Recall, Newton's gravitational law states that any particle of matter in the universe attracts any other particle with a force varying directly as the product of their masses and inversely as the square of the distance between them. It is expressed as

F = Gm1m2/r^2

where

F is the force in Newton

G is gravitational constant = 6.673 x 10^-11 Nm^2/kg^2

m1 and m2 are the masses in kg

r is the distance in meters

From the information given,

r = 1.5

acceleration = 2cm^2/s

Recall, 100 cm = 1m

2 cm = 2/100 = 0.02m

Thus,

acceleration = 0.02m/s^2

Since the masses are identical, then m1 = m2

Each of them is accelerating at 0.02m/s^2

Recall,

Force = mass x acceleration

Force = m1 x 0.02 = 0.02m1 N

By substituting the given values into the formula, we have

0.02m1 = (6.673 x 10^-11 x m1 x m1)/1.5^2

m1 on the left cancels out one m1 on the right. It becomes

0.02 = (6.673 x 10^-11 x m1)/1.5^2

By crossmultiplying,

0.02 x 1.5^2 = 6.673 x 10^-11 x m1

0.045 = 6.673 x 10^-11 x m1

m1 = 0.045/6.673 x 10^-11

**m1 = 6.74 x 10^8 kg**

**The mass of each ball is 6.74 x 10^8 kg**

if you answer i will mark you as brainlist

reaction

Newtons third law states that for every action there is an equal but opposite reaction

and for that object (rocket, missile, jet,) as the engine is ignited the force of the burning fuel pushes backward while the rocket moves forward so the rate at which the fuel burns is equal to the rate at which the object moves forward

the force at which the burning fuel pushes backward is equal but opposite to the rate at which the object moves forward

a 22-g bullet traveling 240 m/s penetrates a 1.7 kg block of wood and emerges going 125 m/s .if the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

**Answer:**

m v1 = M V + m v2 conservation of momentum

V = m (v1 - v2) / M

V = .022 (240 - 125) / 1.7 = 1.49 m/s

determine the tension developed in cable ab for equilibrium of the 440- lb crate. determine the tension developed in cables ac for equilibrium of the 440- lb crate. determine the force developed along strut ad for equilibrium of the 440- lb crate.

In physics, a system is said to be in **equilibrium **when neither its internal energy state nor its state of motion tend to vary over time. If all of the forces operating on a single particle are vector summated to zero, equilibrium occurs.

In physics, tension is the pulling **force **that is conveyed axially by a string, cable, chain, or other similar one-dimensional continuous item, or by either end of a rod, truss member, or other similar three-dimensional object.

**Equilibrium **refers to the state of a system when neither its internal energy state or state of motion tend to change over time. **Equilibrium **for a single particle occurs when the vector sum of all **force **acting on the particle is zero.

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3. Cylinder of radius 5 cm and height 20 cm, is made of iron of density 7800 kg/m³, find

a. The volume of the cylinder in m³

b. The mass of the cylinder in kg.

The volume of the cylinder is 1.57 x 10⁻³ m³.

V = πr²h

V = π(5cm)²(20cm)(1m/100cm)³

V = 1.57 x 10⁻³ m³

The mass of the cylinder is 12.25 kg.

ρ = m/v

m = ρv

m = (7800 kg/m³)(1.57 x 10⁻³ m³)

m = 12.25 kg

a bird flying 3.45 m / s directly north feels a window directly east and accelerates 0.558 m / S^2. What is its velocity 5.25 s later

If a bird flying 3.45 m / s directly north feels a window directly east and **accelerates **0.558 m / s², then its **velocity **after 5.25 seconds would be

There are three equations of motion given by Newton,

v = u + at

S = ut + 1/2×a×t²

v² - u² = 2×a×s

As given in the problem If a bird flying 3.45 m / s directly north feels a window directly east and **accelerates **0.558 m / s²,

By using the first **equation **of motion,

v = u + a × t

v = 3.45 + 0.558 × 5.25

v = 6.537 m / s

Thus, its **velocity **after 5.25 seconds would be 6.537 m / s

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**Answer:**

about 4.53 m/s 40.3° east of north

**Explanation:**

You want the **velocity of a bird** flying **3.45 m/s north** after being **accelerated** by a wind at **0.558 m/s² directly east** for **5.25 seconds**.

