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NemiM [27]
2 years ago
10

You are camping in the breathtaking mountains if Colorado. You spy an unopened diet soda can floating motionless below the surfa

ce of a lake. What is the direction and amount of force the water exerts on it?
A. Zero
B. Down, equal to the can's weight
C. Up, equal to the can's weight
D. Not enough information is given
Physics
1 answer:
Pavel [41]2 years ago
7 0

Answer:

C. Up, equal to the can's weight

Explanation:

You are camping in the breathtaking mountains if Colorado. You spy an unopened diet soda can floating motionless below the surface of a lake. What is the direction and amount of force the water exerts on it?

A. Zero

B. Down, equal to the can's weight

C. Up, equal to the can's weight

D. Not enough information is given

from the principle of flotation which states that a

When a body displaces a weight of water equal to its own weight, it floats. : A floating object displaces a weight of fluid equal to its own weight. ... Archimedes' principle equates the buoyant force to the weight of the fluid displaced.

the upthrust (this is the upward vertical force exerted on an object in fluid)in the water equals the weight of the body in water it floats.

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An automobile moves at a constant speed over the crest of a hill traveling at a speed of 88.5 km/h. At the top of the hill, a pa
Sergio039 [100]

Answer:

r=61.65m

Explanation:

Since the package remains in contact with the car's seat, the package's speed is equal to the car's speed. At the top on the mountain the package's centripetal force must be equal to its weight:

mg=F_c

The centripetal force is defined as:

F_c=ma_c=\frac{mv^2}{r}

Here v is the linear speed of the object and r is the radius of curvature. We need to convert the linear speed to \frac{m}{s}:

88.5\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=24.58\frac{m}{s}

Now, we calculate r:

mg=\frac{mv^2}{r}\\r=\frac{v^2}{g}\\r=\frac{(24.58\frac{m}{s})^2}{9.8\frac{m}{s^2}}\\\\r=61.65m

3 0
2 years ago
What is providing the centripetal force in the following examples?
umka2103 [35]
A yo-yo swung in a circle.
8 0
3 years ago
Read 2 more answers
Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecul
pogonyaev

Hydrogen bonds are too weak to bind atoms together to form molecules, but they do hold different parts of a single large molecule in a specific three-dimensional shape. The given statement is true.

<h3>What are hydrogen bonds?</h3>

A hydrogen bond is an electrostatic force of attraction among a hydrogen atom tightly attached to a more electronegative "donor" atom or group and another electronegative atom bearing a lone pair of electrons, known as the hydrogen bond acceptor.

Hydrogen bonds are too flimsy to connect atoms to form molecules, but they do hold various portions of a single large molecule together in a specific three-dimensional shape.

Thus, the given statement is true.

For more details regarding hydrogen bonding, visit:

brainly.com/question/10904296

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5 0
1 year ago
Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 75.9 N to the north. Jenny’s foot exerts
Lady_Fox [76]

Answer:

Fr^2 = 75.9N+105.8N=181.7

<u><em>Fr = </em></u><u><em>181.7N.</em></u>

6 0
2 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
3 years ago
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