What is the answer to this question?

The forces which act upon the book are Normal, Gravity, Air resistance, and Friction. Therefore, the (A), (B), (D), and (E) options are correct.

The normal force is a contact force that surfaces exert to prevent objects from passing through each other. If two surfaces are not in direct contact, they are able to exert a normal force on each other. The book and the floor are exerting normal on each other.

Gravity is the force by which the earth draws objects toward its center. The force of gravity keeps everything on the surface of the earth. Therefore, the sliding book on the floor experiences the force of gravity.

Air resistance is a force that arises due to air. The force acts in the opposite direction to the motion of an object through the air. As the book slides across the floor, it experiences resistance due to air.

Friction is the resistance caused by the surfaces that are in contact when they slide over each other. Friction arises due to the irregularities on the two surfaces. The book and floor both have irregularities on their surfaces which is caused friction between them.

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Your question is incomplete, but most probably the complete question was,

Consider the situation a book a sliding to a stop while moving across the classroom floor. Of the forces listed identify which act upon the book:

A)Normal,

B) Gravity

C)Applied

D) Air resistance,

E) Friction

F) Tension

Answer:

The average net force exerted on the car and riders by the magnets is 33751.8 newtons.

Explanation:

Let assume that car and its riders accelerate at constant rate, such that acceleration ([tex]a[/tex]), measured in meters per square second, can be found by using the following kinematic equation:

[tex]a = \frac{v_{f}-v_{o}}{t}[/tex] (Eq. 1)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the car and its riders, measured in meters per second.

[tex]t[/tex] - Acceleration time, measured in seconds.

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v_{f} = 45\,\frac{m}{s}[/tex] and [tex]t = 6.8\,s[/tex], the average acceleration of the car is:

[tex]a = \frac{45\,\frac{m}{s}-0\,\frac{m}{s}}{6.8\,s}[/tex]

[tex]a = 6.618\,\frac{m}{s^{2}}[/tex]

By the Second Newton's Law, we find that average force exerted on the car and riders by the magnets ([tex]F[/tex]), measured in newtons, is:

[tex]F = m\cdot a[/tex] (Eq. 2)

Where [tex]m[/tex] is the mass of the car-riders system, measured in kilograms.

If we know that [tex]m = 5100\,kg[/tex] and [tex]a = 6.618\,\frac{m}{s^{2}}[/tex], the net force exerted on the car and riders is:

[tex]F = (5100\,kg)\cdot \left(6.618\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 33751.8\,N[/tex]

The average net force exerted on the car and riders by the magnets is 33751.8 newtons.

Answer:

Explanation:

We know that the relationship for the impulse and momentum is given as

Ft=mv

given data

mass m=0.04kg

t= 26.5ms= 26.5/1000= 0.0265seconds

velocity v= 41.3m/s

substituting in the expression we have

Ft=mv

make F subject of formula we have

F=mv/t

F=(0.04*41.3)/0.0265

F=1.652/0.0265

F=62.33N

Answer:

His de Broglie wavelength is 8.35×10⁻³⁷ m

The uncertainty in his position is 1.15 × 10⁻³⁵ m

Explanation:

First, Convert 227-lb to kg and convert the unit of the speed from mi/h to m/s.

To convert 227-lb to kg,

1-lb = 0.453592 kg

∴ 227-lb = 227 ×  0.453592 kg

227-lb = 102.97 kg

To convert 17.25 ± 0.10 mi/h to m/s

1 mi = 1609.34 m

and 1 h = 3600 s

Therefore,

17.25 mi/h = (17.25 ×1609.34)/3600 m/s = 7.71 m/s

and 0.10 mi/h =  (0.10 ×1609.34)/3600 m/s = 0.044704 m/s

Hence, the speed 17.25 ± 0.10 mi/h = 7.71 ± 0.044704 m/s

Now

(a) To determine the de Broglie wavelength,

De Broglie wavelength is given by

λ = h/mv

Where λ is the de Broglie wavelength

h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)

m is the mass

and v is the speed (velocity)

From the question

m = 102.97 kg

v = 7.71 m/s

Therefore,

λ = 6.626×10⁻³⁴ / (102.97×7.71)

λ = 8.35×10⁻³⁷ m

Hence, his de Broglie wavelength is 8.35×10⁻³⁷ m

(b) To calculate the uncertainty in his position

From

Δx = h/(4πmΔv)

Where Δx is the uncertainty in the position

h is Planck's constant (h = 6.626×10⁻³⁴ kgm²/s)

π is a constant ( π = 3.14)

m is the mass

Δv is the uncertainty in speed

From the question

m = 102.97 kg

Δv = 0.044704 m/s

Hence,

Δx = 6.626×10⁻³⁴  / (4×3.14×102.97×0.044704)

Δx = 1.15 × 10⁻³⁵ m

Hence, the uncertainty in his position is 1.15 × 10⁻³⁵ m.

Answer:

120 north west

Explanation:

Answer:

A. stress

Explanation:

Hope this helped. Have an amazing day/night!!

