Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much work done on the weights? show full working out​

The mass of the wheel is 0.38 kg. It is related to weight, which is the force exerted on an object due to gravity and is proportional to the object's mass.

What is Mass?

Mass is a measure of the amount of matter in an object. It is a scalar quantity and is usually measured in kilograms (kg) in the SI system of units. Mass is a fundamental property of an object and does not change with its location or motion.

Let's start by finding the tension in the wire using the motion of the object. We can use the kinematic equation:

y = (1/2)at^2

where y is the distance the object moves, a is the acceleration, and t is the time. Substituting the given values:

2.95 m = (1/2)a(1.85 s)^2

Solving for the acceleration:

a = 2.47 m/s^2

Now, we can find the tension in the wire using the rotational dynamics of the wheel. The torque due to the tension is equal to the moment of inertia times the angular acceleration:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration. Since the wheel is rotating without friction, the torque due to the tension is the only torque acting on the wheel, so:

Tension * radius = (1/2)mr^2 * α

where m is the mass of the wheel and r is its radius. We can rearrange this equation to solve for the tension:

Tension = (1/2)ma

Substituting the values we found:

Tension = (1/2)(4.3 kg)(2.47 m/s^2) = 5.31 N

Now we can use the torque equation again to solve for the moment of inertia of the wheel:

Tension * radius = (1/2)mr^2 * α

Substituting the values we found and solving for the moment of inertia:

I = 2(Tension * radius) / α = 0.033 kg·m²

Finally, we can use the moment of inertia to find the mass of the wheel using the formula:

I = (1/2)mr^2

Substituting the values we found:

0.033 kg·m² = (1/2)m(0.288 m)^2

Solving for the mass:

m = 2I / r^2 = 0.38 kg

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Answer:

10.0

Explanation:

Mechanical Advantage = distance moves along the plane/distance moved along the vertical

distance moves along the vertical = 0.3m

distance moves along the plane = 3.0m

Mechanical Advantage = 3.0/0.3

Mechanical Advantage = 10

Hence the mechanical advantage is 10.0

We can deduce here that the maximum horizontal force that you can apply to the box is 150 pounds. Thus, the heaviest box that you can push up the ramp at a constant speed is 75 pounds.

Given the following:

Maximum horizontal force = 150 pounds

Coefficient of kinetic friction = 0.50

Weight of the box = 150 pounds

Angle of the ramp = 60°

Normal force = Weight of the box * Cosine of the angle of the ramp

= 150 pounds × Cos(60°)

= 75 pounds.

Force of friction = Coefficient of kinetic friction × Normal force

= 0.50 × 75 pounds

= 37.5 pounds

Maximum force that can be applied to the box = Weight of the box × Cosine of the angle of the ramp - Force of friction

= 150 pounds × Cos(60°) - 37.5 pounds

= 75 pounds

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Answer:

Joule ;)

Explanation:

In the case of work (and also energy), the standard metric unit is the Joule (abbreviated J). One Joule is equivalent to one Newton of force causing a displacement of one meter. In other words, The Joule is the unit of work.

Hope this helps!

Answer:

Explanation:

Joule

In the case of work (and also energy), the standard metric unit is the Joule (abbreviated J). One Joule is equivalent to one Newton of force causing a displacement of one meter. In other words, The Joule is the unit of work.

Answer:

d = √(6² + 8²) = 10 km

Explanation:

displacement is a vector and subject to vector addition.

The length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

Given data: r = 0.2 mm = 0.2 x 10^-3m Resistivity = 3 x 10^-6 ohm R = 15.552 ohm

Formula Used: Resistivity (ρ) = (RA)/L

Where, R is resistance, A is the area of cross-section, L is the length of the wire.

