Starting from rest, a student pulls a (25.0 A) kg box a distance of (4.50 B) m across the floor using a (134 C) N force applied at 35.0 degrees above horizontal. The coefficient of kinetic friction is 0.220 between the floor and the box. Find the work done by the pulling force. Give your answer in joules (J) and with 2 significant figures.

Answer:

here your answer

i am sorry if wrong

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

          d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

         d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

         d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

         sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d

         sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₁ = sin⁻¹ 0.3600

         θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

         sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]  

         θ₂ = sin⁻¹ 0.5401

         θ₂ = 0.571 rad

The angular separation is

         Δθ = θ₂ - θ₁

         Δθ = 0.571 - 0.368

         Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

           [tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

      m = 1

λ₁ = 400 nm

         θ₁ = sin⁻¹  [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]

         θ₁ = 0.120 rad

λ₂ = 600 nm

        θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]

        θ₂ = 0.181 rad

we use the equation of refraction.

         [tex]\theta_r[/tex]  = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )

λ₁ = 400 nm  

       θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]

       θ₁ = 0.090 rad

λ₂ = 600 nm

        θ₂ =sin⁻¹  [tex]\frac{1 sin 0.181}{1.33}[/tex]

        θ₂ = 0.1358 rad

The angular separation is

          Δθ = 0.1358 - 0.090

          Δθ = 0.046 rad.

In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:

       i) The third order is Δθ = 0.203 rad.

      ii) The first order with water is Δθ = 0.046 rad.

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Answer:

the ball will go up 3s and down 3s

v=gt

where t=3s and g=9.8m/s^2

distance=v0(t)+(1/2)gt^2

where initial velocity (v0)=0

Explanation:

The speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.

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 we have to find the speed required to throw a ball straight up and it returns 6 seconds later,

S = ut + 1/2*a*t²

0 = u×6 + 0.5×(-9.81)×6²

0 = 6u - 176.8

6u = 176.8

u = 176.8/6

u = 29.43 meters / seconds

Thus, the speed required to throw a ball straight up and returns 6 seconds later would be 29.43 meters/seconds.

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The mass of the object experimented in the situation will be equal to 20 kg in each case. Hence, option C is correct.

Acceleration is the rate at which the speed and orientation of a moving object change over time. When a spot or object moves faster or slower along a straight line, it is said to be accelerated. Even though the speed is constant, motion on a circle accelerates so because the direction is always shifting. Both by and to acceleration for all other types of motion.

It is a vector quantity because it has both magnitude and direction.

The given data as per the question is,

Force, F = 30 N

Acceleration, a = 1.5 m/s²

[tex]Force = mass*acceleration[/tex]

30 N = m × 1.5

m = 30/1.5

m = 20 kg.

Therefore, the mass of the object is 20 kg.

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(1) As the angle of the ramp is increased, the normal force decreases.

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

Let the angle of inclination of the ramp = θ

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_n = mgcos (\theta)[/tex]

when θ is 0;

[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]

when θ is 90;

[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

[tex]F_x = mgsin(\theta)\\\\[/tex]

when θ is 0;

[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]

when θ is 90;

[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

The force of friction is calculated as follows;

[tex]F_n = \mu F_n[/tex]

[tex]F_k = \mu mgcos(\theta)[/tex]

[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]

Thus, the angle is zero degree

(4)

The net force that would cause acceleration is calculated as follows;

[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]

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Answer:

Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ cao cực đại của vật so với mặt đất?

Explanation:

Explanation:

[tex]V = IR \Rightarrow R = \dfrac{V}{I}[/tex]

Plugging in the numbers, we get

[tex]R=\dfrac{4.5\:\text{V}}{0.5\:\text{A}} = 9.0[/tex] ohms

Answer:

Explanation:

006

They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.

Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6  which is upward.

008

If the two forces act in the same direction, the net force is found by addition.

Net Force = 99.6 + 52.8  = 152.4 N downward.

Answer:nh

0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board.

The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear

Since Force, F = ma where m = mass and a = acceleration. For constant mass, F ∝ a. That is, F is directly proportional to acceleration, a.

Since this is a linear relationship, the graph of force vs acceleration will be linear.

The answer to the question What shape is the graph produced by a force vs acceleration graph is A. Linear

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Answer:

Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.

Answer:

true

Explanation:

Answer:

A solid air freshener loses mass and volume as a result of sublimation, where solid particles skip the liquid state and change directly from a solid to a gas. Sublimation requires that the particles in the solid state gain enough energy to immediately become a gas.

Explanation:

Hope this helps.

Answer:

I think D it's the correct answer!! Let me know if i'm wrong

Explanation:

Answer:

This ability of charged particles to do work is called an electric potential.

Explanation:

When two negative charges are brought close to each other, they also repel. But when a positive and a negative charge are brought close together, they attract each other. When these two opposite charges are combined, they can be used to work. This is why we need a positive (+) and a negative (–) to light a bulb or run any electrical tool, equipment, mobile phone or home appliance.

Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... The energy is transferred to the electrical components in a circuit when the charge carriers pass through them. We use a voltmeter to measure potential difference (or voltage).

Answer:

What is instillation

Explanation:

Answer:

Explanation:

Use kinematic equation v² = u² + 2as

Rearrange for distance

s = (v² - u²) / 2a

Realize that at the top of its flight, the ball has zero velocity and gravity is acting downward in an assumed upward positive reference frame.

s = (0² - 29²) / (2(-9.8))

s = 42.90816...

s = 43 m

Answer:

[tex]-1.99\:\mathrm{m/s}[/tex]

Explanation:

Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:

[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]

Recall the chain rule:

[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]

Therefore,

[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]

Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].

Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:

[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]

Answer:

The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.

hopefully this'll help

have a nice day!!! :D

Answer:

hola como estas hablas español

Explanation:

Answer:

I will use (a / R) for alpha the angular acceleration

T R = I a / R     torque equals angular acceleration for cylinder

M g - T = M a    linear acceleration of center of mass

T = M (g - a) = I a / R^2 from first equation

If I = 1/2 M R^2 then M ( g - a) = M a / 2    from above

or g = 3 a / 2  and a = 2 g / 3

Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3

Substitute I = M R^2 for the hollow cylinder

Looks like a = g/2 for hollow cylinder

Answer: 1.646 m/s²

Explanation:

The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,

   d = Vot + 0.5at²

where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,

  6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²

Determining the value of a,

   a = 1.646 m/s²

Answer:

magnets are magnetic :)

Explanation:

By Newton's second law, the net force on the crate is

∑ F = 300 N - f = (45 kg) (4.44 m/s²)

where f is the magnitude of friction. Solve for f :

300 N - f = (45 kg) (4.44 m/s²)

f = 300 N - (45 kg) (4.44 m/s²)

f = 100.2 N ≈ 100 N (D)

Your answer is → Mass number

A period of a satellite is the time taken by the satellite to travel round a

body.

The comparison between the periods  [tex]T_B[/tex], and [tex]T_A[/tex] is [tex]\underline {T_B = \dfrac{\sqrt{2} }{4 } \cdot T_A}[/tex]

Reason:

The period, T, of a satellite is given as follows;

[tex]T = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \cdot M} }[/tex]

Volume of the planet A = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3[/tex]

Mass of planet A, [tex]m_A[/tex] = [tex]\dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho[/tex]

Volume of the planet B = [tex]\dfrac{4}{3} \cdot \pi \cdot (2 \cdot r)^3 = \dfrac{32}{3} \cdot \pi \cdot r^3[/tex]

Mass of planet B, [tex]m_B[/tex] = [tex]\dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho[/tex]

Period of the satellite on planet A, [tex]T_A[/tex], is given as follows;

[tex]T_A = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{4}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{4}{3} \cdot \pi \times \rho} }[/tex]

Period of the satellite on planet B, [tex]T_B[/tex], is given as follows;

[tex]T_B = 2 \cdot \pi \cdot \sqrt{\dfrac{r^3}{G \times \dfrac{32}{3} \cdot \pi \cdot r^3 \times \rho} } = 2 \cdot \pi \cdot \sqrt{\dfrac{1}{G \times \dfrac{32}{3} \cdot \pi \times \rho} }[/tex]

Therefore, get;

[tex]\dfrac{T_A}{T_B} = \dfrac{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 4 \cdot \pi \times \rho} }}{ 2 \cdot \pi \cdot \sqrt{\dfrac{3}{G \times 32 \cdot \pi \times \rho} }} = \sqrt{\dfrac{32}{4} } = \sqrt{8} = 2 \cdot \sqrt{2}[/tex]

Therefore, [tex]T_A[/tex] = (2·√2)·[tex]T_B[/tex]

[tex]T_B = \dfrac{T_A}{2 \cdot \sqrt{2} } = \dfrac{\sqrt{2} \cdot T_A}{4 }[/tex]

The comparison between [tex]T_A[/tex] and  [tex]T_B[/tex] is therefore;

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Answer:

335 - 30 g

Explanation:

V = Vo + g*T = 335

Vo + 9.8*30 = 335

Vo =

The distance through which the rider at the end of the beam moves is;

L = 15.5 m

We are told that the beam on which the carnival ride is fixed is 12.6m in length.

Since the seat is at the end of the beam with the other end pivoted and the beam sweeps through an angle of 141°, then we can say that the radius of this arc formed by the swing is;

Radius; r = 12.6 m

Also, θ = 141°

The distance through which the driver moves will be the length of the arc formed by the beam at angle of 141°.

Formula for length of arc is given as;

L = 2πr(θ/360)

Plugging in the relevant values gives;

L = 2π × 12.6 × 141/360

L = 15.5 m

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Answer:

TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.