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Andrews [41]
3 years ago
14

A 2 kg object is given a displacement ∆~s = (5 m)ˆı + (3 m) ˆ + (−4 m) ˆk along a straight line. During the displacement, a con

stant force of F~ = (3 N)ˆı + (−1 N) ˆ + (7 N) ˆk acts on the object. Find the work done by F~ for this displacement.
Physics
1 answer:
nadya68 [22]3 years ago
8 0

Answer: 15Nm

Explanation:

Work is said to be done when a force cause a body to move through a distance. Mathematically,

Work = Force (F) × distance (s)

If F = F~ = (3 N)ˆı + (−1 N) ˆj + (7 N) ˆk

S = (5 m)ˆı + (3 m) ˆ + (−4 m) ˆk

According to vector notation,

i.I = j.j =k.k = 1

Multiplication of different component will give 'zero'

F×s = (3 N)ˆi × (5 m)ˆı

Work done = 15Nm

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A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const
avanturin [10]

Answer:

0.17547 m

Explanation:

m = Mass of block = 3.3\times 10^{-2}\ kg

v = Velocity of block = 10.8 m/s

k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m

The amplitude of the resulting simple harmonic motion is 0.17547 m

6 0
3 years ago
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

#SPJ4

5 0
2 years ago
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Consider a horse pulling a buggy. Is the
Degger [83]

Yes, it's true.

But 2nd Newton Law always come to play when the horse is to move forward because obviously the forces interact antagonistically and mass has to be accounted for.

That's what I think. Hope it's right, all the best.

8 0
3 years ago
What is the best unit to use when measuring the mass of a mineral sample?
Ksivusya [100]
Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.
8 0
2 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
2 years ago
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