module 2 question 17

Figure 4.25

(a) What is the mass of the child and basket if a scale reading of 100 N is observed?
kg
(b) What is the tension T in the cord attaching the child to the scale?
N
(c) What is the tension T' in the cord attaching the scale to the ceiling, if the scale has a mass of 0.300 kg?
N
(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. (Do this on paper. Your instructor may ask you to turn in this work.)

Module 2 Question 17Figure 4.25(a) What Is The Mass Of The Child And Basket If A Scale Reading Of 100

Answers

Answer 1

a ) The mass of the child and basket = 5.61 kg

b ) The tension T in the cord attaching the child to the scale = 55 N

c ) The tension T' in the cord attaching the scale to the ceiling = 57.94 N

a ) The mass of the child and basket,

[tex]W_{b}[/tex] = 55 N

g = 9.8 m / s²

W = m g

[tex]m_{b}[/tex] = [tex]W_{b}[/tex]  / g

[tex]m_{b}[/tex] = 55 / 9.8

[tex]m_{b}[/tex] = 5.61 kg

b ) In the cord attaching the child to the scale,

∑ [tex]F_{y}[/tex] = 0

T - W = 0

T = W

T = 55 N

c ) In the cord attaching the scale to the ceiling,

[tex]m_{s}[/tex] = 0.3 kg

[tex]W_{s}[/tex] = [tex]m_{s}[/tex] * g

[tex]W_{s}[/tex] = 0.3 * 9.8

[tex]W_{s}[/tex] = 2.94 kg

∑ [tex]F_{y}[/tex] = 0

T - [tex]W_{s}[/tex] - [tex]W_{b}[/tex] = 0

T = 2.94 + 55

T = 57.94 N

Therefore,

a ) The mass of the child and basket = 5.61 kg

b ) The tension T in the cord attaching the child to the scale = 55 N

c ) The tension T' in the cord attaching the scale to the ceiling = 57.94 N

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Related Questions

what is friction with its SI unit

Answers

When two surface attach each other a force is produced called friction. It's SI unit is newton

An object travels 5m/s in 2 seconds. How far did it travel?​

Answers

Answer:

3

Explanation:

because half I think is the answer

Answer:

10 m

Explanation:

5  m/s   *   2 s = 10 meters     ( see how the  ' s '  cancels out?)

A woman is sitting at a bus stop when an ambulance with a siren wailing at 440 Hz approaches at 31.5 m/s (about 70 mph). Assume the speed of sound to be 343 m/s.
(a)
What frequency (in Hz) does the woman hear?

Answers

As the source moves toward the stationary observer, the frequency the woman will hear the sound from the siren will be 484.5 Hz

What is Doppler Effect ?

Doppler effect is the frequency change due to the relative motion between a sound or light and an observer.

Given that a woman is sitting at a bus stop when an ambulance with a siren wailing at 440 Hz approaches at 31.5 m/s. If the speed of sound is to be 343 m/s, the frequency the woman hear the sound can be calculated by using the formula

Fo = FsC/(C - V)

Where

Fo = Frequency of the observerFs = frequency of the siren = 440 HzV = speed of the car = 31.5 m/sC = speed of the sound = 343 m/s

Substitute all the parameters into the formula

Fo = (440 × 343) / (343 - 31.5)

Fo = 150920 / 311.5

Fo = 484.5 Hz

Therefore, the woman will hear the sound from the siren at the frequency of 484.5 Hz

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A 44-kg child steps onto a scale and reads 430N.

Answers

A 44 kg child steps onto a scale and reads 430 N so the magnitude of the normal force acting on the child will be 430 N.

What is Normal Force?

The normal force is a common force that is experienced when an object is placed on a surface and pushed against it. For instance, the normal force prevents a book from falling through a table when it is placed on one. The book is being pulled downward by gravity, but because it isn't truly falling, something must be pushing it upward. The normal force is what we refer to as. Normal refers to something that is perpendicular to the surface.

According to the question,

Normal Force=Weight of the person

= 430 N

So, the Magnitude of the normal force acting on the child will be 430 N.