The attached table shows the result of the acceleration. The change in "x" velocity is the product of the acceleration and time. The y-velocity is unaffected by the easterly acceleration.

Final velocityThe magnitude of the final velocity is the root of the sum of the squares of its components:

Vf = √(2.9295² +3.45²) ≈ **4.5226 . . . . m/s**

The direction of the final velocity is ...

arctan(3.45/2.9295) ≈ 49.7° . . . . . CCW from east

**The magnitude of the bird's final velocity is about 4.53 m/s. Its direction is about 40.3° east of north, or 49.7° north of east**. **The components of the velocity are about 2.93 m/s east and 3.45 m/s north**.

__

*Additional comment*

The second attachment shows the bird's direction as a "bearing" angle, conventionally measured clockwise from north. The answer above gives the complementary angle, measured CCW from east.

Problem values are given to 3 significant figures, so we have rounded results to that precision.

particle a has a mass m and a speed v, while particle b has a mass 2m and a speed 2v. what is the ratio of the kinetic energy of particle b to the kinetic energy of particle a?

The **ratio of the kinetic energy** of particle b of mass 2m and a speed 2v to the kinetic energy of particle a of mass m and a speed v is 8 : 1

**K = 1 / 2 m v²**

K = Kinetic energy

m = Mass

v = Velocity

Let kinetic energy of particle a be K1 and kinetic energy of particle b be K2.

For particle a,

K1 = 1 / 2 m v²

For particle b,

m = 2m

v = 2v

K2 = 1 / 2 ( 2 m ) ( 2 v )²

K2 = 4 m v²

The ratio of the kinetic energy of particle b to the kinetic energy of particle a,

**K2 / K1 **= ( 4 m v² ) / ( 1 / 2 m v² )

K2 / K1 = 8 / 1

Therefore, the ratio of kinetic energy of particle b to the kinetic energy of particle a is **8 : 1**

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what is the speed of the protons? (note: the speed is high enough that, in principle, we should use a relativistic calculation--something you'll learn about further--but for this problem you should use the formulas you are already familiar with.)

The speed of the **protons** v is 1.089[tex]e^{8}[/tex] m/s.

Every atom has a proton, a subatomic particle, in its nucleus. The particle has an electrical charge that is opposite to the electrons and is positive.

Define potential difference.To establish a potential difference, a unit of positive **electric charge** must be transported from one location to another.

Here we are given a** potential difference **as 62MV= 62[tex]e^{6}[/tex]V

Now, as per classical definition, equation P.E. and the K.E.

So we have,

qV = 1/2[tex]\frac{m}{s^{2} }[/tex]

V is the potential difference and v is the velocity.

v = [tex]\sqrt\frac{2qV}{m}[/tex]

By substituting the values we get,

v = 1.089[tex]e^{8}[/tex] m/s

The speed of the protons v is 1.089[tex]e^{8}[/tex] m/s.

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A patient's tumor is being treated with proton-beam therapy. The protons are accelerated through a potential difference of 62 MV. What is the speed of the protons? (Note: The speed is high enough that, in principle, we should use a relativistic calculation something you'll learn about in Chapter 27 but for this problem, you should use the formulas you are already familiar with.)

an ideal spring of negligible mass is 12.00 cmcm long when nothing is attached to it. when you hang a 3.75 kgkg object from it, you measure its length to be 13.40 cmcm.

If we have an **ideal spring** and attach a mass of 3.75 kg to it such that the spring extends from 12 cm to 13.40 cm, then its **spring constant **is 2678.57 N/m

When a force applied to a spring, then according to the **Hooke's Law**:

F = - k. x

Where:

F = **applied force**

k = spring constant

x = change in spring's length

The minus sign indicates that the displacement and the **restoring force** is in the opposite direction.

The given** parameters**:

x1 = 12 cm

x2 = 13.4 cm

Hence,

x = x2 - x1 = 13.4 - 12 = 1.4 cm = 1.4 x 10⁻² m

F = 3.75 . 10 = 37.5 N

Then,

k = F/x = 37.5 / (1.4 x 10⁻² ) = 2678.57 N/m

Complete question:

An ideal spring of negligible mass is 12.00 cm long when nothing is attached to it. when you hang a 3.75 kg object from it, you measure its length to be 13.40 cm. What is the **spring constant**?

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Suppose an asteroid orbits the sun with a mean radius 11 times that of Earth. Use the simplified version of

Kepler's third law (T2= a³) to find the period of the asteroid? Round your answer to the nearest Earth year.

