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Y_Kistochka [10]

3 years ago

8

A 2.0-kg ball is at rest when a horizontal force of 5.0 N is applied. In the absence of friction, what is the speed of the ball

after it ha gone 10 m?

1 answer:

Rzqust [24]3 years ago

7 0

Answer:

v = 7.1 m/s

Explanation:

Work applied will change kinetic energy

Fd = ½mv²

v = √(2Fd/m) = √(2(5.0)(10) / 2.0) = √50 = 7.07... m/s

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Read 2 more answers

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

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Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

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By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

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6 0

3 years ago

A man of 70 kg climbs a stair. He climbs 15 steps in 1 minute. If the height of 1 step is 15 cm . Calculate his power . (G=9.8 m

UkoKoshka [18]

Answer:

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7 0

2 years ago

A gold wire has a cross-sectional area of 1.0 cm^2 and a resistivity of 2.8 × 10^-8 Ω ∙ m at 20°C. How long would it have to be

Karo-lina-s [1.5K]

Answer:

Length, l = 3.57 meters

Explanation:

It is given that,

Area of cross-section of gold wire, $A=1\ cm^2=0.0001\ m^2$

Resistivity of gold wire, $\rho=2.8\times 10^{-8}\ \Omega-m$

Resistance, R = 0.001 ohms

Resistance in terms of length and area is given by :

$R=\rho \dfrac{l}{A}$

$l=\dfrac{RA}{\rho}$

$l=\dfrac{0.001\times 0.0001}{2.8\times 10^{-8}}$

l = 3.57 meters

So, the length of the wire is 3.57 meters. Hence, this is the required solution.

4 0

3 years ago