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explain why water comes out more slowly from a tap on the first floor than from a tap on the ground floor​

Answers

Answer:

It is because of the pressure exerted by water. As the depth of water increases, the pressure also increases.

Explanation:

What is the frequency of a wave if the velocity is 449.4 m/s and the wavelength is 8.4 meters?

Answers

We will have the following:

First, we remember:

[tex]\lambda=\frac{v}{f}[/tex]

Then we will have:

[tex]\left(8.4m\right?)=\frac{\left(449.4m/s\right)}{f}\Rightarrow f=\frac{\left(449.4m/s\right)}{\left(8.4m\right)}[/tex][tex]\Rightarrow f=53.5Hz[/tex]

So, the frequency of the wave is 53.5 Hz.

The components of a 15 meters per second velocity at an angle of 60 above the horizontal are

Answers

The components of a 15 meters per second velocity at an angle of 60 above the horizontal are 13.05 m / s and 0.435 m / s

Resolving the velocity into its horizontal and vertical components, it makes a right angled triangle.

Hypotenuse = V = 15 m / s

θ = 60°

Opposite side = components = [tex]V_{y}[/tex]

Adjacent side = Horizontal component = [tex]V_{x}[/tex]

In a right angled triangle,

sin θ = Opposite side / Hypotenuse

cos θ = Adjacent side / Hypotenuse

sin 60° = [tex]V_{y}[/tex] / V

[tex]V_{y}[/tex] = 15 * 0.87

[tex]V_{y}[/tex] = 13.05 m / s

cos 60° = [tex]V_{x}[/tex] / V

[tex]V_{x}[/tex] = 0.5 * 0.87

[tex]V_{x}[/tex] = 0.435 m / s

Therefore,

The horizontal component, [tex]V_{x}[/tex] = 0.435 m / sThe vertical component, [tex]V_{y}[/tex] = 13.05 m / s.

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A steel cable carries a 3,000 lbs load which is descending with a speed of 12 ft/s. It slows
down uniformly and comes to a halt in 9ft. Compute the tension in the cable during this motion.

Answers

The tension in the cable during this motion will be 3750 lbs , when a steel cable carries a 3000 lbs load which is descending with a speed of 12 ft/s.

What is tension and what is speed , and how the tension is calculated out to be 3750 lbs?Tension is specifically the pulling force , when we either pull a rope or an object tension is in action.Here in this question, given 3000 lbs load and speed of 12 ft/s and speed is uniform .The halt is given out to be of 9 ft, for calculating the tension in the cable we will first change units, m = 3000 lbs= 1362 kg , final velocity = v= 12 ft/s = 3.66 m/s, distance= d= 9 ft = 2.745 m.Using the kinematics equation , v^2 = u^2 + 2as , 0^2 - 3.66 ^2 = 2a (2.745), a = -2.44 m/s^2.The tension is calculated by using the equation , mg-T = ma , which equals 3750 lbs.

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A uniform plank of weight 200N and length 6m is supported by a rope if breaking tension in the rope is 400N How far can a boy of weight 400N walk towards the support

Answers

The distance the boy can walk to the support = Lm

Board force = Fb = 200 N

Rope breaking stress = Fr = 400 N

Distance to rope = Lr = 6 m

Boy = Fm = 400N

Distance to the center of gravity of the board = Lb = 3m

According to the balance rule:

All moments and forces on the board are 0,

= (Fr)(Lr) - (Fm)(Lm) - (Fb)(Lb) = 0

= (400)(6) - (400×Lm) - (200)(3) = 0

= 2400 - 600 = 400 × Lm

= 400× Lm = 1800

= Lm = 1800

400

= Lm = 4.5m.

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The maximum stress that a material can withstand before it breaks is called breaking stress or breaking stress. Also. Breaking stress is the maximum force that can be applied to a cross-sectional area of ​​material so that it cannot withstand additional stress before it fails.

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Riders on the power tower are launched skyward with an acceleration of 4g, after which they experience a period of free fall. what is a 60 kg rider's apparent weight during the launch?