We have that for the Question "elastic limit is equal to" it can be said that

This speaks to a body retaining it shape after changes occur

From the question we are told

elastic limit is equal to

(a) stress

(b)strain

(c) young's modulus

(d)modulus of rigidity

Where

Where

Stress is the force acting on a material

and

strain

The effect of  force acting  on a material

Young's modulus

This speaks to a body retaining it shape after changes occur

Modulus of rigidity

m=sheer stress/shear strain

Now having that

Elastic limit is a point where an elastic body does not return to its original shape

The option chosen is'

Young's modulus

This speaks to a body retaining it shape after changes occur

Option C

​

Generally the equation for the   is mathematically given as

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Answer:

B. fast-twitch fibers can deliver a quick burst of power. Slow-twitch fibers can maintain a contraction for a longer time.

Answer:

the person above is right, just came for answers lol :')

Explanation:

Answer:

object B has a greater volume and by2 g/cm3

Answer: -0.10m/s

Explanation:

khan academy

Answer:

C. 25 N

Explanation:

w = Weight of box = 80 N

[tex]\mu_k[/tex] = Coefficient of kinetic friction = 0.25

[tex]\mu_s[/tex] = Coefficient of static friction = 0.5

F = Pulling force = 25 N

Static friction

[tex]f_s=\mu_s w\\\Rightarrow f_s=0.5\times 80=40\ \text{N}[/tex]

[tex]F<f_s[/tex]

The force of pulling is less than the static frictional force so this makes the friction force equal to the force of pulling.

So, the friction force on the box is 25 N.

Answer:

the law that the velocity of recession of distant galaxies from our own is proportional to their distance from us.

Explanation:

Hubble's law, also known as the Hubble–Lemaître law, is the observation in physical cosmology that galaxies are moving away from the Earth at speeds proportional to their distance. In other words, the farther they are the faster they are moving away from Earth.

Answer:

Anything that has the most kinetic has the least potential energy.

Explanation:

NEED BRAINLYEST PLZZ!!!!!!!!!!!!!!!

Answer:

0.013 cm/minute

Explanation:

We are given;

Volume rate; dV/dt = 800 cm³/min

Now, volume of a sphere is given by the formula;

V = (4/3)πr³

We want to find the rate at which the radius of the balloon is changing. This is represented by dr/dt.

Now, dr/dt will be gotten from the relation;

dr/dt = dV/dt ÷ dV/dr

Now, dV/dr = 3 × (4/3)πr²

dV/dr = 4πr²

Thus;

dr/dt = 800/(4πr²)

At r = 70 cm

dr/dt = 800/(4π × 70²)

dr/dt = 0.013 cm/minute

Answer:

less in magnitude and opposite in direction to their original velocities

Explanation:

The nature of collisions can be determined by the coefficient of restitution.

Coefficient of restitution is given by

[tex]e=\dfrac{\text{Kinetic energy after collision}}{\text{Kinetic energy before collision}}[/tex]

In the case described here the value of [tex]e[/tex] is in between 0 and 1.

This means that the kinetic energy after collision is less than the kinetic energy before collision.

If the kinetic energy after collision is less this means that the velocity is not balanced with the velocity before the collision however the direction of the masses are opposite Some amount of energy is dissipated as heat or through other forms of energy.

So, just after the collision, their velocities are less in magnitude and opposite in direction to their original velocities.

Answer:

Large igneous provinces are formed by flood basalt, which can flow tens to hundreds of kilometers.

Explanation:

Answer:

The ball would fall vertically for 112.45m

Explanation:

The vertical height of a horizontal throw is affected only by the speed of the throw and the acceleration due to gravity.

At this point, we may use this formula to determine the vertical height of the throw

vertical height = [tex]-v^2/2g[/tex]

The initial throwing speed has to be converted to m/s to ensure uniformity during the calculations. To do this we multiply by 1000 and divide by 3600

169.1km/hr = 46.972m/s

Maximum height = [tex]46.972^2 /2 \times 9.81=112.45[/tex] m

The pitch would make the ball fall vertically for 112.45m by the time it reached the home plate 18.2 m away

Answer:

20 m/s

Explanation:

velocity after 2 sec.  

v=40+(−10)t

=40−10∗2=20m/s

We have that for the Question "A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?" it can be said that the balls velocity after 2 seconds

From the question we are told

A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?

Generally the Newtons equation for the  Motion  is mathematically given as

[tex]v=u+at\\\\Therefore\\\\v=40+9.8*2.8\\\\v=67.44m/s\\\\[/tex]

Therefore

the balls velocity after 2 seconds

v=67.44m/s

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Answer:

Explanation:

The chunk of ice will fall with acceleration of 9.8 m /s²

initial velocity u = 0 ,

v² = u² + 2 gH

H = 343 m

g = 9.8 m /s²

v is final velocity

v² = 0 + 2 x 9.8 x 343

v² = 6723

v = 82 m /s approx .

The velocity of the the chunk of ice as it hits the ground is 82m/s.

