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Natalija [7]
3 years ago
14

The volume of the gas at a temperature of 50 Kelvin on this scale is closest to

Physics
1 answer:
8_murik_8 [283]3 years ago
4 0
<span>500 cubic centimeters</span>
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An incident ray through the focal point of a spherical mirror will:
Delvig [45]
Reflect parallel of the principal axis
3 0
3 years ago
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a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo
12345 [234]

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

W = F\times d

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute W = 80.2\ J\ and\ d =25.0\ m in work done formula.

80.2 = F\times 25

F=\frac{80.2}{25}

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

3 0
3 years ago
Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors?
OlgaM077 [116]

Answer:

Explanation:

The resistors in a unbalanced wheat stone bridge cannot be treated as a combination of series and parallel combination of resistors.

In case of balanced wheat stone bridge, the resistors can be treated as the combination of series and parallel combination.

Here, In the balanced wheat stone bridge

R1 and R2 be in series and Ra and Rx is series and then their combination is in parallel combination.

4 0
3 years ago
A truck initially traveling at a speed of 22 meters per second increases speed at a constant rate of 2.4 meters per second^2 for
Usimov [2.4K]
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:

d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m 
Hope the answer helps. 
8 0
3 years ago
Which is most likely the length of a student’s textbook?
UkoKoshka [18]

For this case, what we must do is to rewrite these measurements in the same unit in order to compare them.

By writing the measurements in meters we have:

30 mm = (30) * (\frac{1}{1000}) = 0.030 m

30 cm = (30) * (\frac{1}{100}) = 0.30 m

30 dm = (30) * (\frac{1}{10}) = 3 m\\30 Hm = (30) * (100) = 3000 m

Therefore, physically the correct measure is:

0.30 m = 30 cm

Answer:

the length of a student's textbook most likely is:

30 centimeters

3 0
3 years ago
Read 2 more answers
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