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Bess [88]
3 years ago
7

How does momentum conservation work for a rocket firing its engines to speed up in deep space?

Physics
1 answer:
Mashutka [201]3 years ago
3 0
Idkhhhhhhhhhhubvgbvcccc xzzz. Bcc bbb
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a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

8 0
3 years ago
What is the formula for calculating the efficiency of a heat engine?
Anton [14]

Answer:

Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.

7 0
3 years ago
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What is a disadvantage of using biomass as an energy resource?
Rama09 [41]

Answer:

hope this helps i think the answer is C

5 0
3 years ago
A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.
vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

           F = B i L

Rod is not necessarily vertical

F_x =i L B_d

F_y= i L B_w

the normal reaction N = mg-F y

static friction       f = μ_s (mg-F y )

horizontal acceleration is zero

F_x-f = 0

iLBd = \mu_s(mg-F_y )

 B_w = B sinθ

 B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}

\theta =tan{-1}{\mu_s}

\theta =tan{-1}{0.6}

\theta = 31^0

B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}

       B = 0.1 T

4 0
3 years ago
PLEASE HELP ASAP WILL GIVE BRAINLIEST!!!!
nydimaria [60]

Explanation:

Please mark me as the brainliest answer

7 0
3 years ago
Read 2 more answers
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