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valina [46]
3 years ago
8

A box is placed on the floor.The area of the box in contact with the floor is 2.4m²Pressure exerted on the floor 16 newtons/m²Wo

rk out the force exerted by the box on the floor
Physics
1 answer:
Rudik [331]3 years ago
7 0

Answer:

The force exerted by the block on the floor is 38.4 Newtons

Explanation:

The given parameters are;

The area of the box in contact with the floor, A = 2.4 m²

The pressure exerted by the block on the floor, P = 16 N/m²

The force exerted by the box on the floor is given as follows;

Pressure, P = \dfrac{Force, F}{Area, A}

∴ F = P × A

From the question, P = 16 N/m², and A = 2.4 m²

∴ F = 16 N/m² × 2.4 m² = 38.4 N

The force exerted by the block on the floor, F = 38.4 N.

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Answer:B

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A voltage of 12 cos(I000t+45) Vis applied to a circuit in which a resistor of 4 .n, aninductor of L H, and a capacitor of 100 μF
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Answer:

0.01 H

Explanation:

V = 12 cos (1000t + 45)

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Let the inductance be L .

When the current and the voltage are in the same phase so it is the condition of resonance.

So capacitive reactance = inductive reactance

Xc = XL

1/ωC = ωL

L = 1 / ω²C

By comparisonV = Vo Cos (ωt + Ф)

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L = 1 / (1000 x 1000 x 100 x 10^-6)

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3 years ago
Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are
Sergio039 [100]

Answer:

21678.47223\ lbf-in^2

383.1109\ lbf-in^2

Explanation:

d = Diameter of column = 0.5 inch

A_c = Area of concrete = 119.4\ in^2

The strain in the system is conserved

\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s

Now

F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf

F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf

Stress is given by

\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2

The stress in the steel is 21678.47223\ lbf-in^2

\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2

The stress in the steel is 383.1109\ lbf-in^2

4 0
3 years ago
Find the change in thermal energy of a 25kg severed clown doll head that heats up from 25°C to 35°C, and has the specific heat o
timama [110]

Answer:

Q = 425 kJ

Explanation:

Given that,

Mass, m = 25 kg

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The specific heat is 1700 J/kg°C

We need to find the internal energy of it. The heat required to raise the temperature is given by the formula as follows :

Q=mc\Delta T\\\\Q=25\times 1700\times (35-25)\\\\Q=425000\ J\\\\Q=425\ kJ

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Answer:

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Explanation:

The average speed is defined as:

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Using the equations for uniformly accelerated motion, we calculate the runner's acceleration:

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v=\frac{7m}{2s}\\v=3.5\frac{m}{s}

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