The **work done** by** friction i**s -1617 J

**Work done** is the product of force and distance moved in the direction of the force.

**Friction** is the opposition to motion between two surfaces

From the lwa of conservation of energy, the **work done **by **friction,** W equals the change in potential energy of the skater, ΔU.

So, ΔU = W

mgΔh = W

mg(h₂ - h₁) = W where

m = mass of skater = 55 kg, g = acceleration due to gravity = 9.8 m/s², h₁ = initial height of skater = 15 m and h₂ = final height of skater = 12 m.So, substituting the values of the variables into the equation for the **work done**, we have

W = mg(h₂ - h₁)

= 55 kg × 9.8 m/s²(12 m - 15 m)

= 55 kg × 9.8 m/s² × (-3 m)

= 539 kgm/s² × (-3 m)

= -1617 J

So, the **work done** is -1617 J

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During which season does the north pole experience constant sunlight?

A. spring

B. winter

C. summer

D. autumn

During the summer **season**, the **north **pole gets constant sunlight.

In the summer **season**, the **north **pole experience constant sunlight. Because in the summer the shadows are short and the sun remains higher in the sky. In the summertime, the sun remains above the horizon at the **north **pole which circles the pole every day.

In the summer the sun's rays reach the **north **pole while in winter **season** the **north** pole is tilted away from the incoming sunshine. Mostly the regions of the **north **pole remain cold because the sunlight does not directly fall on the **north **pole. At the **north** pole, there are only two types of seasons, one is summer and the other is winter **season**. The annual average temperature of the winter **season** at the **north** pole is -40 degrees Celsius while in summer **season** the average temperature is 32 F.

So we can conclude that during the months of the summer **season **the **north **pole gets maximum sunlight.

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A golf ball is hit horizontally at 40 m/s from the top of a hill that is 2.5 m high. If the terrain around the hill is nearly flat,approximately how far will the golf ball fly? Use - 9.81 m/s2 for the acceleration caused by gravity. Ignore air resistance.Round any intermediate calculations to no less than six decimal places, and round your final answer to two decimal places.

**ANSWER**

**28.56 m**

**EXPLANATION**

Let's make a diagram of this situation to understand it better,

The ball has an initial velocity of 40m/s, which is horizontal. This means that there is no vertical initial velocity.

We have to find the horizontal distance the golf ball flies. Since the horizontal velocity is constant - this is because there is no acceleration in that direction, the distance the ball travels is,

[tex]\Delta x=v_{ox}\cdot t[/tex]The horizontal initial velocity is given, but we have to find the time the ball was in the air. To find it, we use the vertical distance the ball travels - which we know is the height of the hill. In this case, we do have vertical acceleration - the acceleration of gravity, so the vertical distance the ball travels, as shown in the diagram, has a parabolic form and it is given by the equation,

[tex]\Delta y=v_{oy}\cdot t+\frac{1}{2}gt^2[/tex]The initial vertical velocity is zero because the ball is hit horizontally,

[tex]\Delta y=\frac{1}{2}gt^2[/tex]Solve for t. Multiply both sides by 2/g,

[tex]\begin{gathered} \Delta y\frac{2}{g}=\frac{1}{2}g\cdot\frac{2}{g}\cdot t^2 \\ \frac{2\Delta y}{g}=t^2 \end{gathered}[/tex]And take the square root to both sides of the equation,

[tex]t=\sqrt[]{\frac{2\Delta y}{g}}[/tex]Δy is the height of the hill, 2.5m, and g = 9.81m/s²,

[tex]t=\sqrt[]{\frac{2\cdot2.5m}{9.81m/s^2}}=\sqrt[]{\frac{5m}{9.81m/s^2}}=\sqrt[]{0.509684s^2}=0.713922s[/tex]This is the time the ball was in the air for. Now we can find the distance it traveled,

[tex]\Delta x=40m/s\cdot0.713922s=28.56m[/tex]The golf ball flew **28.56 m** horizontally.

Which measurements or observations are needed to calculate density

**Answer:**

mass and volume measurements are required to calculate density.

