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Salsk061 [2.6K]
3 years ago
5

If 400 pa of pressure is applied to an area of 55 m2, which is the resulting force?345 n7.27 n22,000 n

Physics
2 answers:
VashaNatasha [74]3 years ago
7 0
The answer should hopefully be: 7.27N
Lady bird [3.3K]3 years ago
6 0
Use formula P=F/a
since pressure is 400Pa and area is 55m2 ..
400=F/55 then take 55 to the other side by multiplying on both side .. we get 400x55
which is equal to 22000.
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From your results on page 29, decide whether you are able to demonstrate that the masses of the 4 different baskets differ meani
Gekata [30.6K]

Yes, it was demonstrated that the masses of the 4 different baskets differ meaningfully.

If we compare the two ranges, the differences are significant because the error range for the same basket was less than the differences between different baskets.

The Error range is a measurement of the degree of uncertainty attached to a number and is defined as the difference between the greatest and lowest error values.

So, Yes the baskets masses differ by more than the range of measurement error that was obtained by repeatedly measuring the mass of one basket.

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6 0
2 years ago
If water is leaking from a certain tank at a constant rate, is it leaking at a rate that is greater than 12 liters per hour?(1 l
Margarita [4]

Answer:

Yes  it is possible because 3.33 is greater than 2.

Explanation:

Given that,

1 liter = 1000 milliliters

Suppose water is leaking from the tank at a rate that is greater than 2 milliliters per second.

We need to change the unit in ml/s

12 lit/hr=\dfrac{12000}{3600}

12 lit/hr=3.33\ ml/s

Hence, Yes  it is possible because 3.33 is greater than 2.

5 0
3 years ago
Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of th
Vilka [71]

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x F_{s} = τ1

F_{s} = the static frictional force (N)

F_{s} = τ1 /r

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7 0
4 years ago
What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?
IRISSAK [1]

Answer:

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction, \vec{F_{up}} = 3.50\times 10^{- 5} N

Charge, Q = - 1.55\micro C = - 1.55\times 10^{- 6} C

Now, we know by Coulomb's law,

F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}

Also,

Electric field, E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}

Thus from these two relations, we can deduce:

F = QE

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\vec{E} = \frac{\vec F_{up}}{Q}

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Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

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6 0
3 years ago
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Zigmanuir [339]
Ernest Rutherford is the answer you are looking for my friend.
5 0
4 years ago
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