**Answer:**

36 years

**Explanation:**

Given the equation of **Kepler's third law** as T² = a³, you want to know the period of an **asteroid** that has an orbital **radius of 11 au**.

Solving the given equation for t, we find ...

T = a^(3/2) . . . . . . take the 1/2 power

Then for a=11, the period in years is ...

T = 11^(3/2) ≈ 36.48

**The period of the asteroid is about 36 years**.

isabella's mom bought her a microscope so she could observe a plant cell.during her observation, she rotates the objective lens from 4x to 10x so she could view the plant cell at a higher magnification. what was the total magnification when she changed the objective lens?

Add the eyepiece's power, typically 10X, to the objective's (4x) power to get the overall **magnification**. Total magnification for 4x lens is **40x **and 10x is **400x**.

Make sure the 4X scanning objective is locked into position and the stage is completely down before viewing a slide through the **microscope**.

Set the slide over the aperture that you want to see, then carefully place the stage clips on top of the slide to secure it in place.

Start with the 4X objective, keep both eyes open while looking through the eyepiece (if necessary, cover one with your palm), and gradually raise the stage using the coarse adjustment knob until the image is clear. The coarse adjustment knob will only need to be used once during the operation. You will be utilizing parfocal **microscopes**, which means that the image. Things appear 40 times bigger than they are using a 40x objective. Comparing objective **magnification **is relative. The capacity of a **microscope **to create an image of an object at a scale bigger (or even smaller) than its real size is known as magnification.

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ANSWER ASAP: Leandra is working with a machine that uses 10,000w of power in 2 hours. Find the amount of work done by the machine.

w means "Watts"

The time must also be converted into seconds.

If Leandra is working with a machine that uses 10,000 watts of **power **in 2 hours, then the amount of **work **done by the machine would be 72000000 **Joules**.

The total amount of **energy** transferred when a force is applied to move an object through some distance.

The **work **done is the multiplication of applied **force **with the displacement.

As given in the problem Leandra is working with a **machine** that uses 10,000 Watts of **power **in 2 hours,

Power = work done / time

10000 = work done /3600*2

Work done = 72000000 Joules

Thus, the **work **done by Leandra would be 72000000 **Joules**.

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What is the pressure when 800N is applied to an area of 0.3m2 ?

**Answer: **[tex]2666.67 \text{ (or in 2 sigfigs, }2.7\times10^{3})[/tex]

**Explanation: **[tex]1\text{ }\dfrac{\text{N}}{\text{m}^2} = 1 \text{ Pa}\\\\P=\dfrac{F}{A}\\\\P=\dfrac{800\text{ N}}{0.3\text{ m}^2} = 2666.\overline6 \text{ N/m}^2 \approx 2666.67 \text{ Pa} \approx 2.7 \text{ kPa}[/tex]

Don't know if you need to convert to other units but typically in physics it's left as N/m^2 or Pa.

For the velocity-time graph shown, which statement describes what happens to the velocity between approximately 24 s and 25 s?

A) The lander's velocity increases away from the reference.

B) The lander's velocity decreases toward the reference.

C) The lander's velocity decreases away from the reference.

D) The lander's velocity increases toward the reference.

*The statement that describes what happens in the ***graph ***is option (B).*

The lander's **velocity** decreases toward the **reference**.

**Velocity time graph** is a type of graph that describes the change in velocity with respect to time of motion of an object.

Below is the basic explanation of the behavior of the **average velocity** of the particle with time.

Thus, we can conclude that between 20 and 25 seconds in the velocity tike graph, the **average velocity** of the object decreased towards the reference point (reference velocity = - 40 m/s.).

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# 14 If you have a tire blowout when driving on the highway, you should not use your brakes until...

If you have a tire **blowout** when driving on the highway, you should not use your brakes until the vehicle has nearly **rolled **to a stop.

If you have a tire** blowout** when driving on the highway, you should not use your brakes until the vehicle has nearly **rolled **to a stop because at high speed the deviation is more.