3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?

**Answer:**

**9.72 x 10^(-11)**

**Explanation:**

The energy stored in a capacitor can be calculated as:

[tex]E=\frac{CV^2}{2}[/tex]Where C is the capacitance and V is the Voltage. So, replacing C by 1.35pF or 1.35 x 10^(-12) F, and V by 12.0 V, we get:

[tex]E=\frac{1.35\times10^{-12}F(12.0V)^2}{2}=9.72\times10^{-11}J[/tex]Therefore, the capacitot would store 9.72 x 10^(-11) J of energy.

The **electrical potential energy **that would be stored in the **capacitor** is 9.72 x 10^(-11) J. It is the **energy** which is present in the object when it is at rest.

**Potential energy** is the energy which is stored inside an object which is at rest.

For the **parallel plate capacitors**, capacitance is dependent upon its geometry, which is given by the formula:

**C=ϵ⋅Ad **or C = ϵ ⋅ A d ,

where, C is the value of the** capacitance**,

A is the** area **of each plate,

d is the **distance** between the plates, and

ϵ is the **permittivity **of the material between the plates of the parallel capacitor.

The **energy** stored in a **capacitor **can be calculated as:

**E = CV²/2**

where, C is the capacitance and V is the Voltage.

So, by replacing C with 1.35pF or 1.35 × 10⁻¹²F, and V by 12.0V, we get:

E = 1.35 × 10⁻¹² × (12)²/2

**E = 9.72 × 10⁻¹¹ J**

Therefore, the **capacitor **would store 9.72 × 10⁻¹¹ J of energy.

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A ball is thrown downwards from a height of 25m with an initial velocity of 10 m/s.

Find the velocity of the ball just before it hits the ground.

•V₁=10mls

h=25m

The **velocity **of the ball just before it hits the ground is 24.29m/s.

**What is acceleration due to gravity?**

The acceleration an object experiences as a result of gravitational force is known as **acceleration due to gravity**. M/s2 is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on earth's surface is 9.8 m/s2.

Given:

Initial velocity u = 10m/s

height h = 25m

we know that the **acceleration due to gravity** g is : 9.8m/s

Therefore using the formula,

v² - u² = 2gs where s is the height in this case and v is the final velocity.

v² - 10² = 2 x 9.8 x 25

=> v² = 490 + 100

=> v = √590

=> v = 24.29m/s

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How do I find if the dog is accelerating or not?

We will have the following:

* The dog walks at a constant speed of 1 m/s in a straight line east. **[Not accelerating]**.

* The dog speeds up from 1 m/s to 7 m/s as he runs away from a cat that scared him. **[Accelerating]**.

* The dog walks at a constant speed of 0.5 m/s in a circle around a bowl of food. **[Not accelerating]**.

* The dog stand in one spot on his bed. **[Not accelerating]**.

* The dog slows down to a stop as he approaches the fence at the edge of the yard. **[Accelerating]**.

if the camera left his hands at 21.9 m/s and took 2.009s to hit the ground, how far down the cliff did it fall?

The camera **falls **down 25.835 meters under **gravity**.

While a frame **falls**, it happens because of **gravity**, and the force of air will increase with speed. This continues till the force of air equals the weight. Now the accelerates but falls at a constant speed known as the** terminal speed**.

A **motion under gravity **refers back to the motion of an item whose vertical movement is laid low with the presence of gravity. The pressure that attracts gadgets **downwards **is gravity. In fact, gravity works towards the center of the Earth.

S = ut - 1/2at²

= 21.9 × 2.009s - 1/2 × 9.8 × (2.009)²

= 43.997 - 18.162

= 25.835 meter.

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A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.

A) What is the potential energy of the spring when it is stretched 0.300 m?

B) What is its potential energy when it is compressed 6.00 cm?

A **force **of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.

