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ra1l [238]
3 years ago
5

PLEASE HELP ME

Physics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

where's the graph?? I need the graph

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Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3
Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

k_{1}=20\ N/m

k_{2}=30\ N/m

k_{3}=15\ N/m

k_{4}=20\ N/m

k_{5}=35\ N/m

According to figure,

k_{2} and k_{3} is in series

We need to calculate the equivalent

Using formula for series

\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}

k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}

Put the value into the formula

k=\dfrac{30\times15}{30+15}

k=10\ N/m

k and k_{4} is in parallel

We need to calculate the k'

Using formula for parallel

k'=k+k_{4}

Put the value into the formula

k'=10+20

k'=30\ N/m

k_{1},k' and k_{5} is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}

Put the value into the formula

k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}

k_{eq}=8.93\ N/m

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0
3 years ago
Please solve this question ​
lesantik [10]

Answer:

88200 Pa

it is because

height =9m

density=1000kg/m(cube)

gravity = 9.8m/s(square)

now,

P=d×g×h

= 1000×9.8×9

=88200pa

8 0
2 years ago
such smut. when he asks how this goal can be accomplished he volunteers to preview every film that will be shown in town. If Ada
juin [17]
What does this all mean sorry ignore me
6 0
2 years ago
To the nearest square foot, how many square feet are there in an area of 4.4 square meters?
vichka [17]

Answer:

    4.4 square meters = 47 square foot

Explanation:

We have

    1 meter = 3.28084 foot

    1 square meter = 3.28084 x 3.28084 square foot = 10.76 square foot

    4.4 square meters = 4.4 x 10.76 = 47.36 square foot = 47 square foot

    4.4 square meters = 47 square foot

3 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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