**Rolling** is a type of motion that combines rotation and translation of that object with respect to a surface, such that, if ideal conditions exist, the two are in contact with each other without sliding. **Rolling** where there is no sliding is referred to as pure **rolling.**

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In coming to a stop, a car has left black tire marks of length 92 m on a road. Assuming

the best rate of slowing down for this car is 7 m/s calculate the velocity of the car just

before braking

Fall

The velocity of the car just before breaking **35.9 m/s**

Braking distance, s = 92.0 m

Deacceleration = 7 m/s²

Speed of the car when stopped, =0

Let the speed of the car just before braking, =0

Using the equation of motion ²=²+2as

² = ² - 2as

² = 0² - 2(-7)(92.0)

² = 1288

≈ 35.9 m/s ← speed of the car just before breaking

What is acceleration?**Acceleration **can be defined as a vector quantity that determines the rate at which an object changes its **velocity**. An object continues to **accelerate **as it changes its speed over and over again.

Acceleration has nothing to do with changing the target's movement **speed**. If an object does not change its **speed**, then the object is not accelerating. The data on the right represents an object **moving north**. The speed of the target changes over **time**. In fact, the speed changes by a **constant amount** - ie. 10 m/s - every second. When the speed of an object changes, the object is said to be in **accelerated motion** or acceleration.

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If an element starts off with an activity of 100Bq, and its half life is 10 minutes, what would the activity be after 5 minutes???

I'm literally giving out 100 points and I really need the help :/

From the calculation, the **activity **of the **isotope **is ** 70.7 Bq.**

We define the term** half life **as the amount of a radioactive **isotope **that is left after a given time. It is in fact the time that it takes for only half of the amount of the radioactive isotopes that is in a sample to remain.

Recall that if a substance is radioactive, this implies that the sample is capable of spontaneous **disintegration**. The amount of the sample that is present tends to diminish in the system as time increases.

Give the fact that;

Ao = initial activity = 100Bq

A = activity after time t = ??

Time taken = 5 minutes

Half life of the isotope = 10 minutes

Now we know that;

A/Ao = (1/2)^t/t1/2

A/100 = (1/2)^5/10

A/100 = (1/2)^1/2

A = (1/2)^1/2 * 100

A =** 70.7 Bq**

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What is the mass of a blow dart that experiences a blowing force of 0.20 Newtons and travels the length of the 1.2 meter barrel in 0.04 seconds? (HINT: first find the acceleration)

The **mass of a blow dart** that experiences a blowing force of 0.20 Newtons and travels the length of the 1.2 meter barrel in 0.04 seconds is 0.26 g

**a = v / t**

**v = d / t**

a = Acceleration

v = Velocity

d = Distance

t = Time

d = 1.2 m

t = 0.04 s

v = 1.2 / 0.04

v = 30 m / s

a = 30 / 0.04

a = 750 m / s²

According to Newton's second law of motion,

**F = m a**

F = Force

m = Mass

F = 0.2 N

m = F / a

m = 0.2 / 750

m = 2.6 * [tex]10^{-4}[/tex] kg

m = 0.26 g

Therefore, the mass of a blow dart is **0.26 g**

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A 1500 kg car travels around a circular track at a constant speed. a) If the radius of the path is 200 m and the car is traveling at 45 m/s, how long does it take the car to do a lap around the track? b) What net force is acting on the car in the horizontal direction? c). What force keeps the car in a circular path? d). What is the coefficient of static friction between the car’s tires and the road?

a) The **time taken** is 4.44 s

b) The **force **that keeps it moving is 15188 N

c) The centripetal force is responsible for the motion in the circular path.

d) The coefficient of the **friction **is 1.03

The** centripetal force** is the force that keep the object moving in a circular path. In this case, we need to find the how long does it take the car to do a lap around the track.

a) The time that is taken is the radio of the distance to the speed of the object. Thus the **time **that is taken = 200 m/45 m/s = 4.44 s

b) The net force that acts on the car as it is negotiating the circular path is given as; mv^2/r = 1500 * (45)^2/200

= 15188 N

c) The centripetal force keeps the car on the circular path

d) The coefficient of static friction is; 15188 N/1500 kg * 9.8 m/s^2

= 1.033

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In the case of falling, the time will double how much the height will increase

In the case of** falling**, the time will double then the **height** will increase by a factor of **4 times.**

In this problem, we are dealing with **free-falling **and** distance**, where distance is The physical quantity that can be defined by how the object or body is seen far away, whereas** displacement **and Distance both are different physical meanings but have the same standard units and whereas a** free falling object** is an object that's falling beneath the sole impact of gravity. Any object that's being acted upon as it were by the **force of gravity **is said to be in a state of free fall.

