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worty [1.4K]
2 years ago
13

PLEASE HELP ME!!!!!!!!!

Physics
1 answer:
dexar [7]2 years ago
3 0
7.5 I think because it can not be 9 because it’s not close to 50
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What is quantitative?
7nadin3 [17]
<span>To relate or measure the by the quantity of something, not against the quantity</span>
5 0
4 years ago
The moon's surface gravity is one-sixth that of the earth. Calculate the weight on the moon of an object that has a mass of 24 k
ad-work [718]
When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
we mean that the acceleration of gravity on the Moon's surface is 1/6 of
the acceleration of gravity on the Earth's surface.

The acceleration of gravity is (9.8 m/s</span>²) on the Earth's surface, so
<span>it would be (9.8/6 m/s</span>²) on the Moon's surface.
<span>
The weight of any object, right now, is

(object's mass) </span>· (acceleration of gravity where the object is located now) .
<span>
If the object's mass is 24 kg and the object is on the Moon right now,
then its weight is 

(24 kg) </span>· (9.8/6 m/s²)

= (24 · 9.8 / 6) kg-m/s²

= 39.2 Newtons
7 0
3 years ago
A bat emitts a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away
Varvara68 [4.7K]

The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.

From the question given above, the following data were obtained:

Velocity of sound (v) = 343 m/s

Distance (x) = 8.42 m

Time (t) =?

We can obtain obtained the time as illustrated below:

v = 2x / t

343 = 2 × 8.42 / t

343 = 16.84 / t

Cross multiply

343 ×  t = 16.84

Divide both side by 343

t = 16.84/343

t = 0.05 s

Thus, the time between  when the bat emits the sound and when it hears the echo is 0.05 s.

<h3>How does a bat know how far away something is?</h3>

A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.

Learn more about time elapses between when the bat emits the sound :

<u>brainly.com/question/16931690</u>

#SPJ4

Correction question:

A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)

8 0
2 years ago
A TV set is pushed a distance of 2 m with a force of 20 N how much work is done on the set ​
vladimir2022 [97]

Answer:

40 J

Explanation:

8 0
3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

4 0
3 years ago
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