Answers

The rider's apparent weight during the launch is 2940 N.

When the rider is rest, the forces acting on the skyward (y-direction) are mg towards earth and F_N opposite to mg. Total force will be,

F = F_N - mg

The mass of the rider is 60 kg. When riders are launched skyward with an acceleration of 4g, the force

F = ma = m*4g

F_N - mg = m*4g

F_N - 60*9.8 = 60*4*9.8

F_N - 588 = 2352

F_N = 2352 + 588 = 2940 N

The rider's apparent weight during the launch is 2940 N.

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An intergalactic spaceship arrives at a distant planet that rotates on its axis with a period of [tex]T=26 hours[/tex]. The mass of the planet is [tex]M=4.9*10^{25} kg[/tex]. The spaceship enters a circular orbit with an orbital period that is equal to the planet's period for the rotation about its axis, T.
A) Enter an expression for the radius of the spaceship's orbit.
(I tried [tex]R=(T^{2} GM/4pi^{2} )^{1/3}[/tex] but its says incorrect and that's all I've been able to find)
B) Calculate the orbital radius in meters.

Answers

The orbital period could be obtained from the question as 8.98 * 10^7 m.

What is the orbital period?

We know that the solar system is composed of the sun and the planets. The planets are known to move around the sun in concentric circles. Following the heliocentric model of the solar system, the sun is at the center of the solar system at all times and all the other planets tend to move round the sun at a good distance that is appropriate for each.

Now we have that;

T = √4π^2r^3/GM

T = period of the orbit

r = radius of the orbit

G = gravitational constant

M = mass of the planet

Hence, we have;

R = ∛T^2GM/4π^2

R =∛ (26 * 60 * 60)^2 * 6.67 * 10^-11 * 4.9 * 10^25/ 4 * (3.142)^2

R = ∛2.86 * 10^25/39.49

R = 8.98 * 10^7 m

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An object is launched at a velocity of 28 m/s in a direction making an angle of 23° upward
with the horizontal.
What is the magnitude of the velocity when it hits the ground?

Answers

Answer:

Explanation:

Angle:

β = - 23°

( From the symmetry condition)

How will the electrical force between two charged particles change if thedistance between them is increased by 4 times the amount?

Answers

Given that the distance between the two charged particles is increased by 4 times the amout, let's determine how the electrical force between them will change.

The electrical force between two charged particles is determined by the formula:

[tex]F=k\frac{q_1q_2}{r^2}[/tex]

Where:

k is the constant

q1 and q2 are the two charged particles

r is the distance between the particles

From the formula, we can see that the electrical force between them is inversely proportional to the the square of the distance between the charged particles.

Thus, in order to increase the electrical force, the distance must be decreased.

If the electrical force is changed by a factor of

module 2 question 15

What net external force is exerted on a 1000-kg artillery shell fired from a battleship if the shell is accelerated at 2.80 ✕ 104 m/s2? (Enter the magnitude.)
N

What is the magnitude of the force exerted on the ship by the artillery shell?
N

Answers

The net external force exerted on a 1000-kg artillery shell accelerated at 2.80 ✕ [tex]10^{4}[/tex] m / s² fired from a battleship is 2.8 * [tex]10^{7}[/tex] N. The magnitude of the force exerted on the ship by the artillery shell is - 2.8 * [tex]10^{7}[/tex] N.

According to Newton's second law of motion,

F = m a

F = Force

m = Mass

a = Acceleration

m = 1000 kg

a = 2.80 ✕ [tex]10^{4}[/tex] m / s²

F = 1000 * 2.80 ✕ [tex]10^{4}[/tex]

F = 2.8 * [tex]10^{7}[/tex] N

This is the net external force exerted by the shell on the battleship.

