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2 years ago

a 4,000 kilogram rocket has accelerates at a rate of 35 m/s2. How much force is required to do this?

Physics

1 answer:

WARRIOR [948]2 years ago

5 0

$\Large{\underline{\sf Given:-}}$

Mass=4000kg
Acceleration=35m/s^2

$\Large{\underline{\sf To\:Find:-}}$

Force

$\Large{\underline{\sf Formula:-}}$

$\boxed{\sf Force=Mass\times Acceleration}$

$\Large{\underline{\sf Solution:-}}$

$\\ \sf\longmapsto Force=4000(35)$

$\\ \sf\longmapsto Force=14000N$

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A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm

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0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

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I = Current

B = Magnetic field

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Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

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2 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a

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Answer:

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b.102nC/m^2

c.3.363pF

d. 77.3pC

Explanation:

Data given

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to calculate the electric field, we use the equation below

V=Ed

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b.the expression for the charge density is expressed as

σ=ξE

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If we insert the values we have

$8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}$

from the expression for the capacitance

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if we substitute values we arrive at

$C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF$

d. To calculate the charge on each plate, we use the formula below

$Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC$

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3 years ago

a 4,000 kilogram rocket has accelerates at a rate of 35 m/s2. How much force is required to do this?​

a 4,000 kilogram rocket has accelerates at a rate of 35 m/s2. How much force is required to do this?