Resistance (R) = ρ (L/A)

Multiplying A on both sides, we get

Resistance (R) x A = ρ L ... equation (1)

Area of the cross-section of a wire of radius (r) is given by, A = πr^2

where, π is a constant whose value is 3.14

Substituting the given values, we get

A = πr^2= π (0.2 x 10^-3m)^2= 1.2566 x 10^-7 m^2

Substituting the values of R, A and ρ in equation (1), we get

Length of wire (L) = (Resistance x Area) / Resistivity= (15.552 ohm x 1.2566 x 10^-7 m^2) / (3 x 10^-6 ohm)= 6.5268 m

Therefore, the length of wire whose resistivity is 3 x 10^-6ohm, and radius is 0.2 mm, with a given value of 15.552 resistance is 6.5268 m.

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Answer:

Gravity

Explanation:

The initial momentum of the 25 kg object is 25 kg * 12 m/s = 300 kgm/s. After the collision, the momentum of the 25 kg object is 25 kg * 8 m/s = 200 kgm/s. According to the conservation of momentum, the momentum lost by the 25 kg object is equal to the momentum gained by the box. Therefore, the resulting momentum of the box is 300 kgm/s - 200 kgm/s = 100 kg*m/s.

Electricity is electrons moving through a conductor.

Voltage pushes electrons through the conductor to create an electric current.

Electricity comes from Power stations , that have generators that produce electricity.

Generators use different sources of energy such as water flow, fossil fuel, etc.

the x- component of the vector is 36.86 m.

Vectors are quantities that have both magnitude and direcion

To calculate the x-component of the vector, we use the formula below.

Formula:

Where;

From the question,

Given:

Substitute these values into equation 1

Hence, the x- component of the vector is 36.86 m.

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Answer:

Should be 400m

Explanation:

Answer:

i) Mr. Dunn arrives to home first.

ii) 3 min

Explanation:

i. To find who arrives first to home you calculate the time, by using the following formula:

\(t=\frac{x}{v}\)

x: distance

v: velocity

Mr. Dunn:

\(t=\frac{64.8km}{48km/h}=1.35h\)

Mrs. Dunn:

\(t=\frac{81.2km}{58km/h}=1.4h\)

Hence, Mr. Dunn arrives to home first.

ii. To calculate the difference in minutes, you convert hours to minutes:

\(1.35h*\frac{60min}{1h}=81min\\\\1.40h*\frac{60min}{1h}=84min\\\\\Delta\ t=(84-81)min=3min\)

the difference between the times is 3min

(i) Mr. Dunn takes less time so he arrives at home first.

(ii) The second person arrives 3 min late.

(i) We have to calculate the time taken to reach home by Mr. Dunn and Mrs. Dunn.

t = x/v

where x is the distance

and v is the velocity

Time taken by Mr. Dunn:

distance x = 64.8 km

speed v  = 48 km/h

t = 64.8 / 48

t = 1.35 h

Time taken by Mrs. Dunn:

distance x = 81.2 km

speed v  = 58 km/h

t' = 81.2 / 58

t' = 1.4 h

Hence, Mr. Dunn arrives at home first.

(ii) To calculate the difference in minutes, you convert hours to minutes:

The time taken by Mr. Dunn in minutes is:

t = 1.35×60 = 81 minutes

The time taken by Mrs. Dunn in minutes is:

t' = 1.4×60 = 84 minutes

the difference between the times is 3min

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The object was dropped from a height of 122.5 meters.

The equation t = √h/4.9 relates the time it takes for an object to fall to the ground (t) to the height from which it was dropped (h). The equation states that the time it takes for the object to reach the ground can be calculated as the square root of the height divided by 4.9, with t measured in seconds.

To solve for h in the equation t = √h/4.9, we can square both sides:

t^2 = h/4.9

Multiplying both sides by 4.9:

4.9 * t^2 = h

So, the height from which an object was dropped can be calculated as:

h = 4.9 * t^2

If an object falls for 5 seconds, then:

h = 4.9 * 5^2

= 4.9 * 25

= 122.5 meters

Key points:

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The instantaneous speed of the ant at t=8s is equal to the slope of the segment between t=7s and t=9s, which is 1.5m/s.

