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4 years ago

70 kg to mg i need to show the work of how i did it

Physics

1 answer:

lyudmila [28]4 years ago

4 0

Answer:

70000000 mg

Explanation:

1 gram (kg) is equal to 1000000 milligrams (mg).

1 kg = 106 mg = 1000000 m

7 kg × 1000000

= 7000000 mg

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In which situation will the lowest resistance occur?

insens350 [35]

Answer:

thick wire and cold temperatures

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In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V = Velocity of actor

$h_i$ = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

$\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s$

As the linear momentum is conserved

$MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s$

Applying conservation of energy again

$\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m$

The maximum height they reach is 1.57772 m

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3 years ago

The type of force measured by grocery store spring scale is

alexandr402 [8]

Answer:

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Explanation:

The spring balance is used to measure weight of an object.

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3 years ago

2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another

worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

P_i = P_f

0 = m_1*V_1 - m_2*V_2

- Evaluate: m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify: 0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify: 0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

(V_1 / V_2)^2 = 2

(V_2 / V_1)^2 = 1 / 2

- simplify: (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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3 years ago

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Answer:

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Explanation:

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3 years ago