All categories

3 years ago

Knowing that 0.80kg object weighs 8.0n find the acceleration of a 0.80kg stone in free fall

Physics

2 answers:

DENIUS [597]3 years ago

7 0

This question can be easily solved with newton's second law of motion.
F=ma
a=F/m
=8/0.8
=10m/s^2

vovikov84 [41]3 years ago

4 0

Answer : $a = 10 m/s^{2}$

Explanation :

Given that,

Mass = 0.80 kg

Force = 8.0 N

According to newton second law

$F = ma$

$a = \dfrac{F}{m}$

$a = \dfrac{8.0 N}{0.80kg}$

The acceleration is

$a = 10 m/s^{2}$

Hence, this is the required solution .

You might be interested in

Air resistance ____ as you move faster. Question 17 options: increases decreases remains the same eventually disappears

Ket [755]

Explanation:

Air resistance drag as you move faster

5 0

3 years ago

Read 2 more answers

I need help fast!!!

Fantom [35]

You could easily help yourSELF if you simply follow the instructions in the question and "Use Ohm's law" to calculate the resistance.

Ohm's Law: Resistance = (voltage) / (current)

Resistance = (10 v) / (5 A)

<em>Resistance = 2 ohms</em>

8 0

3 years ago

You are writing a report about space travel. Where would be the best place to find information? (2 points)

Harrizon [31]

Answer: Contact the National Aeronautics and Space Association

Explanation:

The National Aeronautics and Space Association is an independent organization in US which aims to enable a safe, secure, efficient space journey and launch of new satellites, spacecraft beyond the earth's orbit. It can give better information for the space travel already conducted by the astronauts. Thus will help in writing the report as it is the best place to find the information.

5 0

3 years ago

Read 2 more answers

You sit "at rest" in front of your computer to answer this question. But you sit on the surface of a planet that spins, so even

igomit [66]

Answer: Linear speed is 1,670 Kph.

Explanation:

If we assume that the earth is a perfect sphere, and that is spinning itself once every roughly 24 hr, we can get the angular velocity of the Earth, in magnitude, as follows:

ω = 2π / 24 Hr

Now, by definition, an angle is the relationship between the arc s, and the radius r, so we can replace these values in the angular velocity expression, as follows:

ω = (Δs / r) . 1/Δt ⇒ ω = (Δs/Δt). 1/r

But, by definition, Δs/At, is just the linear velocity, v, so we can conclude the following;

ω = v/r ⇒ v = ω. r

So, we can get v, as follows:

v = 2π /24 hr . 6378 Km = 1,670 Km/hr.

4 0

2 years ago

A scuba diver at 70 m below the surface of a lake, where the temperature is 4 degrees C, releases an air bubble with a volume of

posledela

Answer:

121.3 cm^3

Explanation:

P1 = Po + 70 m water pressure (at a depth)

P2 = Po (at the surface)

T1 = 4°C = 273 + 4 = 277 K

V1 = 14 cm^3

T2 = 23 °C = 273 + 23 = 300 K

Let the volume of bubble at the surface of the lake is V2.

Density of water, d = 1000 kg/m^3

Po = atmospheric pressure = 10^5 N/m^2

P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2

Use the ideal gas equation

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

By substituting the values, we get

$\frac{8\times 10^5\times 14}{277}=\frac{10^{5} \timesV_{2}}{300}$

V2 = 121.3 cm^3

Thus, the volume of bubble at the surface of lake is 121.3 cm^3.

6 0

3 years ago