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3 years ago

6

Amy wants to know whether or not an item will float when placed in a fluid. Which of the comparisons below, when true, will mean

that the item will float?

1 answer:

o-na [289]3 years ago

4 0

A. The object is made of stone or metal, which always sink.

B. The force of gravity is stronger than the buoyant force.

C. The buoyant force is stronger than the force of gravity.

D. The mass of the item is greater than its gravity

The answer is C

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Compute the dot product of the vectors u and v, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and

OLga [1]

Answer:

$\theta = 106.3 degree$

Explanation:

As we know that

$\vec w = -\hat i + 7\hat j$

$\vec v = 7\hat i - \hat j$

also we know that

$\vec v. \vec w = -14$

it is given as

$\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)$

$\vec v. \vec w = - 7 - 7 = -14$

also we can find the magnitude of two vectors as

$|v| = \sqrt{(-1)^2 + (7)^2}$

$|v| = \sqrt{50}$

similarly we have

$|w| = \sqrt{(7^2) + (-1)^2}$

$|w| = \sqrt{50}$

now we know the formula of dot product as

$\vec v. \vec w = |v||w| cos\theta$

$-14 = (\sqrt{50})^2cos\theta$

$\theta = cos^{-1}(\frac{-14}{50})$

$\theta = 106.3 degree$

3 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

a brick is suspended above the ground at a height of 6.6 m. it has a mass of 5.3 kg. what is the potential energy of the brick

Svetradugi [14.3K]

The formula for potential energy is
E(p) = mgh

(Mass x gravity x height)

Therefore energy = (5.3)(9.8)(6.6)
= 342.8 J

How did I get 9.8?
9.8 is the constant for gravity

8 0

3 years ago

HURRY Which change is an example of transforming potential energy to kinetic energy

Jet001 [13]

Answer:

C. changing nuclear energy to radiant energy

Explanation:

Nuclear energy takes atoms in their potential state, split them (fission) or fuse them (fusion) creating chain reactions of radiant energy. Most nuclear electrical power plants use fission, radiant energy heats water making steam to spin turbines.

Or think of the atom bomb. Definitely potential energy until the fuse starts detonation and chain reactions. The radiant kinetic energy and shock waves were horrendous.

3 0

3 years ago

Read 2 more answers

Enumerate the viscosity of magma in different conditions.

Firlakuza [10]

Answer:

The nature of volcanic eruptions is highly dependent on magma viscosity and also on dissolved gas content. ... long it takes the treacle to flow from one end of a boiling tube to the other.

3 0

3 years ago