A student has to work the following problem: A block is being pulled along at constant speed on a horizontal surface a distance d by a rope supplying a force F at an angle of elevation q. The surface has a frictional force acting during this motion. How much work was done by friction during this motion? The student calculates the value to be –Fd sinq. How does this value compare to the correct value?
a. It is the correct value.
b. It is too high.
c. It is too low.
d. The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]

[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]

[tex]I = \frac{ E}{ R + r} \\[/tex]

[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]

Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have

[tex]➺ \: I \: (R + r) = E \times 1[/tex]

[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]

[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]

[tex]\boxed{ Hence\:proved. }[/tex]

[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]

It does not change the chemical composition of water.

Answer: D. F represents the activation energy

Explanation:

The activation energy is the energy required to get the reactants to begin reacting with one another such that products are created. This energy ranges from the minimum to the maximum energy required.

F is therefore the activation energy because it shows the range between the minimum energy it took for the reaction to start and the maximum energy that was required to continue the reaction.

Answer:

[tex]R1/R2=\frac{2}{3}[/tex]

Explanation:

From the question we are told that:

1st's Length [tex]l=L[/tex]

1st's radius [tex]r=r[/tex]

2nd's Length [tex]l=6L[/tex]

2nd's radius [tex]r=2r[/tex]

Generally the equation for Resistance R is mathematically given by

 [tex]R=\frac{\rho L}{\pi r^2}[/tex]

Therefore

 [tex]R_1=\frac{\rho L}{\pi r^2}[/tex]

And

 [tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]

Therefore

 [tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]

 [tex]R1/R2=\frac{2}{3}[/tex]

Answer:

Natae Si Jordan Kaya Sya Napaihe

Explanation:

haha

Answer:

Graph B express the magnetic relationship of magnetic flux and electronic flow

Answer:

increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.

Explanation:

Answer:

The correct answer will bs "2.41 m".

Explanation:

According to the question,

M = 50 g

or,

   = 0.050 kg

[tex]\Theta = 25^{\circ}[/tex]

k = 25.9 N/m

Δx = 0.200 m

Let the traveled distance be "x".

By using trigonometry, the height will be:

⇒ [tex]h = l Sin \Theta[/tex]

hence,

⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]

                                       [tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]

By putting the values, we get

             [tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]

                                              [tex]l=2.41 \ m[/tex]      

Answer:

Explanation:

a)

Below  the worker , the tension in cable  is pulling  the crate . Let the tension be T₁ .

weight of  crate is acting downwards .

Total weight  1510  N.

Net force acting on both = T₁ - 1510

Applying second law of Newton ,

T₁ - 1510 = 1510 / 9.8 x 0.62                [ 1510 / 9.8 = mass of  crate ]

T₁ - 1510 = 95.5

T₁  = 1605.5  N.

b )

Above the worker , the tension in cable  is pulling both the worker and the crate . Let the tension be T₂ .

weight of both worker and crate is acting downwards .

Total weight = 965 + 1510 = 2475 N.

Net force acting on both = T₂ - 2475

Applying second law of Newton ,

T₂ - 2475 = 2475 / 9.8 x 0.62  [ 2475 / 9.8 = mass of both worker and crate ]

T₂ - 2475 = 156.6

T₂  = 2631.6 N.

Answer:

P =40.69 atm

Explanation:

We need to find the approximate pressure at a depth of 400 m.

It can be calculated as follows :

P = Patm + ρgh

Put all the values,

[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]

So, the approximate pressure is equal to 40.69 atm.

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

Answer:

periodic

Explanation:

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

Answer:

Approximately [tex]0.97\; \rm N[/tex]. This force would point away from the center of the square (to the left at [tex]45^\circ[/tex] above the horizontal direction.)

Explanation:

Coulomb's constant: [tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].

By Coulomb's Law, the electrostatic force between two point charges [tex]q_1[/tex] and [tex]q_2[/tex] that are separated by [tex]r[/tex] (vacuum) would be:

[tex]\displaystyle F = \frac{k \cdot q_1 \cdot q_2}{r^2}[/tex].

Consider the charge on the top-left corner of this square.

Apply Coulomb's Law to find the electrostatic force between this charge and the charge on the lower-left corner.

Convert quantities to standard units:

[tex]q_1 = q_2 = 3 \times 10^{-6}\; \rm C[/tex].

[tex]r = 0.40\; \rm m[/tex].

[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(0.40\; \rm m)^{2}} \\ &\approx 0.506\; \rm N\end{aligned}[/tex].

As the two charges are of the same sign, the electrostatic force on each charge would point away from the other charge. Hence, for the charge on the top-left corner of the square, the electrostatic force from the charge below it would point upwards.

Similarly, the charge to the right of this charge would exert an electrostatic force with the same magnitude (approximately [tex]0.506\; \rm N[/tex]) that points leftwards.

For the charge to the lower-right of the top-left charge, [tex]r = \sqrt{2} \times 0.40\; \rm m[/tex]. Therefore:

[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(\sqrt{2} \times 0.40\; \rm m)^{2}} \\ &\approx 0.253 \; \rm N\end{aligned}[/tex].

This force would point to the top-left of the top-left charge, which is [tex]45^\circ[/tex] above the horizontal direction. Decompose this force into two components that are normal to one another:

Consider the net force on the top-left charge in two components:

Combine these two components to find the magnitude of the net force on this charge:

[tex]\begin{aligned}F &= \sqrt{{F_x}^{2} + {F_y}^{2}} \\ &\approx \sqrt{0.685^2 + 0.685^2 }\; \rm N \\ &\approx 0.97\; \rm N\end{aligned}[/tex].