To find the instantaneous speed of the ant at time t=8s, we need to calculate the derivative of the displacement function with respect to time. Since the displacement of the ant is given by a piecewise function, we need to differentiate each segment of the function separately and then piece them together. From 0s to 4s, the displacement of the ant decreases linearly from 6m to 2m. The slope of this segment is -1m/s. From 4s to 7s, the displacement of the ant is constant at 2m. The slope of this segment is zero. From 7s to 9s, the displacement of the ant increases linearly from 2m to 5m. The slope of this segment is 1.5m/s. Finally, from 9s to 10s, the displacement of the ant is constant at 5m. The slope of this segment is zero.

Therefore, the instantaneous speed of the ant at t=8s is equal to the slope of the segment between t=7s and t=9s, which is 1.5m/s. This means that the ant is moving upwards at a constant speed of 1.5m/s at time t=8s. It's important to note that the instantaneous speed of the ant tells us how fast it's moving at a specific moment in time. It's not the same as the average speed over a given time interval, which would be calculated by dividing the total displacement by the total time elapsed.

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Answer:

Photovoltaic detection is a technique that converts light into electrical energy. It is a process that involves the use of a photovoltaic cell, which is made up of semiconductor materials, to generate an electric current when exposed to light.

The photovoltaic cell absorbs the photons of light, which then knock electrons out of their orbits, creating a flow of electricity. The amount of electricity produced is proportional to the intensity of the light. The photovoltaic cell is commonly used in solar panels to generate electricity from sunlight. The efficiency of the photovoltaic cell is dependent on several factors, including the type of semiconductor material used, the purity of the material, and the thickness of the cell.

The photovoltaic cell has many applications, including in solar power generation, telecommunications, and remote sensing. The technique of photovoltaic detection is an important area of research, as it has the potential to provide a clean and renewable source of energy that can help mitigate climate change.

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Explanation:

thermodynamics is the study of heat.

Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie

Explanation:

Answer:

\(n=3.8\)

Explanation:

From the question we are told that:

Magnetic Field \(B=0.01T\)

Current \(I=1.00\)

Wire Diameter \(d_w=0.5*10^3m\)

Layers Diameter \(d_l=1*10^2m\)

Length \(l=0.2m\)

Generally the equation for number of layers is mathematically given by

\(n=\frac{Bd_w}{\mu_o I}\)

Where

\(Vacuum\ permeability=\mu_0\)

\(n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }\)

\(n=3.8\)

Answer:

where u put it last time or retrace ur steps to where u last put it

The primary difference is that electromagnetic waves can propagate through a vacuum or empty space, while mechanical waves require a physical medium to transmit energy.

A primary difference between an electromagnetic wave and a mechanical wave is the medium through which they propagate.

Electromagnetic waves can propagate through a vacuum or empty space without requiring a material medium. They are generated by the oscillation and interaction of electric and magnetic fields.

Examples of electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. These waves can travel through space, air, or other materials, as they do not rely on physical particles to transmit energy.

On the other hand, mechanical waves require a physical medium to propagate. They are disturbances that travel through a material medium, transferring energy from one location to another. Mechanical waves rely on the interaction and displacement of particles within the medium to transmit energy.

Examples of mechanical waves include sound waves, water waves, seismic waves, and waves on a string. These waves cannot travel through a vacuum as they depend on the physical presence and interaction of particles within the medium.

In summary, the primary difference is that electromagnetic waves can propagate through a vacuum or empty space, while mechanical waves require a physical medium to transmit energy.

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The frictional force of the wind on water produces surface currents that travel in a pattern analogous to "wind currents".

Wind, temperature, water density, and the moon's gravitational pull may all affect ocean currents.

Wind-driven waves, sometimes referred to as surface waves, are created by friction between the wind and the water's surface. When wind constantly disturbs the surface of an ocean or lake, a wave crest is created.

Friction slows the wind, which also changes its direction. This phenomenon might lead to surface roughness variations along regional boundaries or turbulence in winds near the surface. Turbulence stirs the lower atmosphere.

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Answer:

The mass of the ice added = 16.71 g

Explanation:

The heat gained by the ice is equal to the heat lost by the calorimeter cup and the water in the cup.

