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1 year ago

A 400g mass object at rest is acted on by a 200 N net force for 12 seconds. What is it’s final velocity?

Physics

1 answer:

Kisachek [45]1 year ago

3 0

Explanation:

do we can say that the first step is finding the acceleration of the object. We do so by saying F = ma

so a = F/m

a = 500 m/s^2

we now have the time, acceleration and initial velocity.

so we can find the final velocity by using one of newtons equations which is: vf = vi + at

so:

vf = 0 + 500 × 12

vf = 6000

note that our initial velocity was zero since the object was initially at rest.

if you still have any doubt dont hesitate to ask for further help.

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Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might

Tanya [424]

<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>

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3 years ago

A 1000 kg elevator is rising and its speed is increasing with an acceleration of 3 m/s^2. What is the resulting tension in the v

Alchen [17]

Answer:

12800 N

Explanation:

7 0

2 years ago

A person is pulling their 20 kg luggage using the luggage handle. The handle is at an angle of 56 degrees above the horizontal.

rusak2 [61]

Answer:

The answer to your question is: a = 1.99 m/s²

Explanation:

Data

mass = 20 kg

angle = 56°

Force = 71 N

horizontal acceleration = ?

Process

Find the horizontal force

cos Ф = adjacent side / hypotenuse

adjacent side = hypotenuse x cosФ

adjacent side = 71 x cos 56

a.s. = 39.70 N

Newton's second law

F = ma

a = F/m

a = 39.7 / 20

a = 1.99 m/s²

4 0

3 years ago

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$

Now

$F_c+F_s=50000\\\Rightarrow 10.74658F_s+F_s=50000\\\Rightarrow F_s=\dfrac{50000}{11.74658}\\\Rightarrow F_s=4256.55807\ lbf$

$F_c=10.74658F_s\\\Rightarrow F_c=10.74658\times 4256.55807\\\Rightarrow F_c=45743.44182\ lbf$

Stress is given by

$\sigma_s=\dfrac{4256.55807}{\pi \dfrac{1}{2^2}}{4}\\\Rightarrow \sigma_s=21678.47223\ lbf-in^2$

The stress in the steel is $21678.47223\ lbf-in^2$

$\sigma_c=\dfrac{45743.44182}{119.4}\\\Rightarrow \sigma_s=383.1109\ lbf-in^2$

The stress in the steel is $383.1109\ lbf-in^2$

4 0

3 years ago

In July 2005, NASA's "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the s

eimsori [14]

Question:

(a) What change in the comet’s velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet’s velocity? Would this change be noticeable? (The mass of the earth is 5.97×1024kg.)

Answer:

The answers to the question are;

(a) The change in the comet’s velocity produced by the collision is 2.86 × 10⁻⁶km/h or 7.944 × 10⁻⁷ m/s

(b) It would change our planet’s velocity by 6.70× 10⁻⁸ km/h or 1.86× 10⁻⁸ m/s

Change is too small to be noticeable

Explanation:

We not that the question is about conservation of liner momentum

Therefore we have, by listing out the known parameters

m₁ = Mass of "Deep Impact" = 372 kg

m₂ = Mass of Tempel 1 comet = (0.1 to 2.5) × 10¹⁴ kg,

v₁ = Vaelocity of "Deep Impact" = 37000 km/h

v₂ = Velocity of Tempel 1 comet = 40000 km/h

From the principle of linear momentum, we have, for both bodies moving in opposite direction;

m₁×v₁ + m₂×v₂ = m₁×v₃ + m₂×v₃ since it was a crash, it is assumed that they both have the same final velocity

This gives

372 kg ×37000 km/h - 0.1 × 10¹⁴ kg × 40000 km/h = (372 kg + 0.1 × 10¹⁴ kg )×v₃

13764000 kg·km/h - 4.0 × 10¹⁷ kg·km/h = 10000000000372×v₃

v₃ = (-399999999986236000 kg·km/h)/10000000000372 = -39999.999997 km/h ≈ - 40000 km/h in the direction of Deep Impact

Change in comet velocity

= 40000 km/h - 39999.999997 km/h

= 2.86 × 10⁻⁶km/h = 7.944 × 10⁻⁷ m/s

(b) If the colission is with earth, we have

m₃ = Mass of earth

From the principle of conservation of linear momentum, we have

m₂v₂+m₃ v₃ = (m₂ + m₃) v₄

v₃ = Initial velocity of Earth = 0 km/h

m₃ = Mass of Earth = 5.97 × 10²⁴ kg

Therefore, pluggin in the vaalues gives

0.1 × 10¹⁴ kg × 40000 km/h + 5.97 × 10²⁴ kg × 0 km/h = (0.1 × 10¹⁴ kg + 5.97 × 10²⁴ kg) × v₄

Therefore,

v₄ = (4.0 × 10¹⁷ kg·km/h + 0 kg·km/h)/ (5970000000010000000000000 kg)

= 6.70× 10⁻⁸ km/h = 1.86× 10⁻⁸ m/s

Change is too small

5 0

3 years ago