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grin007 [14]
2 years ago
13

You just found a metallic object in a creek bed. you have found its mass to be 96.5 g, and it displaces 5mL of water. what is th

e substance?
A.) aluminum
B.)iron
C.)gold
D.)none of the above
Physics
2 answers:
MaRussiya [10]2 years ago
4 0
Judging by the density (19.3 g/ml) there's a very good chance that it could be gold.
Amanda [17]2 years ago
4 0

Answer:

C) gold

Explanation:

Given:

Mass of the object = 96.5 g

Volume of water displaced= 5 ml

Formula:

Density = \frac{Mass}{Volume} \\Density of metal =\frac{96.5 g}{5ml} = 19.3 g/ml

The known densities of Al, Fe and Au are:

Density of Al = 2.7 g/ml

Density of Iron (Fe)= 7.87 g/ml

Density of gold (Au) = 19.3 g/ml

Therefore, the metallic object is gold

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What is the biggest similarity between sports in the United States and sports in other cultures?
leva [86]

Answer:

The first difference is fairly straightforward: other countries prefer other sports. While some aspects of American sports culture are universal there are other things that are sports-specific, such as the 7th-inning stretch at a baseball game or counting the steps of a basketball player after he fouls out.

Explanation:

4 0
2 years ago
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You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon
Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

u_{1} = 0m/s

t_{1} = t

u_{2} = 43.12m/s

t_{2} = (t-2.2)s

The distance by the first balloon is

D = u_{1} t_{1}  + \frac{1}{2} at_{1}^2

where

a = 9.8m/s2

Inputting the values

D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2

The distance traveled by the second balloon

D = u_{2} t_{2}  + \frac{1}{2} at_{2}^2

Inputting the values

D = (43.12)(t-2.2)  + \frac{1}{2} (9.8)(t-2.2)^2

simplifying

D = 4.9t^2 + 21.56t -71.148

Substituting D of the first balloon into the D of the second balloon and solving

4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

D = 4.9(3.3)^2\\ D = 53.361 m

7 0
3 years ago
A moving object is in equilibrium. Which best describes the motion of the object if no forces change?
Alchen [17]
If the object is in equilibrium that means that the sum of the forces on it is zero and the net force is zero. If none of the forces changes then the object continues in constant uniform motion. That means constant speed in a straight line.
8 0
3 years ago
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A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
IgorLugansk [536]

Answer:

Approximately 0.077\; {\rm m\cdot s^{-1}} (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Momentum of the t-shirt:

\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

p(\text{t-shirt}) + p(\text{cannon}) = 0.

p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}.

Rewrite p = m\, v to obtain v = (p / m). Since the mass of this cannon is m(\text{cannon}) = 33\; {\rm kg}, the velocity of this cannon would be:

\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}.

8 0
1 year ago
A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor di
Marrrta [24]

Answer:

2625 m deep

Explanation:

Let the sound speed in sea water be 1500 m/s. If he hears the echo 3.5s after the strike, then the sound would have traveled a distance of 1500 * 3.5 = 5250 m to the bottom and back. This would mean the ocean is 5250 / 2 = 2625 m deep.

5 0
3 years ago
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