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14 days ago

luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that

acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is μk = 0.300. The suitcase travels 3.90 m along the ramp. Part A Calculate the work done on the suitcase by F⃗ . Express your answer with the appropriate units. WF = nothing nothing Request Answer Part B Calculate the work done on the suitcase by the gravitational force. Express your answer with the appropriate units. Ww = nothing nothing

Physics

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The cluster of decisions that managers make to assist the organization to achieve its goals is known as:

Softa [3030]

The answer is option C. SWOT analysis. The compilation of decisions made by managers to aid the organization in reaching its objectives is referred to as SWOT Analysis.
Here are the options. A. Strategy B. Scenario planning C. SWOT analysis D. Diversification E. Related diversification

8 0

2 months ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, $m_{p} = 1.67\times 10^{- 27} kg$

Initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

2 months ago

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

Keith_Richards [3271]

25.82 m/s Explanation: Given: Force applied by the baseball player; F = 100 N Distance the ball travels; d = 0.5 m Mass of the ball; m = 0.15 kg To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved. It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows: F × d = ½mv² 100 × 0.5 = ½ × 0.15 × v² Solving gives: v² = (2 × 100 × 0.5) / 0.15 v² = 666.67 v = √666.67 v = 25.82 m/s.

4 0

2 months ago

A large box of mass m sits on a horizontal floor. You attach a lightweight rope to this box, hold the rope at an angle θ above t

inna [3103]

Answer:

The answer to the specified question will be " $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$ ".

Explanation:

Referring to the question,

$\sum F_{x}$

⇒ $TCos \theta-F_{s}=0$

⇒ $T_{m}Cos \theta =F_{s}$ ...(equation 1)

$\sum F_{y}$

⇒ $TSin \theta+F_{N}=m_{g}$

⇒ $M_{g}-TSin \theta=F_{N}$ ...(equation 2)

Now,

From equation 1 and equation 2, we conclude

⇒ $T_{m} Cos \theta = \mu_{s}F_{N}$

By substituting the value of $F_{N}$ , we derive

⇒ $T_{m} Cos\theta = \mu_{s}(M_{g}-T_{m}Sin \theta)$

⇒ $\mu_{s}=\frac{T_{m}Cos\theta}{M_{g}-T_{m}Sin\theta}$

4 0

2 months ago

A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a

inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

$C=\frac{\epsilon_0A}{d}$

In this equation, $\epsilon_0$ refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

$E_1=\frac{Q^2}{2C_1}$ ......(3)