t + (1/2)(-9.8)t^2

t = 0.35s

Now that we have the time of flight, we can use the kinematic equation that will relate the known variables in the x-direction:

x = (1/2)(vix+vfx)t

(12) = (1/2)(vix+(0))(0.35)

68.6m/s = vix

Of the following given choices;

The answer is; gravitational.

When the delivery man pushes the doorbell, he is utilizing mechanical energy. This energy is later converted to electrical energy that travels through a circuit to the inside of the house. An actuator then converts the electrical energy to sound that is heard inside the house.

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

[tex]E = mc^2[/tex]

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

[tex]E = mc^2[/tex]

the answer is B i just took it

Answer and explanation;

According to twin studies attitudes may be genetically based, research has supported that Identical twins are more likely to share similar attitudes. Attitudes change as the result of behavior and not the other way around

Genetic factors may influence general dispositions and condition-ability that may influence formation of more specific attitudes. Controlled twin studies in US and Sweden reveal that identical twins share more similar attitudes than fraternal twins.

The correct answer is average speed.

Since we are referring to distance expressed over time taken we are referring to average speed.

If it is displacement expressed over time taken, then we get the average velocity.

Also the velocity expressed over time will give us average acceleration.

Note that distance is a scalar quantity so it goes with another scalar quantity which is speed.

The displacement which is also a vector quantity will go with velocity which is also a vector quantity.

The correct option is B. Which is average speed.

When you are on a swing, and at the lowest point of your motion, is your apparent weight greater than to your true weight.

In general, an object's apparent weight is equal to its mass multiplied by the vector difference between gravitational acceleration and the object's acceleration.

According to this definition, apparent weight is a vector that can act in any direction, not just vertically.

The weight of an object is how gravity pulls on it, but apparent weight is a measure of downward force. Weight, on the other hand, is a measure of the force exerted by gravity.

We can pretend you're in a uniform circular motion at the bottom of the swing trajectory, so the net force points toward the center of the circle.

As a result, the swing must exert a greater force on you than the earth does in the opposite direction. As a result, your apparent weight is greater than your true weight.

Thus, the true weight will be less than the apparent weight.

For more details regarding apparent weight, visit:

https://brainly.com/question/14323035

#SPJ2

The average atomic mass of an element is the mass found on the periodic table under the element symbol.  It is called the average atomic mass because it is actually a calculated sum of the masses of the element's isotopes (same element with different number of neutrons), each multiplied by their natural abundances.

The average atomic mess of an element is the sum of masses of its isotopes, each multiplied by its natural abundance .

Answer:

C. 6.14 m

Explanation:

The maximum horizontal distance travelled by a projectile is given by:

[tex]d=\frac{v^2}{g}sin (2\theta)[/tex]

where

v is the initial velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

[tex]\theta[/tex] is the angle of launch

From the equation, we see that the maximum range is achieved when the projectile is fired at [tex]\theta=45^{\circ}[/tex].

The dart in the problem has an initial velocity of

v = 7.76 m/s

Substituting into the formula, we find the maximum horizontal distance:

[tex]d=\frac{7.76^2}{9.8}sin (2\cdot 45^{\circ}) = 6.14 m[/tex]

Answer:

acceleration = 0.12 m/s^2  

Explanation:

Using Second Law of Motion,

F = m*a

μ = f/Fn

Fn = m*g – 90*sin(30)

Fn = (20)*(9.81) – 90*sin(30)

Fn = 151.2 N

f = μ*Fn

f = (0.5)*(151.2)

f = 75.6 N

F = 90cos(30) – 75.6

F = 2.34 N

a = F/m

a = 2.34/20  

a = 0.12 m/s^2  

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

[tex]\vec v_{ra} = 12 m/s[/tex] North

so it will be

Let North direction is along Y axis and East direction is along X axis

[tex]\vec v_{ra} = 12\hat j[/tex]

also it is given that speed of air is 6.7 m/s relative to ground

[tex]\vec v_a = 6.7 \hat i[/tex]

now as we know by the concept of relative motion

[tex]\vec v_{ab} = \vec v_a - \vec v_b[/tex]

[tex]\vec v_{ra} = \vec v_r  - \vec v_a[/tex]

now by rearranging the terms

[tex]\vec v_r = \vec v_{ra} + \vec v_a[/tex]

[tex]\vec v_r = 12 \hat j + 6.7 \hat i[/tex]

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

[tex]v_r = \sqrt{12^2 + 6.7^2}[/tex]

[tex]v_r = 13.7 m/s[/tex]

so the net speed of Robin with respect to ground will be 13.7 m/s

 Vi =0  ( as car starts from rest )

Acceleration= a = 5m/s²

t =12sec

Vf =Vi + at

Vf= 0 + (5m/s²) (12s)

Vf=60 m/s

A. anything less than 3.0 magnitude on a richters scale usually can't be felt by humans but instruments can pick it up.

Answer:

A. Less than 2.9

Explanation:

Richter scale is a measure of the intensity of the earthquakes designed and developed by Charles Richter in 1935. Its value ranges from 1.0 to 9.0 and greater. The earthquakes having a Richter magnitude of less than 2.9 are minor earthquakes which are slightly felt by a few humans but can be recorded on a seismograph. Earthquakes with magnitude above 2.9 can easily be felt by humans.

Let say the basket ball is tossed upwards with initial speed "v"

now when it moves up in the air its speed will decrease due to acceleration of gravity in opposite direction

So here we can say that its final speed at the top position will be zero

here it will take t = 2.5 s to reach the top

now we can use equation of kinematics

[tex]v_f = v_i + at[/tex]

[tex]0 = v + (-9.8)*2.5[/tex]

now we have

[tex]v = 9.8 * 2.5[/tex]

[tex]v = 24.5 m/s[/tex]

so the ball is tossed up initially with speed 24.5 m/s

If both nuggets have the same mass and different densities then it can be concluded that the volume with the highest density is smaller.

For, so to speak, density is a relation of how much mass is there in a given volume. So the more the mass and the smaller the volume, the greater the density.

To verify this, let us calculate the volumes of iron pyrite and gold pyrite.

For the iron pyrite nugget:

density = mass / volume

volume = mass / density

volume = 50/5

volume = 10cm3

For the gold nugget:

volume = mass / density

volume = 50 / 19.3

volume = 2.59cm3

Therefore it is found that the nugget with the highest density (gold) is the one with the lowest volume.