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3 years ago

12. The resistivity of a given wire of cross-sectional

Physics

1 answer:

luda_lava [24]3 years ago

3 0

R = resistivity × (l/A), A = πr^2= πd^2/4

2 =( 4.9 × 10^ -4) * (l/π0.7^2/4)

make l the subject and calculate.

i assumed we are looking for the length if the wire.

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Sam moves a box with a with a force of 400N a distance of 5 meters. How long did it take him to move the box if 20 Watts of powe

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100s

Explanation:

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An automobile with tires that have a radius of 0.260 m travels 76000 km before wearing them out. How many revolutions do the tir

Tomtit [17]

Answer:

Explanation:

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3 years ago

The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A

steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

$\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}$

$\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}$

Net force

$\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}$

$\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}$

$\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}$

$\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}$

$\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}$

Work done is defined as

$W = \overrightarrow{F}.\overrightarrow{S}$

$W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k} \right )$

W = -1120 + 450 + 70

W = - 600 J

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3 years ago

A skater has outstretched arms preparing for a turn with a moment of inertia of

Helen [10]

Answer:

The final angular speed is 16.1 rad/s

Explanation:

Given;

initial moment of inertia, I₁ = 2.56 kg.m²

final moment of inertia, I₂ = 0.40 kg.m²

initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s

Apply the principle of conservation of angular momentum;

I₁ω₁ = I₂ω₂

where;

ω₂ is the final angular speed

ω₂ = (I₁ω₁) / (I₂)

ω₂ = (2.56 x 2.514) / (0.4)

ω₂ = 16.1 rad/s

Therefore, the final angular speed is 16.1 rad/s

7 0

3 years ago