But for this question, the cup is said to be perfectly insulated, hence, there is no loss of heat from the calorimeter cup.

Heat gained by the ice = Heat lost by the 326 g of water.

Let the mass of ice be m

The heat gained by the ice = (Heat gained by ice in temperature from -5.5°C to 0°C) + (Heat used by the ice to melt at 0°C) + (Heat required for the melted ice to rise in temperature from 0°C to 15°C)

Heat gained by ice in temperature from -5.5°C to 0°C = mCΔT

m = unknown mass of ice

C = Specific Heat capacity of ice = 2.108 J/g°C

ΔT = change in temperature = 0 - (-5.5) = 5.5°C

Heat gained by ice in temperature from -5.5°C to 0°C = m×2.108×5.5 = (11.594m) J

Heat used by the ice to melt at 0°C = mL

m = unknown mass of ice

L = Latent Heat of fusion of ice to water = 334 J/g

Heat used by the ice to melt at 0°C = m×334 = (334m) J

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = mCΔT

m = unknown mass of water (which was ice)

C = Specific Heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 15 - 0 = 15°C

Heat required for the melted ice or water now, to rise in temperature from 0°C to 15°C = m×4.186×15 = (62.79m) J

Total heat gained by the ice = 11.594m + 334m + 62.79m = (408.384m) J

Heat lost by the water in the calorimeter cup = MCΔT

M = mass of water in the calorimeter cup = 326 g

C = specific heat capacity of water = 4.186 J/g°C

ΔT = change in temperature = 20 - 15 = 5°C

Heat lost by the water in the calorimeter cup = 326×4.186×5 = 6,823.18 J

Heat gained by the ice = Heat lost by the 326 g of water.

408.384m = 6,823.18

m = (6,823.18/408.384)

m = 16.71 g

Hope this Helps!!!

In "The Argonauts," Robinson effectively utilizes language techniques such as metaphor, repetition, and sentence structure to develop his argument to his younger self and engage young readers in the present. Through these techniques, Robinson creates a powerful and relatable narrative that resonates with his audience.

Robinson employs metaphors to convey complex ideas in a compelling and accessible manner. For instance, he compares his struggle with identity and gender to the mythical journey of the Argonauts, making it relatable and captivating for young readers. This metaphorical language enables readers to grasp the profound emotions and challenges he faced during his own personal journey.

Repetition is another technique Robinson employs to reinforce key ideas and create a rhythmic and memorable reading experience. By repeating certain phrases or concepts, he emphasizes their significance and invites readers to reflect on them. This repetition serves to engage young readers by encouraging them to contemplate their own experiences and perspectives.

Furthermore, Robinson carefully structures his sentences to create a sense of rhythm and flow, enhancing the overall readability and impact of his argument. Short, concise sentences create moments of clarity and emphasis, while longer, more descriptive sentences evoke a contemplative and introspective tone. This varied sentence structure adds depth and nuance to his narrative, captivating young readers and keeping them engaged throughout.

In conclusion, through the effective use of metaphor, repetition, and sentence structure, Robinson engages and captivates young readers, inviting them to reflect on their own identities and experiences. His language choices not only develop his argument to his younger self but also establish a connection with present-day young readers, making his work both impactful and relatable.

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Given:

Mass, m = 600000 g

Height, h1 = 38 m

Height, h2 = 14 m

Let's solve for the following:

• (a). Gravitational potential energy

To find the gravitational potential energy, apply the formula:

Where:

m is the mass in kg = 600000 g = 600 kg

g is the acceleration due to gravity = 9.8 ms/s²

h2 is the height in meters over the second hill = 14 m

Input the values into the formula and solve for PE:

Therefore, the gravitational potential energy at the second hill is 82320 Joules

• (b),. Kinetic energy at the second hill

To find the kinetic energy, apply the formula:

Where:

m is the mass

h1 = 38 m

h2 = 14 m

Thus, we have:

• (c). Mechanical Energy.

To find the mechanical energy, apply the formula:

Mechanical Energy = Potential energy + Kinetic energy

Mechanical Energy = 82320 + 141120 = 223440 J

• (d). Speed of the car when it goes over the second hill which is 14 m.

