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KiRa [710]
3 years ago
14

A force F acting on an object of mass m causes a displacement d. The angle between these two vectors is Θ. The work W done on th

at object is expressed by
Physics
1 answer:
777dan777 [17]3 years ago
4 0
Work:
W= \int\limits^b_a {F.} \, ds

F: force
ds: displacement

Assuming the work is constant:

W= \int\limits^b_a {F.} \, ds =  F . d = F * cos\theta * d

d displacement


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A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria
skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}

Differentiate both sides with respect to d.

\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

\frac{d^{2}A}{dt^{2}}= + ve

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0
3 years ago
a stone is projected vertically up from the top of a tower 73.5m with velocity 24.5 m/s . find the time taken by the stone to re
hoa [83]

The stone's altitude at time t is given by

y=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\dfrac{\rm m}{\mathrm s^2} is the acceleration due to gravity. The stone reaches the ground when y=0:

0=73.5\,\mathrm m+\left(24.5\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=7.11\,\rm s

6 0
3 years ago
In a circuit a battery is used to
Anarel [89]

Produce power through the circuit.

7 0
4 years ago
Read 2 more answers
A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
4 years ago
Proposed Exercise - Circular Movement
notka56 [123]

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

5 0
3 years ago
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