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Naddik [55]
3 years ago
11

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

he impact: one travels in the air and the other in the concrete, and they are 0.90s apart. The speed of sound in air is343 m/s, and in concrete is 3000 m/s.How far away did the impact occur?
Physics
1 answer:
Jet001 [13]3 years ago
4 0
<h2>The impact is occurred at a distance 348.55 m far from person.</h2>

Explanation:

Let s be the distance to the impact position and t be the time when he hears the sound though concrete after impact.

Time when he hears impact through air = t + 0.90

The distance traveled by sound wave in both concrete and air is s.

Speed of sound in air = 343 m/s

Speed of sound in concrete = 3000 m/s

Distance traveled in air = Distance traveled in concrete

343 x Time when he hears impact through air = 3000 x Time when he hears impact through concrete

        343 x (t + 0.90) = 3000 x t

                  3000t - 343 t = 343 x 0.90

                  2657t = 308.7

                      t = 0.116 s

Distance traveled in concrete = 3000 x t = 3000 x 0.116 = 348.55m

So the impact is occurred at a distance 348.55 m far from person.

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Which of these properties of an object best quantifies its inertia: velocity, acceleration, volume, mass, or temperature?
LuckyWell [14K]

Answer:

Mass

Explanation:

Inertia is essentially an object's tendency to stay in motion or at rest unless it is forced to do otherwise (pun intended). It only makes sense to me that mass would best quantify an object's inertia, because an object with more mass would be harder to move and/or stop from moving.

3 0
3 years ago
Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at t
il63 [147K]

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

6 0
3 years ago
Will give brainliest
NNADVOKAT [17]

I love science! if you need any more help let me know i cant guarantee i can help but i will try!

The correct answer is  "Some substances must be dissolved in water before they can be used".  

5 0
3 years ago
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
I am Lyosha [343]

Answer:

The density of this object is approximately 1.36\; {\rm kg \cdot L^{-1}}.

The density of the oil in this question is approximately 0.600\; {\rm kg \cdot L^{-1}}.

(Assumption: the gravitational field strength is g =9.806\; {\rm N \cdot kg^{-1}})

Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

Conversely, if the weight of an object is (\text{weight}) in a gravitational field of strength g, the mass m of that object would be m = (\text{weight}) / g.

Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}.

When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.

When this object was suspended in water, the buoyancy force on this object was (63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}. Hence, the weight of water that this object displaced would be 47.0 \; {\rm N}.

The mass of water displaced would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}.

The volume of that much water (which this object had displaced) would be:

\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}.

Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately 4.793\; {\rm L}.

The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately 4.793\; {\rm L}.

The weight of oil displaced would be equal to the magnitude of the buoyancy force: 63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}.

The mass of that much oil would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}.

Hence, the density of the oil in this question would be:

\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}.

(Rounded to \text{$3$ sig. fig.})

7 0
2 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ball B has a mass of 3 kilograms. The initial v
ratelena [41]
1. The balls move to the opposite direction but the same speed. This represents Newton's third law of motion.
2. The total momentum before and after the collision stays constant or is conserved.
3. If the masses were the same, the velocities of both balls after the collision would exchange.
4 and 5. Use momentum balance to solve for the final velocities.
8 0
4 years ago
Read 2 more answers
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