A) The potential energy of the spring when it is stretched 0.300 m will be 81 J

B) The potential energy when it is compressed 6.00 cm will be 3.24 J

**Hooke's law**, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.

To find spring constant , using formula

k= F/x

where

F = Force

k = spring constant

x = displacement

k = 540 / .300 = 1800 N/m

A) **Potential energy **of the spring when it is stretched 0.300 m

U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.300^{2}[/tex] = 81 J

B ) Potential energy of the spring when it is stretched 0.06 m

U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.06^{2}[/tex] = 3.24 J

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Can anyone tell me the final displacement of the puck

The final **displacement **of the puck is 4.14i + 14.14j

**Displacement** is the change in position of an object.

Since for the first **displacement **of the puck from the goal is 20 m at 45°, we have that its** displacement **vector is d = (20cos 45°)i + (20sin45°)j.

= (20 × 0.7071)i + (20 × 0.7071)j.

= 14.14i + 14.14j

Also, for the second **displacemen**t is 10 m to the left. Its displacement vector is d' = -10i

So, we see that the total **displacement** is D = d + d'

So, adding the displacements, we have D = d + d'

= 14.14i + 14.14j + (-10i)

= 14.14i - 10i + 14.14j

= 4.14i + 14.14j

So, we see that the final **displacement **of the puck is 4.14i + 14.14j

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7. Determine the resistance of a lamp that draws 3 amperes when connected to a120-volt supply

A lamp draws **3** amperes of current and is connected to a **120** volt supply.

We are asked to find the resistance of the lamp.

Recall the Ohm's law is given by

[tex]V=IR[/tex]Where **V** is the voltage, **I** is current, and **R** is the resistance.

Re-arrange the equation for resistance

[tex]R=\frac{V}{I}[/tex]Substitute the given values

[tex]\begin{gathered} R=\frac{120}{3} \\ R=40\; \Omega \end{gathered}[/tex]Therefore, the resistance of the lamp is **40** ohms.

A railroad locomotive is at rest with its whistle shrieking and then it starts moving toward you. Describe what happens to the frequency and wavelength of the sound that hits your ear when:A) It is approaching you.B) It pulls away from you.What happens to the amplitude in both of these scenarios?

a) Frequency increases , the time between waves decreases , wavelenghts decrease

b) Frequency decreases , wavelenght increases.

The doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. AS the source approaches the listener , the sound waves get closer together, increasing their frequency and decreasing their wavelenghts.

Amplitude remains the same.

7 The distance-time graph for a motorcyclist riding off from rest is shown in Figure 1.2.18.

a Describe the motion.

b Calculate how far the motorbike moves in 30 seconds.

c Calculate the speed.

The **distance **travelled by the motorbike is** 600 m** and the **speed **of the motorbike is **20 m/s.**

The **motion **of the motorbike can be described as **linear **type of **motion **since the distance travelled by the motorbike increases with increase in the time of motion.

There is directly **linear **relationship between the distance travelled by the motorbike and the time of motion of the motorbike.

The **distance **travelled by the motorbike is calculated as follows;

from the given graph, at time 30 seconds, the corresponding the **distance **of the motorbike on the vertical axis is 600 m.

distance = 600 m

The **speed **of the motorbike is calculated as follows;

v = d/t

v = 600 m / 30 s

v = 20 m/s

Thus, the **motion **of the motorbike is a **linear **type of motion, the **distance **travelled by the motorbike increases with increase in the time of motion and the **speed **of the motorbike depends on the **distance **and time of motion.

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The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity. Use the appropriate kinematics equation to show that this is true.

WILL MARK BRAINLIEST LOTS OF POINTS!

The** rocket’s velocity** just before it hits the ground is the same magnitude as the initial **velocity** because change in length in y direction will zero.

if u and v are the initial and final** velocities **of the **rocket** then from kinematic equations it may be written as

[tex]s=ut+\frac{1}{2} at^{2}\\0=ut+\frac{1}{2} at^{2}\\\frac{1}{2} at^{2} =-u t\\at=-2u[/tex]

since

v=u + at

v=u - 2u

v = - u

Hence the **rocket** hit the ground with same **velocity** as it projected.

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Jason is pulling a box across the roompulling with a force of 16 newtons and his arm is making a 68 anglet with the horizontal what is the horizontal component of the he is pulling with?