According to Newton's third law of motion,

[tex]F_{BS}[/tex] = - [tex]F_{SB}[/tex]

[tex]F_{BS}[/tex] = Force exerted by the battleship on the shell

[tex]F_{SB}[/tex] = Force exerted by the shell on the battleship

[tex]F_{BS}[/tex] = 2.8 * [tex]10^{7}[/tex] N

[tex]F_{SB}[/tex] = - [tex]F_{BS}[/tex]

[tex]F_{SB}[/tex] = - 2.8 * [tex]10^{7}[/tex] N

Therefore,

The net external force exerted on the artillery shell = 2.8 * [tex]10^{7}[/tex] NThe magnitude of the force exerted on the ship by the artillery shell = - 2.8 * [tex]10^{7}[/tex] N

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Calculate the orbital radius of a geostationary satellite orbiting Saturn, where a day is 10.6 hrs? (MS = 5.7x1026 kg, rS = 5.4x107 m)

Answers

In order to find the orbital radius, we need to use Kepler's third law's equation

The formula states:

T^2/R^3 = 4pi^2/G(m)

Since the satellite is geostationary, weknow that the period of the satelitte is going to be 10.6 hours

Plugging in everything we know already

(10.6)^2/R^3 = 4pi^2/G(5.7*10^26)

(10.6)^2/4pi^2/G(5.7*10^26) = R^3

solve for R^3 = 1.08*10^17

R = 476523 m from the surface of Saturn

5.4*10^7 + 476523 = 54476523 meters for the orbital radius

Which one is the answer

Answers

The graph that shows the relationship between the speed and the kinetic energy is option B.

What is the relationship between kinetic energy and speed?

Recall that the kinetic energy is referred to as the speed that a body possess by virtue of its motion. Thus an object that is in motion is said to posses the kinetic energy.  The kinetic energy is the energy of an object that has a velocity

The kinetic energy is linearly related to the speed of the object. This implies that the speed is directly proportional to the kinetic energy of the object. If there is a direct relationship, it the follows that the graph of the speed against the kinetic energy of the object ought to be a straight line graph.

As such, the relationship between the kinetic energy and the speed of the object can be seen from the image in option B.

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pls help, it's due tomorrow!

An electron travels 2.68 m in 6.11 × 10−8 s. How fast does it travel in cm/s? Answer in units of cm/s.

Answers

The speed (in cm/s) of the electron, given that it travels 2.68 m in 6.11×10⁻⁸ s is 4.39×10⁵ cm/s

How to determine the speed

We'll begin by defining speed. This is given below.

Speed is defined as the distance travelled per unit.

Speed = distance / time

With the above formula, we can obtain the speed of the electron. Details below

The following data were obtained from the question give above:

Distance = 2.68 m = 2.68 / 100 = 0.0268 cmTime = 6.11×10⁻⁸ sSpeed =?

Speed = distance / time

Speed = 0.0268 cm / 6.11×10⁻⁸ s

Speed = 4.39×10 cm/s

Therefore, the speed is 4.39×10⁵ cm/s

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Compute the density in g/cm3 of a piece of metal that has a mass of 0.350 kg and a volume of 62 cm3.answer in g/cm3

Answers

We will haev the following:

[tex]\begin{gathered} \rho=\frac{350g}{62cm^3}\Rightarrow\rho=\frac{175}{31}g/cm^3 \\ \\ \Rightarrow\rho\approx5.6g/cm^3 \end{gathered}[/tex]

So, the density of the material is 175/31 g/cm^; that is approximately 5.6 g/cm^3.

help!!!!!!
Why are the accelerations due to gravitational force on the moon and the Earth different? Do you think you could shield a gravitational field in a vacuum?

Answers

The accelerations due to gravitational force on the moon and the Earth different due to difference in mass of its body and its radius. No, gravitational field cannot be shielded in a vacuum.

Acceleration due to gravity = Local gravitational field strength = g

g = G M / r²

G = Gravitational constant

M = Mass of the body

r = Radius of the body

G = 6.67 * [tex]10^{-11}[/tex] N m²

From the equation,

g ∝ M

g ∝ 1 / r²

Since the Earth is much bigger and heavier than moon, the acceleration due to gravity is much lesser in moon in comparison with Earth.

