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2 years ago

How to calculate that

Physics

2 answers:

kaheart [24]2 years ago

7 0

Explanation:

Acceleration= 15-3/2

=6m/s²

erastova [34]2 years ago

3 0

<u>Answer:</u>

acceleration = $\bf 6 m/s^2$

<u>Explanation:</u>

We can calculate the acceleration of an object using the following formula:

$\boxed{acceleration = \frac{Resulatant \space\ force}{Mass}}$ .

The resultant force is the effective force that is being applied to an object. In this case, there is a 15N force to the right, and a 3N force to the left.

∴ Resultant force = 15N - 3N

= 12N

Now we can use the formula for above formula to calculate acceleration:

$acceleration = \frac{12}{2}$

= $\bf 6 m/s^2$

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A fillet weld has a cross-sectional area of 25.0 mm2and is 300 mm long. (a) What quantity of heat (in joules) is required to acc

HACTEHA [7]

Answer:

77362.56 J

163730.28571 J

Explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant = $3.33\times 10^{-6}$

$\eta$ = Heat transfer factor = 0.75

$f_m$ = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

$V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3$

The unit energy for melting is given by

$U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3$

Heat would be

$Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J$

Heat required to weld is 77362.56 J

Amount of heat generation is given by

$Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J$

The heat generated at the welding source is 163730.28571 J

7 0

4 years ago

A nonconducting solid sphere of radius 8.40 cm has a uniform volume charge density. The magnitude of the electric field at 16.8

mina [271]

Answer:

The sphere's volume charge density is 2.58 μC/m³.

Explanation:

Given that,

Radius of sphere R= 8.40 cm

Electric field $E= 2.04\times10^{3}\ N/C$

Distance r= 16.8 cm

We need to calculate the sphere's volume charge density

Using Gauss's law

$\int{\vec{E}\cdot\vec{da}}=\dfrac{Q_{enc}}{\epsilon_{0}}$

$E\times 4\pi r^2=\dfrac{1}{\epsilon_{0}}\times\dfrac{4}{3}\piR^3\rho$

$E=\dfrac{\rho R^3}{3\epsilon_{0}r^2}$

$\rho=\dfrac{3\times E\times\epsilon_{0}r^2}{R^3}$

Put the value into the formula

$\rho=\dfrac{3\times2.04\times10^{3}\times8.85\times10^{-12}\times(16.8\times10^{-2})^2}{(8.40\times10^{-2})^3}$

$\rho=2.58\times10^{-6}\ C/m^3$

$\rho=2.58\ \mu C/m^3$

Hence, The sphere's volume charge density is 2.58 μC/m³.

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