The horizontal component of a force is defined as follows

[tex]F_x=F\cdot\cos \theta[/tex]Where F = 16 N, and theta = 68. Let's use these values to find the horizontal component of F.

[tex]F_x=16N\cdot\cos 68\approx6N[/tex]Therefore, the horizontal component of the pulling force is 6 Newtons.Graphing is a way to represent data visually to represent the motion of a rolling ball students craft data on A-line graph which of the following would accurately label the X ax

Such an object lacks a velocity since it lacks a **slope**. Such an object must have a **variable **velocity. It's quickening.

Starting off, let's graph a few instances of motion with **constant speed**. The graph on the right shows three alternative curves, each with a zero initial point. The first thing to note is how **straight **each graph is. (On a **graph**, a curve can be any type of line that is drawn.

It is not possible to determine the velocity from a **position**-**time** **graph's **slope when the graph is curved. Only straight lines have slope as a characteristic. Such an object lacks a velocity since it lacks a **slope**. Words like "the" and "a" are highlighted in this text to emphasize that there isn't a single **velocity **for these conditions. Such an object must have a variable velocity. It's quickening.

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Please help me

how law of motion applied in basketball explain

**Explanation:**

According to the first law of motion, the basketball is always moving in one direction, unless acted on by another force.

What would happen if an unbalanced force was acting on an object? Would it move at a constant speed? Explain why or why not.

According to **Newton's Second Law of Motion**, if a net force acts on an object, the object accelerates acording to the equation:

Where **m** is the mass of the object.

Then, if an unbalanced force is acting on an object, the net force is different from 0, which means that the object will accelerate in the direction of the net force.

If an object has a uniform acceleration, it won't have a constant velocity, but it may have a constant speed anyway if the force is always perpendicular to the velocity, in that case the object would be moving in a circular trajectory.

In other words, **an unbalanced force will cause an acceleration**, this can either change the speed of an object or the direction of its velocity (or both).

Then, it might be possible for an object to have a constant speed under the influence of an unbalanced force if that object follows a circular trajectory.

A= 1/2 W (R+S) for W

The change of the **subject **in the **formula **is given by; W = 2A/(R + S)

When we have a **formula**, each of the components of the formula are the **parameters **of the formula. In this case, we have a formula that says A= 1/2 W (R+S). The parameters that are in this formula are A, W, R and S.

According to the question, we have been asked to make W the **subject **of the formula.

The** first step **is to multiply both sides by 2

2A = W(R + S)

And **secondly**, we divide both sides by the sum of R and S and we have;

W = 2A/(R + S)

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Which is true of the force pair of Newton's third law?a.)The two forces are in the same direction.b.)The two forces never produce an acceleration. c.)The two forces act on different objects.

**Given:**

Force pair of Newton's third law.

**Required:**

To find the true statement for the force pair of Newton's third law.

**Explanation:**

According to Newton's third law

This can be understood by taking an example.

Suppose a person hits the wall with his hand, the person hitting the wall is the action.

This action results in pain in the person's hand which is the reaction.

Here, the hand exerted force on the wall and the wall exerted an equal force in opposite direction.

The conclusion can be drawn as

Action and reaction act on two different objects.

The forces act in opposite direction.

**Final Answer: **The two forces act on different objects.

A cell of a constant e. m.f is connected in series with a parallel arrangement of 8ohms coil and a uniform wire of length (Lcm). The total external resistance in the circuit is found to be 6ohms. calculate the resistance of the wire and its length. Resistivity of the wire material = 6.0×10^-5 ohm/meter, cross-sectional area of the wire = 6.0×10^-3cm².

