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Vikki [24]
4 years ago
15

A vertical rectangular wall with a width of 20 m and a height of 12 m is holding a 7-m-deep water body. The resultant hydrostati

c force acting on this wall is (a) 1370 kN (b) 4807 kN (c) 8240 kN (d) 9740 kN (e) 11,670 kN
Physics
1 answer:
denis23 [38]4 years ago
5 0

Answer:

(c) 8240 kN

Resultant hydrostatic force F = 8240 kN

Explanation:

Force can be expressed as;

F = ma

F = mg ......1

Where,

F = force

m = mass

a = acceleration

g = acceleration due to gravity

And we know that mass of an object is the product of

Density and Volume

m = pV

And volume V = area × height = Ah

m = pAh .....2

Substituting m into equation 1

F = pAhg = pghA ......3

Given;

Density of water p = 1000kg/m^3

g = 9.81m/s^2

Area A = 20m × 12m = 240m^2

h = depth/2 = 7/2 = 3.5m

Substituting the values into equation 3, we have;

F = 1000×9.81×3.5×240

F = 8240400 N

Resultant hydrostatic force F = 8240 kN

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Explanation:

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3 years ago
A radar transmits a signal towards an aeroplane. The velocity of the signal is 3.0 x 10^8 m/s. After 0.0036 s, the radar detects
avanturin [10]

The distance of the airplane from the radar will be 1.08×10⁶ m. Distance can refer to a physical length.

<h3>What is distance?</h3>

Distance is a numerical representation of the distance between two objects or locations.

The distance of the airplane from the radar is found by the formula;

Distance = velocity × time

Distance =  3.0 x 10^8 m/s× 0.0036 s,

Distance =  1.08×10⁶ m.

Hence, the distance of the airplane from the radar will be 1.08×10⁶ m.

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3 years ago
A 13.3 kg box sliding across the ground
adelina 88 [10]

Answer:

<em>0.25</em>

Explanation:

According to newtons law of motion

\sum F_x = ma

F_f =  ma

nR = ma

nmg = ma

ng = a

n = a/g

g is the acceleration due to gravity

Given

a = 2.42m/s²

g = 9.8m/s²

Substitute into the formula;

n = 2.42/9.8

n = 0.25

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5 0
3 years ago
A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s
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Answer:

(a) I (Moment of inertia)=0.0987 kgm^{2}

(b) W(Angular Speed)=2.66 \frac{rad}{s}

Explanation:

Given data

m (Monkey mass)= 1.80 kg

d=2.50 m

T (Time Period)=0.940 s

Angle= 0.400 rad

(a) I (Moment of Inertia)=?

(b) W (Angular Speed)=?

For part (a) I (Moment of Inertia)=?

Time Period Formula is given as

T=2(3.14)\sqrt{\frac{I}{mgd} }

After Simplifying we get

I=\frac{mgdT^{2} }{4*(3.14)^{2}}

I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }

I=0.0987 kgm^{2}

For Part (b) Angular Speed

From Kinetic Energy we get

KE=\frac{1}{2}IW^{2}

Pontential Energy

PE=mgd(1-Cosa)

KE=PE

\frac{1}{2}IW^{2}=mgd(1-Cosa)

W^{2}=\frac{2mgd(1-Cosa)}{I}

W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }

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