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alekssr [168]
3 years ago
8

A closed cylindrical tank of radius 3.5 m and height 2m is made from

Physics
2 answers:
I am Lyosha [343]3 years ago
7 0

Explanation:

Answer:

Given :-

A closed cylindrical tank of radius 3.5 m and height 2 m is made from a sheet of metal.

To Find :-

How much sheet of metal is required.

Formula Used :-

{\red{\boxed{\large{\bold{T.S.A\: of\: cylinder\: =\: 2{\pi}r(r + h)}}}}}

where,

r = Radius

h = Height

Solution :-

Given :

Radius = 3.5 m

Height = 2 m

According to the question by using the formula we get,

⇒ \sf T.S.A =\: 2 \times \dfrac{22}{7} \times 3.5(3.5 + 2)

⇒ \sf T.S.A =\: 2 \times \dfrac{22}{7} \times 3.5(5.5)

⇒ \sf T.S.A =\: 2 \times \dfrac{22}{7} \times 3.5 \times 5.5T.S.A=2×722×3.5×5.5⇒ \sf T.S.A =\: 2 \times \dfrac{22}{7} \times 19.25

⇒ \sf T.S.A =\: 2 \times 60.5

➠ \sf\bold{\purple{T.S.A =\: 121\: {m}^{2}}}

∴ 121 m² of sheet of metal is required.

Dahasolnce [82]3 years ago
3 0

Explanation:

Total surface area of cylinder:

2\pi \times r(h + r) \\ 2 \times \pi \times 3.5(3.5 + 2) \\ 2\pi \times 19.25 = 120.951317 \\ 120.95  \: {cm}^{2}

sheet of metal required = 120.95 cm^2

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Rom4ik [11]

Acceleration of the ball is 50 m/s^2

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

F=ma

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F is the net force

m is the mass

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For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

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3 years ago
Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

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A force of 100N is applied to an area of 100mm².what is the pressure exerted on the area in N/m².​
Dmitry_Shevchenko [17]

Answer:

P = 1000000[Pa] = 1000 [kPa]

Explanation:

To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

P=F/A

where:

P = pressure [Pa] (units of pascals)

F = force = 100 [N]

A = area = 100 [mm²]

But first we must convert the units from square millimeters to square meters.

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Now replacing:

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3 0
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