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Vladimir79 [104]
3 years ago
7

A cat has a mass of 4.6 kg. What is its weight on Earth's surface?

Physics
2 answers:
kap26 [50]3 years ago
6 0

Answer:

Choice b is correct answer.

Step-by-Step explanation:

We have given the mass of a cat.

mass = m = 4.6kg

We have to find the  weight of cat on earth' surface.

The formula to find weight is:

weight = mass × gravity

w = mg

The value of gravity on earth surface is 9.8 m/sec².

Putting values in above formula, we have

w = 4.6 × 9.8 (kgm/sec²)              ∴   kgm/sec²  =  Newtons

w = 45 Newtons which is the answer.

andreyandreev [35.5K]3 years ago
4 0
<h3>✽ - - - - - - - - - - - - - - -  ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Earth's gravity is approximately 9.81

weight = mass x gravity

weight = 4.6 x 9.81

weight = 45.126

Answer is B. 45N

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

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1104 km/hour

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A space station has a large ring-like component that rotates to simulategravity for the crew. This ring has a massM= 2.1×105kg a
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Answer:

Each thruster has to applied a force of 294.5N in tangential direction

Explanation:

Mass of the ring ,M =2.1×105kg

Mass of the ship ,m = 3.5×104kg

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distance of ship from center of the ring, r =31.0 m

Let force applied by each thruster be F

Time taken to reach  gravity ,t = 3hrs = 3600× 3 =10800sec

The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.

Centrifugal force = weight of the object on earth

Assume the ring is moving with angular speed ω

Centripetalforce of the object kept at ring

m₁R ω²=m₁g  (m₁=mas of object)

⇒Rω² = g

⇒ω = √g/R

The ring start from 0 angular speed with constant angular acceleration

Let the constant angular acceleration be ∝

∝ = ω  / t

(ω = √g/R)

∝ = \frac{1}{t } \sqrt{\frac{g}{R} } }

Consider Torque on the ring and ship system

T = FR + FR = 2FR

Moment of inertia of ring ship system

I = I(ring)+I(ship)+I(ship)

= MR² + mr² + mr² = MR² + 2mr²

angular acceleration of the ring ship system

∝ = \frac{2FR}{MR^2 + 2mr^2}

Now we have ,

∝ = \frac{1}{t} \sqrt{\frac{g}{R} }  , ∝ = \frac{2FR}{MR^2+2mr^2}

equating both values

We have,

F = \frac{MR^2+2mr^2}{2Rt} \sqrt{\frac{g}{R} }

where,

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r = 31m, g = 9.8m/s² , t = 10800sec

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