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Korolek [52]
3 years ago
14

What is the five things to do in P.E

Physics
2 answers:
Nimfa-mama [501]3 years ago
6 0

Answer:

Fucd6

Explanation:

Cuufufcuvjgjvug7fuguguguguf7

sergejj [24]3 years ago
6 0

Answer:

1: Stretches

2: Mile run

3: Sit ups

4:Push ups

5:Planks

Explanation:

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A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th
Readme [11.4K]

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = .250\times 10^{- 3}\ m

Now,

Applying the equilibrium condition at the left end for torque:

T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}

T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}

T_{C} = 341.357\ Nm

The weight of the wire balances the tension in both the wires collectively:

W = T_{Al} + T_{C}

531 = T_{Al} + 341.357

T_{Al} = 189.643\ Nm

Now,

The fundamental frequency is given by:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

\mu = A\rho = \pi R^{2}\rho

(a) For the fundamental frequency of Aluminium:

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}

where

\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz

(b)  For the fundamental frequency of Copper:

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}

where

\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz

7 0
4 years ago
Potassium and chlorine are most likely to form what type of bond? ionic covalent metallic polar
fiasKO [112]

Answer:

The correct answer is the IONIC bond.

Explanation:

4 0
4 years ago
Read 2 more answers
(a) Calculate the total force of the atmosphere acting on the top of a table that measures 1.5 m 2.2 m.
frozen [14]
A)
Atmospheric pressure is 101325 Pa.
Pressure = Force / Area
Area of table = 1.5 x 2.2
= 3.3 m²
Force = 101325 x 3.3
= 334 kN

B)
According to Newton's third law, every action has an equal and opposite reaction. Thus, the upward force is equal to the force acting downward on the table and is being balanced. The force is 334 kN in the upward direction.
8 0
4 years ago
What is the equilibrium line of a glacier?
drek231 [11]
The equilibrium line<span> separates the ablation zone and the accumulation zone; it is the altitude where the amount of new snow gained by accumulation is equal to the amount of ice lost through ablation. The upper part of a </span>glacier<span>, where accumulation exceeds ablation, is called the accumulation zone.</span>
8 0
4 years ago
You throw a ball straight up. The ball has an initial speed of 11.2 m/s when it leaves your hand What is the maximum height the
Jet001 [13]

Answer:

Explanation:

Given

initial velocity u=11.2\ m/s

At maximum height velocity of ball is zero

using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

(0)^2 -(11.2)^2=2\times (-9.8)\times (s)

s=\frac{11.2^2}{2\times 9.8}

s=6.4

time taken by the ball to reach the maximum height

v=u+at

0=11.2-9.8\times t

t=\frac{11.2}{9.8}

t=1.142\ s

At t=2\ s height of ball is

h=ut+\\frac{1}{2}at^2

h=11.2\times 2-\frac{1}{2}9.8\times (2)^2

h=22.4-19.6

h=2.8\ m

i.e. ball is moving downward

height at v=5\ m/s

v^2-u^2=2as

s=\frac{25-125.4}{2\times (-9.8)}

s=5.12\ m    

8 0
3 years ago
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