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Ksju [112]
2 years ago
13

Find the force necessary to pull a 6 kg object 3 m/s2

Physics
1 answer:
mars1129 [50]2 years ago
6 0

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 6 × 3

We have the final answer as

<h3>18 N</h3>

Hope this helps you

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What are stepdown transformers used for
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3 years ago
If velocity is in m/s and time is in sec, what is the unit of acceleration?​
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2 years ago
A 1.60 m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the
Viefleur [7K]

Answer:

Explanation:

Given

mass of book(m)=2.1 kg

height up to which book is lifted is (h)2.2 m

height of person (h_0)1.6 m

Potential energy of book relative to ground=mgh

PE=2.1\times 9.8\times 2.2=45.276 J

(b)PE w.r.t to person head =mg(h-h0)

=2.1\times 9.8\times (2.2-1.6)=12.348 J

work done by person in lifting box 2.2 m w.r.t floor

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6 0
3 years ago
A 7.75-l flask contains 0.482 g of hydrogen gas and 4.98 g of oxygen gas at 65°c. What is the partial pressure of oxygen in the
dimulka [17.4K]

Answer:

0.558 atm

Explanation:

We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT

Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:

P₀ = P₁ + P₂ + ....

P₀= total pressure

P₁=P₂= is the partial pressure of each gass

If we can consider that each gas is an ideal gas, then:

P₀= (nRT/V)₁ + (nRT/V)₂ +..

Considering the molecular mass of O₂:

M O₂= 32 g/mol

And also:

R= ideal gas constant= 0.082 Lt*atm/K*mol

T= 65°C=338 K

4.98 g O₂ = 0.156 moles O₂

V= 7.75 Lt

Then:

P°O₂=partial pressure of oxygen gas=  (0.156x0.082x338)/7.75

P°O₂= 0.558 atm

3 0
3 years ago
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