Since we are given that the falling time is doubled, since the formula for the distance by the **equation of motion** is

s=1/2gt²

where g is the **acceleration due to gravity** and t is the** time**, so if the time is increased by 2 times, then the distance increase by the value of 4 times as distance is directly proportional to the square of the **falling time**.

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two forces parallel to the x axis do 11.0 j of work on a small tray while moving it 16.7 m in the x direction across a gym floor. one of the forces has a value of 3.71 n in the x direction. what is the other force?

The other **force **has **magnitude **of 2.41 N and **direction opposite **to that of force 3.71 N **parallel **to x axis

Let the other force=P

**work done=total force*displacement**

16.7=(P+3.71)*11

P+3.92=1.51

P=-2.41

**Newton **is the abbreviation for the SI **unit **of absolute **force **known as newton (N). It is described as the amount of force required to accelerate a kilogram of mass by one meter per second. In the foot-pound-second (English, or customary) system, one **newton **is equivalent to around 0.2248 pounds of force or 100,000 dynes in the centimeter-gram-second (**CGS**) system.

The **newton **was given its name in honor of Sir** Isaac Newton,** whose second rule of **motion **outlines the modifications that a **force **may cause in a body's motion.

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a football kicker can give the ball an initial speed of 25m/s. what are the a) least and b) the greatest elevation angles at which he can kick the ball to score a field goal from a point 50m in front of goalposts whose horizontal bar is 3.44m above the ground?

tan0=1.95 and tan0=0.605 are two solutions. 0 = 63 and theta 0 = 31 are the **corresponding **(first quadrant) **angles**.

The **elevation** angle is the angle between the horizontal line of sight and the object when standing and looking at something. The **angle** of depression is the angle between the horizontal line of sight and the object if the person is standing and looking down at the object. The rise or rise of something is called elevation. Mountains have an elevation based on their height, and your mood rises the happier you are. **Elevation** is a noun that describes the height of something above a surface or **ground** line. It is also a term of measurement for things like temperatures or degrees.

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Which best describes the velocity of a rolling pen?

4 s2

4 cm/s

4 cm west

4 cm/s west

[tex]\huge\underline\mathcal{Answer \: - }[/tex]

velocity is a vector quantity that tells us about the speed as well as direction of an object.

Thus , the best description of velocity will be the one which tells us the speed as well as direction of the rolling pen.

therefore ,

the last option **4**** ****cm/****s ****west **is correct!

hope helpful ;-;

a train slows down as it rounds a sharp horizontal turn, going from 90.0 km/h to 50.0 km/h in the 15.0 s it takes to round the bend. the radius of the curve is 150 m. compute the acceleration at the moment the train speed reaches 50.0 km/h. assume the train continues to slow down at this time at the same rate

The **acceleration** at the moment the train speed reaches 50.0 km/h is -0.74 m/s².

The **acceleration** of an object is the change in the **velocity** of the object with time.

a = Δv/Δt

where;

Δv is change in velocity of the objectΔt is change in time of motion of the objecta = (vf - vi) / t

where;

vf is the final velocity of the train = 50 km/h = 13.89 m/svi is the initial velocity of the train = 90 km/h = 25 m/sSubstitute the given parameters and solve for the **acceleration** of the train.

a = (13.89 m/s - 25 m/s) / (15 s)

a = -0.74 m/s²

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a 0.11 kgkg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed vv . the length of the string is 0.62 mm , and the tension in the string when the ball is at the top of the circle is 4.0 nn . what is vv ?

The **speed **of the ball during **centripetal motion **is** 4.75 m/s**.

We need to know about **centripetal force** to solve this problem. When an object moves in a circular motion, the object is experiencing centripetal force. The magnitude of **centripetal force** is

Fc = m . v²/R

where Fc is the **centripetal force**, m is mass, v is velocity and R is the radius.

From the question, we know that

m = 0.11 kg

F = 4 N

R = 0.62 m

The tension in the string is equal to **centripetal force**

T = Fc

4 = m . v²/R

4 = 0.11 . v² / 0.62

v² = 22.55

v = 4.75 m/s

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A piston contain oxygen at 280k occupying a volume of 0.25m^3.the cylinder is compressed adiabtically to 0.14m^3 find the increase in tempratur of the gas

The gas's internal **energy **rises by 250 J when the value is positive. So, the increase in temperature of the gas is 250 J.