To find the speed of the car, apply the formula:

Where:

KE is the kinetic energy = 223440 J

m = 600 kg

v is the velocity.

Let's solve for v:

The speed when it goes over the second hill is 27.29 m/s

ANSWER:

• (A). 82320 J

• (b). 141120 J

• (C). 223440 J

• (d). 27.29 m/s

Answer:

the study of the past

Explanation:

dogs and cats

Answer:

Explanation:

The answer is **(c) 9.39 m/s** for the velocity of the bullet-block system after it's hit, and **(g) 209.39 m/s** for the bullet's velocity before it hit the box.

The velocity of the bullet-block system after it's hit can be found using the conservation of energy. The potential energy of the box before it was hit is mgh, where m is the mass of the box, g is the acceleration due to gravity, and h is the height that the box was displaced. After the bullet hits the box, the potential energy of the box is zero, but the kinetic energy of the bullet-block system is mv^2/2, where m is the total mass of the bullet-block system and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mgh = mv^2/2

```

Solving for v, we get:

```

v = sqrt(2mgh)

```

In this case, we have:

* m = 0.05 kg + 1.3 kg = 1.35 kg

* g = 9.8 m/s^2

* h = 4.5 m

So, the velocity of the bullet-block system after it's hit is:

```

v = sqrt(2 * 1.35 kg * 9.8 m/s^2 * 4.5 m) = 9.39 m/s

```

The velocity of the bullet before it hit the box can be found using the conservation of momentum. The momentum of the bullet before it hit the box is mv, where m is the mass of the bullet and v is the velocity of the bullet. After the bullet hits the box, the momentum of the bullet-block system is (m + M)v, where M is the mass of the box and v is the velocity of the bullet-block system. Setting these two expressions equal to each other, we get:

```

mv = (m + M)v

```

Solving for v, we get:

```

v = mv/(m + M)

```

In this case, we have:

* m = 0.05 kg

* M = 1.3 kg

* v = 9.39 m/s

So, the velocity of the bullet before it hit the box is:

```

v = 0.05 kg * 9.39 m/s / (0.05 kg + 1.3 kg) = 209.39 m/s

```

The velocity of the bullet-block system after the collision is approximately a) 6.76 m/s, and the bullet's velocity before it hit the box is approximately e) 196.76 m/s.

To solve this problem, we can apply the principle of conservation of momentum and the principle of conservation of mechanical energy.

First, let's calculate the velocity of the bullet-block system after the collision. We can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Let m1 be the mass of the bullet (0.05 kg) and m2 be the mass of the box (1.3 kg). Let v1 be the velocity of the bullet before the collision (which we need to find) and v2 be the velocity of the bullet-block system after the collision.

Using the conservation of momentum:

m1 * v1 = (m1 + m2) * v2

0.05 kg * v1 = (0.05 kg + 1.3 kg) * v2

0.05 kg * v1 = 1.35 kg * v2

Now, let's calculate the velocity of the bullet-block system (v2). Since the system goes up by a height of 4.5 m, we can use the principle of conservation of mechanical energy.

m1 * v1^2 = (m1 + m2) * v2^2 + m2 * g * h

0.05 kg * v1^2 = 1.35 kg * v2^2 + 1.3 kg * 9.8 m/s^2 * 4.5 m

Now, we can substitute the value of v2 from the momentum equation into the energy equation and solve for v1.

By solving these equations, we find that v1 is approximately 196.76 m/s.

Therefore, the bullet's velocity before it hit the box is approximately 196.76 m/s. (e) 196.76 m/s

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Answer:

a. democracy

Explanation:

beacouse the government control of their members

The work done by the man against friction is 4,192.86 J.

The given parameters;

The work done by the man against friction is calculated as follows;

\(W = F \times d \times cos(75)\\\\W = 540 \times 30 \times cos(75)\\\\W = 4,192.86 \ J\)

Thus, the work done by the man against friction is 4,192.86 J.

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