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mamaluj [8]
3 years ago
9

What is the acceleration of an object with a mass of

Physics
1 answer:
aev [14]3 years ago
3 0

Heya!!

For calculate aceleration, lets applicate second law of Newton:

                                                    \boxed{F=ma}

                                                 <u>Δ   Being   Δ</u>

                                             F = Force = 183 N

                                            m = Mass = 367 kg

                                             a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

\boxed{a=183 \ N / 367\ kg}

⇒ Resolving

\boxed{ a = 0.49 \ m/s^{2}}

Result:

The aceleration is <u>0,49 meters per second squared (m/s²)</u>

Good Luck!!

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A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
Murljashka [212]

The work done on the ship is 1.5\cdot 10^7 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F=5.00\cdot 10^3 N (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

\theta=0^{\circ} (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
3 years ago
Neil has 3 partially full cans of white pants. they contain 1/3 gallon, 1/5 gallon,and 1/2 gallon of paint About how much paint
Oliga [24]
He has 1 1/30 gallons, or 31/30 gallons, you can find this by setting all the fractions to a common denominator and adding them
7 0
3 years ago
What is the average speed of a boy who jogs 250 meters in 110 seconds
Ede4ka [16]

2.27 mps repeating.

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8 0
3 years ago
Read 2 more answers
A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r
elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

V_1=26.8 L

P_1=744mm Hg

T_1=31.2^{\circ} C=31.2+273=304.2K

P_2=385mmHg

T_2=-14.8^{\circ}=273-14.8=258.2K

We know that

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{P_2T_1}

Substitute the values

V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}

V_2=43.96 L

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0
3 years ago
Please Help !!!!!!!!!!!!!!!!!!!!!!!!!!
larisa [96]

As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.

Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.

As per Planck's quantum theory -

                                                      hf =E'-E

          Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.

The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.

We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

       v=\lambda f

[here \lambda is the wavelength of the wave ]

It is given that wavelength =21 cm=0.21 m

 The frequency is calculated as -

                                                         f=\frac{v}{\lambda}

                                                                 =\frac{c}{\lambda}

                                                                 =\frac{3*10^{8} m/s }{0.21 m}

                                                                  =14.2857*10^{8} s^{-1}

 Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz

Here Hertz[ Hz] is the unit of frequency.

7 0
3 years ago
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