A cannonball and a marble roll smoothly from rest down an incline. Is the cannonball’s(a) time to the bottom and(b) translationa

A parachutist, after opening her parachute, finds herself gently floating downward, no longer gaining speed. She feels the upwar

Sav [3153]

As the parachutist is descending at a steady rate

we can conclude that

$a = \frac{dv}{dt}$

Acceleration indicates the change in velocity

given the constant velocity in this scenario

$a = 0$

Thus, in this situation, we find the acceleration to be zero

It’s understood from Newton's second law

$F_{net} = ma$

where a is equal to 0

$F_{net} = 0$

$F_{net} = F_g - F_b$

Here, the force due to gravity

equals the force due to buoyancy

$F_g$ Hence, we can deduce

$F_b$

therefore

$F_g - F_b = 0$

as such the upward force is counteracted by the downward force.

5 0

2 months ago

In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2. Calculate the amount of force at

Ostrovityanka [3204]

The force can be determined using the equation F (force) = mass * acceleration. The unit of measurement, N (Newton), is equivalent to kilogram-meter/seconds2.
Thus, F= 1300 kg * 1.07 m/s2 = 1391 N.
The resultant value is 1391 N.

7 0

2 months ago

Read 2 more answers

Two thin lenses with a focal length of magnitude 12.0cm, the first diverging and the second converging, are located 9.00cm apart

kicyunya [3294]

b ) The first lens is a concave lens with a focal length of f₁ = - 12 cm and an object distance of u = - 20 cm. Using the lens formula, 1 / v - 1 / u = 1 / f, we get 1 / v + 1 / 20 = -1 / 12. This leads to 1 / v = - 1 / 20 - 1 / 12, which simplifies to 1 / v = -0.05 - 0.08333, yielding v = -7.5 cm. Consequently, the first image is formed before the first lens, near the object side, which becomes the object for the second lens with a distance of 16.5 cm from the second lens. c ) For the second lens, object distance is u = -16.5 cm, and focal length f₂ = + 12 cm (convex lens). Using the lens formula leads to 1 / v + 1 / 16.5 = 1 / 12, and this results in 1 / v = 1 / 12 - 1 / 16.5, which simplifies to 1 / v = 0.08333 - 0.0606. Finally, we find v = 44 cm (approximately). This image will be formed on the other side of the convex lens, which is 53 cm from the first lens. Magnification by the first lens is v / u = -7.5 / -20 = 0.375. For the second lens, it is v / u = 44 / - 16.5 = -2.67. d ) The total magnification becomes 0.375 x - 2.67 = - 1.00125. The height of the final image is then calculated as 2.50 mm x 1.00125 = 2.503 mm. e ) The final image will be inverted compared to the object since the total magnification is negative.

6 0

2 months ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

kicyunya [3294]

Answer:

$tan \theta = \mu_s$

Explanation:

Aby obiekt był w spoczynku na nachyleniu, wynikowa siła działająca na niego musi wynosić zero. Równanie sił działających w kierunku równoległym do nachylenia jest następujące:

$mg sin \theta - \mu_s R =0$ (1)

gdzie

$mg sin \theta$ to składowa ciężaru równoległa do nachylenia, przy czym m oznacza masę obiektu, g oznacza przyspieszenie grawitacyjne, a $\theta$ to kąt nachylenia

$\mu_s R$ to siła tarcia, z $\mu_s$ jako współczynnikiem tarcia oraz R jako reakcją normalną nachylenia

Równanie sił w kierunku prostopadłym do nachylenia to

$R-mg cos \theta = 0$

gdzie

R to reakcja normalna

$mg cos \theta$ to składowa ciężaru prostopadła do nachylenia

Obliczając R,

$R=mg cos \theta$

I podstawiając do (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Rearanżując równanie,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

To jest warunek, przy którym równowaga jest zachowana: gdy tangens kąta staje się większy niż wartość $\mu_s$ , siła tarcia nie jest w stanie zrównoważyć składowej ciężaru równoległej do nachylenia, dlatego obiekt zaczyna zsuwać się w dół.

4 0

3 months ago