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frozen [14]
4 years ago
14

Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown

made of pure (100%) gold? The density of water is rhow=1.00 grams per cubic centimeter. The density of gold is rhog=19.32 grams per cubic centimeter.
Physics
1 answer:
9966 [12]4 years ago
4 0

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

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tresset_1 [31]

(20 miles) x ( 1/45  hour/mile) = 

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3 years ago
A stream of doubly ionized particles (2 protons) moves at a velocity of 3.0 × 104 m/s perpendicularly to a magnetic field of 0.0
max2010maxim [7]

The force on a charged particle on a magnetic field is given by:

F=Bqv\sin\theta

where B is the magnitude of the field, q is the charge of the particle, v is its velocity and θ is the angle between the field and the velocity of the particle.

In this case we have that:

• The magnitude of the field is 0.09 T.

,

• The charge of the particle is 3.2 x 10 –19C .

,

• The velocity is  3.0 × 10 4 m/s

,

• The angle between the field and the velocity is 90°, since they are perpendicular.

Plugging these we have:

\begin{gathered} F=(0.09)(3.2\times10^{-19})(3\times10^4)\sin90 \\ F=8.64\times10^{-16}\text{ N} \end{gathered}

Therefore, the force on the charge is:

F=8.64\times10^{-16}\text{N}

3 0
2 years ago
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4 0
3 years ago
Read 2 more answers
⦁ A baseball is struck by a bat 46 cm from the axis of rotation when the angular velocity of the bat is 70 rad/s. If the ball is
photoshop1234 [79]

Answer:

No, the ball will not clear the fence.

Solution:

Angular velocity, \omega = 70\ rad/s

Height, h = 1.2 m

Angle, \theta = 45^{\circ}

Distance covered by the ball, d = 110 m

Length of the fence, l = 1.2 m

Radius of the axis, R = 46 cm = 0.46 m

Now,

To calculate the linear velocity of the ball, v:

v = \omega R = 70\times 0.46 = 32.2\ m/s

Total time taken:

t = \frac{2vsin\theta}{g} = \frac{2\times 32.2sin45^{\circ}}{9.8} = 4.646\ s

The distance at which the ball falls, with a = 0 is given by:

x = vt + \frac{1}{2}at^{2} = 32.2cos45^{\circ}\times 4.646 = 105.78\ m

Since, the ball has to clear a fence 1.2 m long and a t a distance 110 m away, clearly it will not be able to cross it.

6 0
3 years ago
Suppose you have a 113-kg wooden crate resting on a wood floor. (μk = 0.3 and μs = 0.5) (a) What maximum force (in N) can you ex
AURORKA [14]

Answer:

554.27N

Explanation:

(a)  The max frictional force exerted horizontally on the crate and the floor is,

Substitute the values,

μs=0.5

mass=113kg

g=9.81m/s

Ff=μsN

   =μsmg

   =(0.5 x 113 x 9.81)

Ff=554.27N

3 0
3 years ago
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