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3 years ago

15

Can you help me in any of these questions!!

2 answers:

LiRa [457]3 years ago

8 0

A and B are right but what grade are you in

Oxana [17]3 years ago

5 0

1.
a. The one with the greater volume will rise slower than the one of the left. b. It would make the readings high than they were before.

b. it will make the readings higher than they were before

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39. I also use it when I do the homework. (par 1 line 3).

Yuri [45]

Answer:

A

Explanation:

4 0

2 years ago

Read 2 more answers

In the picture above, what is the period of the pendulum?<br> (Remember to include a unit)

-BARSIC- [3]

69 i agree with her hope this helps

6 0

2 years ago

a desktop computer and monitor together draw about 2 A of current they plug into a wall outlet that is 120 V what is the Resista

Delvig [45]

Answer:

$60 \Omega$

Explanation:

the relation between current, voltage and resistance in an electrical circuit is given by Ohm's law:

$V=IR$

where V is the voltage, I is the current and R is the resistance. In this problem, the current is I=2 A, the voltage is V=120 V, therefore we can arrange the previous equation and find the resistance:

$R=\frac{V}{I}=\frac{120 V}{2 A}=60 \Omega$

7 0

3 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

3 years ago

A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr

Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

brainly.com/question/15365772

#SPJ4

7 0

2 years ago