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3 years ago

What is the acceleration of an object with a mass of 367 kg when a net force of 183 N is applied to it

Physics

1 answer:

GarryVolchara [31]3 years ago

4 0

Answer:

mass=367kg

force=183N

acceleration =?

Now,

force=m×a

183=367×a

183÷367=a

0.49=a

acceleration =0.49m/s^2 answer!!!!

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A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

3 0

3 years ago

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Amanda [17]

Can you speak English pls?

5 0

3 years ago

Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at

BARSIC [14]

Answer: Option (c) is the correct answer.

Explanation:

Physical properties are the properties in which there is no change in chemical composition of a substance. On the other hand, chemical properties are the properties which change the chemical composition of a substance.

For example, when water boils at $100 ^{o}C$ then it changes into vapor state whereas when water freezes at $0^{0}C$ then it changes state from liquid to solid.

This means only physical state of water is changing and there is no change in chemical composition of water.

Hence, we can conclude that best option describing given information is that these are the physical changes water undergoes.

4 0

3 years ago

What waves shown on the electromagnetic spectrum disturb the medium it passes through at answer?

pychu [463]

Electromagnetic waves do not require material medium for propagation. Radiowaves have the longest wavelength but the least frequency.
Gamma ray has the shortest wavelength but the highest frequency.

The gamma rays and x-rays due to their high frequency, they can disturb the medium they pass through. They have high energy that can cause damage to the matter the come across.

4 0

3 years ago

3. Alpha Centauri A and B are Sun-like stars, and together they form the binary star Alpha Centauri AB. Alpha Centauri A has 1.1

IgorLugansk [536]

Answer:

8722.8 m/s, average speed of B relative to A (approximate)

Explanation:

The total mass of the two stars,

M = M₁ + M₂ = 2.007 M☉ = 3.99101985e+30 kg

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

GM = 2.663726378e+20 m³ sec⁻²

The orbital period,

P = 79.91 years = 2.521767816e+9 sec

The semimajor axis of the orbit,

a = ∛[P²GM/(4π²)]

a = 3.50090e+12 meters

Assuming that the orbit is circular, with radius equal to the calculated semimajor axis, the average speed in orbit,

V = C/P

where

C = 2πa = 2.19968e+13 meters

V ≈ 8722.8 m/s, average speed of B relative to A (approximate)

However, we can look up the eccentricity of the orbit of α Cen A relative to α Cen B, and then we can calculate the average speed in orbit more accurately.

e = 0.5179

C = 4a ∫(0,π/2) √(1−e²sin²θ) dθ

C = 2.04379e+13 meters

V = 8104.6 m/s, average speed of B relative to A (more accurate)

As you can see, the more nearly correct value is considerably different than the approximation obtained when the orbit is assumed circular.

Incidentally, we can find that

The periapsis distance is 1.68779e+12 meters.

The apoapsis distance is 5.31402e+12 meters.

The semilatus rectum is 2.56189e+12 meters.

The semiminor axis is 2.99482e+12 meters.

The focal parameter is 4.94669.

We can also calculate the separation of A and B when the orbital speed has this average value:

r = 2a { 1 + [ (2/π) ∫(0,π/2) √(1−e²sin²θ) dθ ]² }⁻¹

r = 3.75778349e+12 meters

Moreover, that separation occurs when the stars reach the part of their orbit having true anomalies of

θ = ± arccos{ [ 1 − (a/r) (1+e²) ] / e }

θ = 110.518° and 249.482°

3 0

3 years ago