The resistance of the wire will be **24ohm** and the length of the wire will be** 24×10⁴cm.**

As given in the question,

**Rₜₒₜₐₗ **= 6ohm

**Resistivity **= 6×10⁻⁵ohm/m = 6×10⁻⁵/100ohm/cm

Area(A)= 6×10⁻³cm²

Let R = resistance of the wire, therefore;

Rₜₒₜₐₗ = R×8 / R+8

6 = R×8 / R+8

6R + 48 = 8R

48 = 2R

R = **24ohm**

R = 24ohm

we know that **resistance **(R)=rho×l/A

Therefore,

rho×l/A = 24

6×10⁻⁵/100 × l/6×10⁻³ = 24

10⁻⁴ × l = 24

l = **24×10⁴cm**

So, the resistance of the wire will be R=24ohm and the length of the **wire **will be 24×10⁴cm.

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A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle α (Figure 1). Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is μr.

What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.

Express your answer in terms of m , α , v0 , L , g , β and μr .

((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ) is the **distance **that the truck moves up the ramp before coming to a halt.

Let the distance the truck moves up the ramp be by x.

The** kinetic energy** of the truck on an icy road is given by,

K1 = (1÷2)mv²

The potential energy of the truck on an icy road is given by,

U1 = mgLSinα

The kinetic energy of the truck on the tuck ramp is given by,

K2 = 0

The **potential Energy** of the truck-on-truck ramp is given by,

U2 = mgxSinβ

Work done is given by,

W(others) = -µ×mg×cosФ

Hence, by using the work-energy theorem,

W(others) = (K2 + U2)(K1 + U1)

Therefore, by putting the values we get,

((1÷2)mv² + mgLSinα)(0 + mgxSinβ) = -µ×mg×cosФ

x = (K1 + mgLSinα) ÷ (mg(Sinβ + µcosβ))

x = ((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ)

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1) speed of bob?2)Show calculation for the speed of the bob:(write formula first, then insert value with unit, then find the result with unit)

Speed of bob:

[tex]\begin{gathered} \omega=2\pi f \\ _{\text{ }}where\colon \\ f=\frac{1}{T} \\ \omega=\frac{2\pi}{T} \end{gathered}[/tex]so:

[tex]\begin{gathered} \omega=\frac{2\pi}{1.293} \\ \omega=4.83\frac{rad}{s} \end{gathered}[/tex][tex]\begin{gathered} v=\omega\cdot r \\ so\colon \\ v=4.83\cdot0.22 \\ v=1.06\frac{m}{s} \end{gathered}[/tex]

I don't know what I'm doing wrong I just can't seem to get the answers right. Help please.

The **current **and **potential difference **in the lumped element model of **electrical circuits** are dealt with by **Kirchhoff's circuit laws**, which are two equalities. German scientist Gustav Kirchhoff published the first account of them in 1845. This was done before James Clerk Maxwell and generalized Georg Ohm's work.

Because the sum of the currents entering and leaving the junction is equal, **Kirchhoff's first law** is based on the **conservation of charge**. The algebraic sum of potential drops in a closed circuit must equal zero, according to **Kirchhoff's second law**. Therefore, it is based on **energy conservation.**

A stream of charged particles—such as **electrons **or ions—moving through a **conductor **for electricity or into empty space is known as an **electric current**. It is determined by measuring the net rate of **electric charge **flow through a surface or into a control volume.

The amount of **resistance **in an electrical circuit represents the resistance to current flow. The Greek letter **omega **(**Ω**), which stands for **resistance**, represents **ohms**.

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At room temperature, most metals are

A. Solids

B. Liquids

C. Gases

D. Plasma

**Answer:**

A. Solids

**Explanation:**

only Mercury is liquid soo .... solid

An Earth satellite moves in circular orbit 602 km above the earths surface with a period of 96.53 min. What are the speed and the magnitude of the centripetal acceleration of the satellite.

The **speed **and the** magnitude **of the **centripetal acceleratio**n of the satellite are 7.568 km/s and 8.205 x 10^-3 m / s^2

Here this problem we are dealing with the** speed** and c**entripetal acceleration** where the speed of the satellite is the speed required to attain adjustment between **gravity's pull** on the **satellite** and the **inertia **of the satellite's motion whereas centripetal acceleration is referred to as the **acceleration** of a body navigating a circular way.Since we are given the orbital radius which is 602 km and the period which is 96.53 min.