Therefore, the accelerations due to gravitational force on the moon and the Earth different due to difference in mass of its body and its radius. No, gravitational field cannot be shielded in a vacuum.

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Jerry is unable to pick up the picnic
basket, then he eats spinach and
becomes stronger. Now he can easily lift
up the basket.

This is an example of F=ma
because.

Answers

This is an example of F = ma because the net force applied is directly proportional to the weight (mass) of the basket.

What is Newton's Second Law of Motion?

Newton's Second Law of Motion states that the acceleration of an object is inversely proportional to its mass and directly proportional to the net force acting on the physical object.

Mathematically, Newton's Second Law of Motion is given by this expression;

F = ma

Where:

F represents the net force.m represents the mass of a physical object.a represents the acceleration of a physical object.

In conclusion, the above scenario is an example of the demonstration of Newton's Second Law of Motion.

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A young researcher wants to test how well extremophiles, organisms that thrive in extreme conditions, live in different environments. She will do this by putting extremophiles of different species into different solvents
Because of limited lab space, she can run only one sample at a time, and each sample must remain undisturbed for 7 days. To reduce the total time needed to process all samples, the researcher decides to use two acidic
solutions along with a control. The information below includes the extremophiles and the solvents the researcher will use in her tests.
Solvents
Species of
Extremophiles
T. prosperus
A. aceti
Based on the tables, how many days will all of the researcher's experiments take?
You may use the calculator.
OA. 7
OB. 21
OC. 49
OD. 84
A brierleyi
H. acidophilum
Name
hydrochloric acid
ascorbic acid
water
PH
1.3
4.2
7.0

Answers

It will take exactly 84 days with all the researcher's experiments , even after reducing the total time needed to process all samples.

What is an extremophile and how to take the samples and notice their living condition?As is explained in the question as well an extremophile is the kind of organism can better thrive or live in harsh or extreme conditions.Some examples of extremophilic species are polar bears living in extreme cold regions .Here in this question the researcher has taken two samples for reducing the time of results, in which one extremophile takes 7 days to grow.Taking two samples would count to 14 days for one extremophile , and the researcher in the question took 4 samples and three solvents.In the way total of 84 days would be required by the researchers to obtain the result of the change in the living environment of extremophilic species.

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The smooth ER differs from the rough ER in that:
A. The rough ER contains no ribosomes, the smooth ER contains ribosomes.
B. The smooth ER produces carbohydrates and lipids, rough ER produces proteins.
C. The smooth ER produces proteins, rough ER produces lipids.
D. The rough ER is the site of mRNA transcription, smooth ER is the site of protein translation.

Answers

The answer should be B

⚠️Please help me!!!⚠️Vector V is 448 m long in a 224 direction. vector W is 336 m long in a 75.9 direction. Find the direction of their vector sum.

Answers

The magnitude of the vector sum of the two vectors V and W is determined as 240.84 m.

Resolution of the vectors in x and y axis

The given vector can be resolved into x and y axis as follows;

Vx = Vcosθ

Vx = 448cos(224)

Vx = -322.26

Vy = Vsinθ

Vy = 448 x sin224

Vy = -311.2

Wx = 336 x cos(75.9)

Wx = 81.86

Wy = 336 x sin75.9

Wy = 325.88

∑X = -322.26 + 81.86 = -240.4

∑Y = -311.2 + 325.88 = 14.68

Resultant vector

R = √[(-240.4)² + (14.68)²]

R = 240.84 m

What is vector?

In physics, a vector is a quantity with both magnitude and direction. It is often represented by an arrow whose length is proportional to the magnitude of the quantity and whose direction is the same as that of the quantity. A vector does not have position, while having magnitude and direction. In other words, a vector's shape is unaltered if it is shifted parallel to itself as long as its length is unaltered.

Ordinary quantities with a magnitude but no direction are referred to as scalars in contrast to vectors. For instance, while speed, time, and mass are scalars, displacement, velocity, and acceleration are vector values.