Given the data in the question;

Before the chunk of ice to fell off the roof, its was initially at rest, so

Final velocity as the ice hits the ground; [tex]v = \ ?[/tex]

To determine the final velocity, we use the third equation of motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is final velocity, u is the initial velocity, s is distance or height of the building and a is acceleration due to gravity{ since the ice will be under gravity as it falls, ([tex]a = g = 9.8m/s^2[/tex]) }

We substitute our values into the equation

[tex]v^2 = [0^ 2] + [ 2\ *\ 9.8m/s^2\ *\ 343m]\\\\v^2 = 2\ *\ 9.8m/s^2\ *\ 343m\\\\v^2 = 6722.8m^2/s^2\\\\v = \sqrt{6722.8m^2/s^2}\\\\v = 81.99m/s\\\\v = 82m/s[/tex]

Therefore, the velocity of the the chunk of ice as it hits the ground is 82m/s.

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Answer:

The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].

Explanation:

After a careful reading of the statement, we notice that merry-go-round-child system is a system with no external force exerting on it, such that the Principle of Angular Momentum Conservation can applied to analyze the system:

[tex]m_{C}\cdot \left(\frac{v}{R} \right) = (I+m_{C}\cdot R^{2})\cdot \omega[/tex] (Eq. 1)

Where:

[tex]m_{C}[/tex] - Mass of the child, measured in kilograms.

[tex]v[/tex] - Initial speed of the child, measured in meters per second.

[tex]R[/tex] - Radius of the playground merry-go-round, measured in meters.

[tex]I[/tex] - Moment of inertia, measured in kilograms per square meter.

[tex]\omega[/tex] - Angular velocity of the merry-go-round-child system, measured in radians per second.

Then, we clear the final angular speed:

[tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex]

The angular velocity of the merry-go-round is [tex]\omega = \frac{m_{C}\cdot v}{R\cdot (I+m_{C}\cdot R^{2})}[/tex].

Answer:

Explanation:

The question is incomplete. Here is the complete question.

At a construction site, a crane lifts a steel beam with a mass of 330 kg.

i. What is the gravitational potential energy added to the steel beam when the beam reaches a height of 18.4 m? (1 point)

Gravitational potential energy GPE = mgh

m is the mass of the steel beam = 330kg

g is the acceleration due to gravity = 9.81m/s²

h is the height the beam reaches = 18.4m

Substitute

GPE = 330 * 9.81 * 18.4

GPE = 3237.3 * 18.4

GPE = 59,566.32Joules

Hence the gravitational potential energy is 59,566.32Joules

Answer:

The increase in gravitational potential energy is approximately 60,000 J.

Explanation:

GPE=mgh

m=330

g=9.8^2

h=18.4

GPE=330*9.81*18.4

GPE=59,566.32 J

Hope I could help! :)

Answer:

Explanation:

For the mass m1;

The sum of forces acting on the body is expressed according to Newton's second law as;

\sum Fx = m1ax

T - Ff = m1a .... 1

T is the tension

Ff is the frictional force acting on m1

For the mass m2:

\sumFy = m2a

W - T = m2a

m2g - T = m2a.... 2

W is the weight

g is the acceleration due to gravity

If the acceleration of the system is 0, the equation becomes;

From 2:

m2g - T = m2a.... 2

a = m2g-T/m2

From 1:

a = T-Ff/m1

Equate both accelerations

m2g-T/m2 = T-Ff/m1

Cross multiply

m1m2g - Tm1 = m2T-m2Ff

m1m2g+m2Ff = m2T+m1T

m1m2g+m2Ff = T(m2+m1)

T = m1m2g+m2Ff/m2+m1

Hence the tension in strings is expressed as

T = m1m2g+m2Ff/m2+m1

Explanation:

We are expected to solve for the total momentum before collision

the expression for momentum is given as

p = mv

p =momentum

m=mass

v=velocity

For small compact truck

m1=1000kg

v1=15m/s

For small truck

m1=2000kg

v1=10m/s

the total momentum can also be expressed as

Ptotal=m1v1+m2v2

Ptotal=1000*15+2000*10

Ptotal=15000+20000

Ptotal=35000kg•m/s

The direction is east because the momentum of the truck is more than the car

Answer:

Did he wash the shoes?-Were the shoes washed by him?

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

From the question we have

force = 0.5 × 50

We have the final answer as

Hope this helps you

We know, Current is given by :

[tex]I = \dfrac{EA}{\rho}[/tex]

Here, [tex]\rho[/tex] is resistivity and for iron.

[tex]\rho=9.71\times 10^{-8}\ \Omega\ m[/tex]

Putting [tex]\rho[/tex] in above equation, we get :

[tex]I = \dfrac{0.062\times \pi\times (2.3\times 10^{-3})^2}{4\times 9.71\times 10^{-8}}[/tex]

[tex]I=2.65\ A[/tex]

Therefore, the electron current is 2.65 A.

Hence, this is the required solution.

Answer:

Since velocity is considered a vector quantity and vectors show you in what direction it goes the object wont be moving forward but backwards because its a negative and negatives go to the left and that would be considered backwards.

Explanation:

Hope this helps