The** radius** of the satellite's orbit, r = earth radius + orbit radius

=>r =6378 km + 602 km = 6980 km

Since the formula for the orbital** circumference** is :

**2πR **

=>2 • 3.14 •6980 = 43,834.4 km

Now divide the circumference by the time period we get the orbital speed which is,

43,834.4 km / (96.53 min • 60) = 7.568 km/s

the formula we are referring for calculating the centripetal acceleration is:

**a = v²/ r** ,

=> a = (7.568 ^2) / 6980 = 8.205 x 10^-3 m / s^2

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thunder can make objects inside a room vibrate explain what causes the objects to vibrate

**Thunder **can make objects inside a room vibrate by which our house columns in **underground **that causes the objects to **vibrate**.

**Thunder **is caused by the rapid expansion of **air **around the path of **lightning**. It only takes a few millionths of a second for lightning to burst through the air from a cloud to a nearby tree or roof. The loud thunder that follows a lightning strike is generally said to come from the lightning itself. However, the **rumbling **and roaring heard during** thunderstorms **is actually due to the rapid expansion of air around the lightning.

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Provide an example of a situation that can be explained by each of Newton's Three Laws of Motion.

**ANSWER and EXPLANATION**

First, let us state Newton's three laws of motion.

The first law of motion states that:

The second law of motion states that:

The third law of motion states that:

Let us take an example of a ball that is stationary on a frictionless floor. As long as no force acts on the ball, it will continue to remain at rest but once a person kicks the ball, it begins to move and stay in motion (since the floor is frictionless, hence, no resistive force).

That is an application of the first law of motion.

The acceleration of this ball depends on the force with which it was kicked and the mass of the ball, as shown in the force equation:

[tex]F=ma[/tex]where m = mass, a = acceleration

That is an application of the second law of motion.

Assuming that this ball then comes in contact with a wall and stops moving. The force that the ball exerts on the wall is equal to the force that the wall exerts on the ball, but both forces act opposite one another.

That is an application of the third law of motion.

a 10 kg block of ice slides a ramp 20 m long. inclined at 10 degrees to the horizontal, if the ramp frictionless what is the acceleration of the ice

The **acceleration **of the **ice **is obtained as 0.97 m/s².

We have to recall that the **acceleration **has to do with the rate at which the **velocity **is change per unit time. Now, we know that from the parameters that have been given in the question that is before us here;

μmgcos(θ) = ma

μgcos(θ) = a

Hence;

μ = coefficient of friction

g = acceleration due to gravity

θ = the angle of inclination of the plane

We can now find the **acceleration **by the use of the formula as shown above hence we have;

a = (0.1)(9.8) cos(10)

a = 0.97 m/s²

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When you drop a 0.36 kg apple, Earth ex- erts a force on it that accelerates it at 9.8 m/s² toward the earth's surface. According to New- ton's third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 x 1024 kg, what is the magnitude of the earth's acceleration toward the apple? Answer in units of m/s². Answer in units of m/s^2

The** magnitude** of the** earth's acceleration** toward the apple is** 5.7x10^-25 m/s^2**

Given:

Mass if the apple **(m)**= **0.36**

Mass of the earth **(M)**=**5.98x10^24**

Gravitational acceleration **(g)**= **9.8**

we know that the formula is:

**Ma=mg**

So, according tot the formula

a=(m/M)g

=(0.36/5.98x10^24)9.8

=**5.7x10^-25 m/s^2**

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A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. Howfar will the spring stretch if the mass is motionless?

[tex]\begin{gathered} \text{Spring constant = K= 8N/m} \\ \text{mass}=\text{ m = }2\operatorname{kg} \\ \text{gravity = }g=9.81m/s^2 \\ stretch\text{ing = x =?} \\ \text{spring force = Kx} \\ Kx=mg \\ \text{Solving x} \\ x=\frac{mg}{K} \\ x=\frac{(2kg)(9.81m/s^2)}{\text{8N/m}} \\ x=2.4525m\approx2.45m \\ \text{The stretching is 2.45m} \end{gathered}[/tex]

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