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3 years ago

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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The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

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3 years ago

A small boat sailed straight north out of a harbor in strong east wind (blowing from west to east). After sailing for 120 minute

Amiraneli [1.4K]

it can be said that the speed of the east wind is

v=0.3608m/s

From the question we are told

A small boat sailed <u>straight </u>north out of a harbor in <em>strong </em>east wind (blowing from west to east).

After sailing for 120 minutes, it ended up hitting a buoy 60^\circ60 ∘ to the north-east of the harbor. If the straight-line distance between the buoy and the harbor is 3 km,

what is the speed of the east wind?.

<h3> the speed of the east wind</h3>

Generally the equation for the distance is mathematically given as

BA=3000sin60

BA=2598.07m

Therefore

the speed of the east wind

$V_w=\frac{BA}{120*60}\\\\V_w=\frac{2598.07}{120*60}$

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For more information on this visit

brainly.com/question/22568180

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