Any device addressing the relationship between heat and work must be designed with a working knowledge of thermodynamics. This method ignores actions that take place at the molecular level in favor of considering a particular amount of matter.

In macroscopic thermodynamics, the effects are based on observable, quantifiable quantities, and the attributes of the system are given to the system as a whole.

Work cannot "store" **energy **like the aforementioned forms can. It is merely a momentary **manifestation **process of energy transmission. Heat cannot transport **energy **either; it is merely a temporary **manifestation **process. Therefore, other types of energies in transition include heat and work.

by setting a limit, identifies the subject of the **manifestation **analysis. Either "a distinct collection of substance" or "a region in space" must be handled consistently.

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When the lever is pulled, 2 kg of carbon dioxide is ejected at a speed of 60 m/s. The remaining mass of the person, chair, and cylinder is 80 kg. After the ejection, how fast will the chair be moving?.

The **combined velocity** of the car and the chair is 1.5 m/s.

By the use of the law of the conservation of **linear momentum** we know that the total momentum before collision is equal to the sum of the final momentum after collision.

We have to first obtain the momentum of the system before and after the collision has occurred as we see below;

Then we have;

Momentum before collision = 2 Kg * 60 m/s = 120 Kgm/s

Momentum after Collison = 80v Kgm/s

Momentum = mv

m = mass of the object

v = velocity of the object

Using the principle of the conservation of momentum;

120 = 90v

v = 120/80

v = 1.5 m/s

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A 0.10kg mass is oscillating at a small angle from a light string with a period of 0.63s. What is the length of the pendulum?

**Answer: 00.99m**

**Explanation:**

because

**Answer: 0.099**

**Explanation: Khan**

A 0.12-kg apple falls off a tree branch that is 2.3 m above the thick grass. The apple sinks 0.067 m into the grass while stopping. Determine the average contact force that the grass alone exerts on the apple while stopping it. Ignore air resistance.

**Answer:**

**F = 20 N**

**Explanation:**

Given:

m = 0.12 kg

H = 2.3 m

S = 0.067 m

g = 9.8 m/s²

____________

F - ?

1)

**Potential energy over grass:**

Wp = m*g*H = 0.12*9.8*2.3 ≈ 2.7 J

2)

**Work against the friction force of the grass:**

A = F₁*S

3)

**According to the law of conservation of energy:**

A = Wp

F₁*S = Wp

F₁ = Wp / S = 2.7 / 0.067 ≈ 40 N

4)

**Average grass resistance strength:**

F = F₁ / 2 = 40 / 2 = 20 N

when rebuilding her car’s engine, a physics major must exert 240 n of force to insert a dry steel piston into a steel cylinder. what is the magnitude of the normal force between the piston and cylinder in newtons if the coefficient of kinetic friction between the piston and cylinder is 0.3?

If the coefficient of **kinetic friction **between two sliding surfaces is μ and the **normal force** between them is N, then the kinetic friction between them is given by μ = F/N . Therefore** N= 800**.

Given

**kinetic friction** μ = 0.3

and kinetic friction μ = F/N

N = normal force

where F =** frictional force **

F = 240 n

N = F/μ

= 240 n/0.3

N = 800 n

There are two types of friction **kinetic friction**, also known as dynamical friction, and static friction. A **force **acting in opposition to the direction of a moving body on the surface is felt. The two materials' coefficient of kinetic friction will determine how much **force **is applied. Kinetic friction occurs whenever two surfaces slide past one another. The application of brakes to tyres, the movement of an object across the ground, such as a box, or the rubbing of sandpaper all cause kinetic friction. The coefficient of friction, which measures the relationship between the frictional **force **and the normal force, provides a good description of friction. It obstructs the motion of several different items.

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Imagine a black hole lacking any accretion disk whatsoever. Would this black hole produce any light? why?.

We wouldn't be able to observe the **black hole **if the accretion disc in it didn't create any light. Due to the **black hole's **intense gravitational pull, these particles spin at incredibly high speeds, producing an accretion disc and radiation in the process.

Therefore, there is no light when the **accretion disc **is absent.

Please refer to the solution for this step.

So, option an is the proper response. All other **possibilities **are false.

Please refer to the solution for this step.

A is the right response in this case.

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find moles of fe from 392.2 g of fe(nh4)2(so4)2
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