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djverab [1.8K]
3 years ago
8

Find the mass of an object if a 40 N of force causes the object to accelerate at 5.5 m/s/s

Physics
1 answer:
Murrr4er [49]3 years ago
7 0

Answer:

<h3>The answer is 8 kg</h3>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m =  \frac{40}{5}  \\

We have the final answer as

<h3>8 kg</h3>

Hope this helps you

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Explanation:

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What makes a substance, such as<br> aluminum, a good conductor?
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Answer:

For a material to be a good conductor, the electricity passed through it must be able to move the electrons; the more free electrons in a metal, the greater its conductivity.

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3 years ago
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3. A coil of 100 turns encloses an area of 100 cm2. It is placed at an angle of 700 with a
sasho [114]

Explanation:

Given that,

Number of turns in the coil, N = 100

Area of the coil, A = 100 cm² = 0.01 m²

It is placed at an angle of 70°.

Magnetic field, B = 0.1 Wb/m²

We need to find the magnetic flux through the coil and the emf is induced in the coil after 10⁻³ s.

Magnetic flux is given by :

\phi =BA\cos\theta\\\\\text{For N turns},\\\phi =NBA\cos\theta \\\\\phi=100\times 0.1\times 0.01\times \cos(70)\\\\=0.034\ Wb

So, the magnetic flux through the coil is 0.1 Wb.

Emf induced in the coil is :

\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{0.034}{10^{-3}}\\\\=34\ V

So, 34V of emf is induced in the coil.

7 0
3 years ago
The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

3 0
3 years ago
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