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2 years ago

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A lever is used to lift a heavy rock. The rock has a mass of 800 N. It requires 200 N of force to lift the rock using the lever.

What is the MA of the lever?

1 answer:

inna [77]2 years ago

5 0

Answer:

The answer for this solution is 4

Explanation:

MA=L/E

800/200

=4

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gayaneshka [121]

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2 years ago

A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on

9966 [12]

F=ma
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3 years ago

Read 2 more answers

Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.

mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, $\rho=10^{-6}\ \Omega -m$

Density, $d=8.31\times 10^3\ kg/m^3$

(A) The length of the wire is given by using the definition of resistance as :

Volume,

$V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3$

Area,

$A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}$ ....(1)

$R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m$

(b) Equation (1) becomes :

$A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m$

Hence, this is the required solution.

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3 years ago

How long does it take the projectile to reach its highest point in its trajectory?

NeTakaya

Answer:

It takes about 88 seconds for the cannonball to reach its maximum height (ignoring air resistance).

Explanation:

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3 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

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3 years ago