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As shown in the figure, a square insulating slab 5.0 mm thick measures 2.0 m ×
2.0 m. It has a charge of 8.0 × 10^-11 C distributed uniformly throughout its volume. Use Gauss’s law to determine the electric field at point P, which is located within the slab, 1.0 mm from the top face.
Assume P is far from the (front, back, left, and right) sides of the slab. (ε0 = 8.85 × 10^-12 C2/N ∙ m2)

Answers

The electric field at point P  is mathematically given as

E=-6.69n-m^2/c

This is further explained below.

What is the electric field?

Generally, the equation for magnetic Flux is mathematically given as

[tex]Flux $=\mathrm{E}.2 \mathrm{~A}\\=\mathrm{Q}_{\text { enc }} / \varepsilon_0$[/tex]

[tex]=\mathrm{E} 2 \mathrm{~A}\\\\=\mathrm{p} * \mathrm{Vol} / \varepsilon_0$[/tex]

[tex]=E 2\left(d x^* d y\right)\\\\=p^*\left(d x^* d y^* 2^* d z\right) / \varepsilon O$[/tex]

E=p^* d z /

Taking into account the fact that Let the permittivity of the slab is approximate:

[tex]$\varepsilon O=8.85^* 10^{\wedge}-12 \mathrm{~F} / \mathrm{m}$[/tex]

Distance from the origin to the point in question:

dz=0.0015m

Charge distribution over the slab is uniform:

[tex]&\mathrm{p}=\mathrm{Q} / \mathrm{vol}=\left(-7.9 * 10^{\wedge}-11\right) \mathrm{C} /\left(0.005^* 2 * 2\right) \mathrm{m}^{\wedge} 3 \\[/tex]

p=-3.95^* 10^-9

[tex]\mathrm{E}=\mathrm{p}^* \mathrm{dz} / \varepsilon 0\\\\=\left(3.95^* 10^{\wedge}-9 * .0015\right) / 8.85^* 10^{\wedge}^{-11} \\[/tex]

E=-6.69n-m^2/c

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When light falls on a soap bubble or on the oily film spread on a water puddle, a spectrum of colors is observed. How is this spectrum of colors formed?1) separation of white light by a prism2) absorption of colors in a pigment present3) constructive and destructive interference of light waves4) reflection of light by soap bubble or oil film

Answers

When the light falls on a soap bubble spread on a water puddle, and a spectrum of color is observed due to constructive and destructive of light waves which means option (3) is correct. Light waves interfere into each other to produce a spectrum which consists of bright and dark fringes.

PLEASE HELP!

1. Terry kicks a soccer ball that is sitting motionless on the field. What is the best description of the energy transfer? (1 point)

A. Kinetic energy is transferred from the soccer ball to Terry's foot

B. Kinetic Energy is Transferred from Terry’s foot to the soccer ball.

C. Kinetic Energy in Terrys Foot is changed to potential energy in the Soccer ball

D. Potential energy is transferred from Terry’s foot to the Soccer ball

2. Which of the Following scenarios best shows the transfer of kinetic energy?

A. a Baseball bat strikes a baseball

B. A baseball player holds a baseball at their side

C. a baseball sits on a bench

D. a baseball flies through the air

3.

A bowling ball is traveling at 7 6 meters per second when it hits a pin. If the bowling ball has a mass of 6 kilograms, how much kinetic energy does it have when it hits the pin?

KB = m²

A. 173.28 J

B. 228 J

C. 346.65 J

D. 693 12 J

4. What happens when a tennis racket hits a ball?

A. Kinetic energy is created.

B. Potential energy in the racket is transferred to the ball.

C.Kinetic energy is transferred from the ball to the racket.

D.Kinetic energy is transferred from the racket to the ball.

Answers

Answer:

1. B. Kinetic energy is transferred from Terry's foot to the soccer ball.

2. A. A baseball bat strikes a baseball.

3. A. 173.28J

4. D. Kinetic energy is transferred from the racket to the ball.

Explanation:

I took the quiz and got 100%

Answer:

1. Kinetic energy is transferred from Terry's foot to the soccer ball.

2. A baseball  bat stricks a baseball

3. 173.28

4. kinetic energy is transferred from the racket to the ball

5. 1,102.5 J

remember to keep the decimal on the last one

Explanation: They added another question so I decided to give the answer to those who had it.

