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Vikentia [17]
3 years ago
5

How is the smell from a stinky diaper similar to the electric field produced by a charge

Physics
1 answer:
sukhopar [10]3 years ago
6 0
Both of them are unpleasant!
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A spring scale exerts a net force of 8.5 N on an object. What is the object's mass if it has an acceleration of
Amiraneli [1.4K]

Answer:

mass of the object is 2.18 kg

Explanation:

Given

Force (F) = 8.5 N = 8.5 kg.m/s^{2}

acceleration (a) = 3.9 m/s^{2}

Mass (m) = ?

We know that the newton's second law of motion gives the relation between mass of ab object. force acted upon and the amount the object is accelerated. It is expressed in the form of an equation:

F = ma

mass, m = F/a

               = \frac{8.5 kgms^{-2} }{3.9 ms^{-2} }

               = 2.18 kg

4 0
3 years ago
The moon is significantly smaller than the sun. Why is the moon able to fully block the sun from our view on
makvit [3.9K]

Answer:

The moon is 400x smaller but it's also 400x closer so it looks the same size even though it's not

Explanation:

5 0
3 years ago
Which advantage of stored digital data could be a disadvantage in some
Lesechka [4]

Answer:B- All types of information can be pressed on to receivers of the data.

Explanation: hope this helps

4 0
3 years ago
Bobo, the clown, can swim at 2.0 m/s. he must make a landing directly across to the north side of the styx river, which is 100.
victus00 [196]
We are given the following:

Bobo's swimming speed = 2.0 m/s
Width of the river = 100 m
Flowrate of the river = 6.0 m/s due east

First, we need to illustrate the problem. Draw the river with a width of 100 meters. Then, the flow of the river, east at 6 meters per second. Lastly, draw Bobo at one side of the river facing north and an arrow representing swimming speed at 2 meters per second.

Now, we can use the Pythagorean theorem to solve this rate problem.

c^2 = a^2 + b^2 

c = speed of Bobo needed
a = speed of Bobo facing north
b = flow rate of the river going east

c^2 = 2^2 + 6^2 

c = 6.32 m / s should be his speed to overcome the current and make a landing at the desired location. 
6 0
3 years ago
9) An object with a height of 18 cm is placed in front of a converging lens. The image has a
vovikov84 [41]

Answer:

Explanation:

a) Magnification = image height / object height = -9 / 18 = -0.5

b) Magnification = - image distance / object distance = -0.5

so image distance = 0.5 object distance

1/focal length = 1/image distance + 1/object distance

1/6 = 1/(0.5 object distance) + 1/object distance

object distance = 18.0 cm

c) Image appears behind the lens.

6 0
3 years ago
Read 2 more answers
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