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mina [271]
2 years ago
9

12 points , just need help with this question

Physics
1 answer:
tigry1 [53]2 years ago
3 0

Answer:f 30

Explanation: I am not really sure but try this

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A science class puts a balloon containing 1.25 l of air at 101 kpa into a bell jar. using an air pump, the class removes some of
storchak [24]
The first thing you should know in this case is the following definition:
 PV = nRT
 Then, as the temperature is constant, then:
 PV = k
 Then, we have two states:
 P1V1 = k
 P2V2 = k
 We can then equalize both equations:
 P1V1 = P2V2
 Substituting the values:
 (1.25) * (101) = (2.25) * (P2)
 Clearing P2:
 P2 = ((1.25) * (101)) /(2.25)=56.11Kpa
 answer:
 the new pressure inside the jar is 56.11Kpa
8 0
3 years ago
If an object's velocity changes from 25 meters per second to
damaskus [11]

Answer:

B

Explanation:

the way to get b you have divion each other

3 0
2 years ago
Read 2 more answers
Who can do my worksheet for me
timurjin [86]

Answer:

what  is it on? like name one of the questions

Explanation:

5 0
3 years ago
A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0
3 years ago
Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
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