We used a “dot” to represent the Sun’s location in the picture. Is this dot too small, too large or just the right size to represent the size of the Sun on the picture?

Answers

The dot used to represent the Sun’s location in the picture is too large to represent the size of the Sun on the picture.

1 cm = 10000 light-years

1 mm = 1000 light-years

Dot = 0.5 to 1 mm

Diameter of the Sun = 1.5 * [tex]10^{-7}[/tex] light-years = 1.5 * [tex]10^{-10}[/tex] mm

On comparing, the Sun should not be visible to a human eye in that picture. Even using an Electron microscope, we cannot spot the Sun according to the given scale.

A light-year is a measurement of distance. It is the distance travelled by light in one year.

Light speed = 300000 km / s = 9.46 * [tex]10^{12}[/tex] km / yr

1 light year = 9.46 * [tex]10^{12}[/tex] km

Diameter of sun = 1.39 * [tex]10^{6}[/tex] km

Therefore, the dot used to represent the Sun’s location in the picture is too large to represent the size of the Sun on the picture.

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1.00 kg of ice at -24.0°C is placed in contact with a 1.00 kg block of a metal at 5.00°C. They come to equilibrium at -8.88°C. What is the specific heat of the metal?

Ice: c = 2000 J/(kg*C)

(Unit = J/(kg*C))

Answers

The specific heat of the metal is  2178.7 J/kg⁰C.

What is specific heat capacity?

The specific heat capacity of a substance is the quantity of heat needed to raise a unit mass of a substance by 1 kelvin.

The specific heat capacity of the metal is calculated by applying the principle of conservation of energy.

Heat lost by the metal = Heat gained by the ice

m₁c₁(T₁ -T) = m₂c₂(T - T₂)

where;

m₁ is mass of the metalm₂ is mass of the icec₁ is specific heat capacity of the metalc₂ is specific heat capacity of the iceT is the equilibrium temperatureT₁ is the initial temperature of the metalT₂ is the initial temperature of the water

(1)(c₁)(5 - -8.88) = (1)(2000)(-8.88 - - 24)

(1)(c₁)(5 + 8.88) = (1)(2000)(-8.88 + 24)

13.88c₁ = 30,240

c₁ = 30,240/13.88

c₁ = 2178.7 J/kg⁰C

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Answer:

2180

Explanation:

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a crate is accelerated at .035 m/s for 28.0 s along a conveyor belt. If the crates initial speed is 0.76 m/s what is it's final speed?

Answers

The final speed of the crate with an initial speed of 0.76 m / s accelerated at 0.035 m/s for 28.0 s along a conveyor belt is 1.74 m / s

v = u + at

v = Final speed

u = Initial speed

a = Acceleration

t = Time

a = 0.035 m / s²

y = 28 s

u = 0.76 m / s

v = 0.76 + ( 0.035 * 28 )

v = 0.76 + 0.98

v = 1.74 m / s

The equation used to solve the problem is an equation of motion. An equation of motion describes the position of an object using acceleration, displacement, time, initial and final velocities.

Therefore, the final speed of the crate is 1.74 m / s

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6. What horizontal force is required to slide a 150 N crate with uniform motion across a
floor where the coefficient of kinetic friction is 0.20?

Answers

Horizontal force required to slide a 150 N crate with uniform motion across a floor where the coefficient of kinetic friction is 0.20 is calculated to be 30 N .

A force that is applied in parallel to the horizon is known as horizontal force. The force exerted on a body has two components, that is the horizontal component and a vertical component.

Two types of horizontal force are :

Static friction which is the maximum horizontal force that can be applied before an object starts to move. Kinetic friction which the horizontal force that acts in the opposite direction of the motion of an object.

Given , force = 150 N

Coefficient of kinetic friction, μ  = 0.20

Horizontal force, F = μmg

=0.20* 150

Horizontal force = 30 N

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