This force would point to the top-left of this charge (also at [tex]45^\circ[/tex] above the horizontal direction, away from the center of the square) because its horizontal and vertical components have the same magnitude.

6.07 m

Explanation:

Given:

[tex]v_0=24.5\:\text{m/s}[/tex]

[tex]\theta_0 = 35.5°[/tex]

First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that

[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]

or

[tex]t = 1.29\:\text{s}[/tex]

To find the vertical height where the ball hit the wall, we use

[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]

[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]

[tex]\:\:\:\:=6.07\:\text{m}[/tex]

Answer:

The expected radius of the Earth is 3.883 meters.

Explanation:

The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:

[tex]U = K[/tex] (1)

Where:

[tex]U[/tex] - Gravitational potential energy, in joules.

[tex]K[/tex] - Translational kinetic energy, in joules.

Then, we expand the formula by definitions of potential and kinetic energy:

[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

Where:

[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.

[tex]M[/tex] - Mass of the Earth. in kilograms.

[tex]m[/tex] - Mass of the rocket, in kilograms.

[tex]r[/tex] - Radius of the Earth, in meters.

[tex]v[/tex] - Escape velocity, in meters per second.

Then, we derive an expression for the escape velocity by clearing it within (2):

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)

If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:

[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]

[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]

[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]

[tex]r = 3.883\,m[/tex]

The expected radius of the Earth is 3.883 meters.

Answer:

I DON'T UNDERSTAND

Explanation:

GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.

Answer:

A) nm, um, mm, cm

B) 1mm, 1cm, 1m, 1km

A) 3500g, B) 2500m, C) 7200 seconds

Answer:

a)   W_total = mg (2h + d)   , b)     E_total = - mg (h + d)

Explanation:

a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed

Let's look for work for the part that is in free fall

        y = y₀ + v₀ t - ½ g t²

when he jumps out of a plane his vertical speed is zero

        y =y₀ - ½ g t²

        dy = 0 - ½ g 2t dt

the work in this first part is

        W₁ = ∫ F dy

        W₁ = mg ∫ g t dt

        W₁ = m g² t² / 2

the time it takes to travel the distance y₀-y = h is

         y₀-y = ½ g t²

         t =[tex]\sqrt{2h/g}[/tex]

we substitute

          W₁ = m g² 2h / g

          W₁ = m g 2h

now we look for the work for the part with constant speed

since the velocity is constant let's use the uniform motion ratio

          W₂ = F d

           W₂ = mg d

the total work is

           W_total = W₁ + W₂

           W_total = 2mgh + m gd

           W_total = mg (2h + d)

b) The change in gravitational potential energy

           U = mg Δy

in the part with accelerated movement

           U₁ = mg h

in the part with uniform movement

            U₂ = mg d

the total potential energy is

           E_total = U₁ + U₂

           E_total = - mg (h + d)

Answer:

The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.

Weight can be calculated using the equation:

weight = mass × gravitational field strength

This is when:

weight (W) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

Answer:

the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Explanation:

Given;

the speed of the nerve impulse, v = 90.1 m/s

To convert this speed in meters per second to miles per hour, we use the following method;

1,609 meter = 1 mile

3,600 s = 1 hour

[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]

Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Explanation:

Given

Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]

time taken to reach ground is given by

[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]

(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]

From the figure, it can be written

[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]

Considering horizontal motion

[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]

(c) The vertical velocity with which it strikes the ground is given by

[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]

Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]

Explanation:

Given

Ball is projected horizontally from a building of height  

time taken to reach ground is given by

(b) Line joining the point of projection and the point where it hits the ground makes an angle of  

From the figure, it can be written

Considering horizontal motion

(c) The vertical velocity with which it strikes the ground is given by

Thus, the ball strikes with a vertical velocity of

Answer:

A. (b)

B. (a)

Explanation:

The electric dipole moment is the product of charge and the length of the dipole.

The torque on the dipole placed in the external electric field is given by

torque = p E sin A

where, p is the electric dipole moment, E is the electric field, A is the angle between the field and dipole moment.

When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium.

When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.

So, the option (b) is correct.

Teh energy is given by

U = - p  E cos A

When the angle A is zero , the potential energy is negative and it is minimum.

In this exercise we have to use the knowledge about dipole to be able to mark the correct alternative for each question, in this way we find that:

A) Letter b

B) Letter a

So knowing that the electric dipole moment is the product of charge and the length of the dipole and the torque on the dipole placed in the external electric field is given by:

[tex]torque = p E sin (A)[/tex]

where:

When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium. When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.

Now the energy is given by:

[tex]U = - p E cos (A)[/tex]

We can say that when the angle A is zero , the potential energy is negative and it is minimum.

See more about dipole at brainly.com/question/12757739

Answer:

Average Velocity = 3.63 m/s

Explanation:

First, we will calculate the total displacement of the quarterback, taking forward direction as positive:

Total Displacement = 15 m - 3 m + 24 m = 36 m

Now, we will calculate the total time taken for this displacement:

Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s

Therefore, the average velocity will be:

[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]

Average Velocity = 3.63 m/s

Answer:

t= 100s

Explanation:

use v=v0+at

plug in givens and solve for t

30=20+0.1*t

t= 100s

Answer:

SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.

Answer:

In 1 second the amount of charge flowing through the conductor is 2.5 Q.

Explanation: