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2 years ago

Question 3 of 10

Physics

1 answer:

Eva8 [605]2 years ago

5 0

Answer:

D. Newton's second law

Explanation:

Newton's second law of motion states that force of an object is a product of its mass and its acceleration.

Mathematically, F= ma where m is mass and a is acceleration

So from the statement above : The acceleration of an object is proportional to the force applied to it and inversely proportional to its mass , it can be seen from the formula variation as;

F= ma -----making a the subject of the formula

a= F/ m

a= 1/m * F --------- a is inversely related to m as you can see from 1/m but directly related to F hence;

Increase in mass with the same force applied causes the body to accelerate slower where as when force increases, the body accelerates faster.

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Can anyone solve this?

love history [14]

Answer:

F = 39.2 N

Explanation:

Since, the object is in uniform motion. Therefore, the frictional force on object will be:

Frictional Force = μk N = μk mg

where,

μk = coefficient of kinetic friction = 0.2

m = mass of crate = 10 kg

g = 9.8 m/s²

Therefore,

Frictional Force = (0.2)(10 kg)(9.8 m/s²)

Frictional Force = 19.6 N

The horizontal component of force must be equal to this frictional force to continue the uniform motion:

F Sin 30° = 19.6 N

F = 19.6 N/Sin 30°

3 0

3 years ago

An electric field of intensity 3.25 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.

jekas [21]

Answer:

$\varphi_1= 796.25 N m^2/C$

$\varphi_2= 0 N m^2/C$

$\varphi_3=686.1 N m^2/C$

Explanation:

From the question we are told that

Electric field of intensity $E= 3.25 kN/C$

Rectangle parameter Width $W=0.350 m$ Length $L=0.700 m$

Angle to the normal $\angle=30.5 \textdegree$

Generally the equation for Electric flux at parallel to the yz plane $\varphi_1$ is mathematically given by

$\varphi_1=EA cos theta$

$\varphi_1=3.25* 10^3 N/C * ( 0.350)(0.700) cos 0$

$\varphi_1= 796.25 N m^2/C$

Generally the equation for Electric flux at parallel to xy plane $\varphi_2$ is mathematically given by

$\varphi_2=EA cos theta$

$\varphi_2=3.25* 10^3 N/C * ( 0.350)(0.700) cos 90$

$\varphi_2= 0 N m^2/C$

Generally the equation for Electric flux at angle 30 to x plane $\varphi_3$ is mathematically given by

$\varphi_3=EA cos theta$

$\varphi_3=3.25* 10^3 N/C * ( 0.350)(0.700) cos 30.5$

$\varphi_3=686.072219 N m^2/C$

$\varphi_3=686.1 N m^2/C$

7 0

3 years ago

A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$