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3 years ago

A box of mass 115 kg sits on an inclined surface with an angle of 59º. What

Physics

1 answer:

LenKa [72]3 years ago

8 0

Answer:

966 N

Explanation:

The component of the weight along the direction perpendicular to the plane is:

$mg cos \theta$

where

m is the mass of the box, g the acceleration of gravity, $\theta$ the angle of the inclined plane.

Similarly, the component of the weight of the box along the surface is

$mg sin \theta$

acting down along the incline.

Here we have:

m = 115 kg

g = 9.8 m/s^2

$\theta=59^{\circ}$

Substitting in the last equation, we find the component of the weight of the box along the surface:

$mg sin \theta = (115)(9.8)(sin 59)=966 N$

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Describe the movement of air over 2 nearby land areas, one of which is heated more than the other.

77julia77 [94]

In this case, the air from the warm area will always start moving towards the colder areas because as the temperature in both lands should be equal. This is one of the laws of thermodynamics.

7 0

4 years ago

An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the

OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

$m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2$

Through the aforementioned formula we will have to

$v_f^2-v_i^2 = 2ax$

The particulate part of the rest, so the final speed would be

$v_f^2 = 2gx$

$v_f=\sqrt{2(9.8)(3.1)}$

$v_f = 7.79m/s$

Now from Newton's second law we know that

$F = ma$

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

$F = m\frac{dv}{dt}$

Replacing we have that,

$F = (89)\frac{7.79}{0.5}$

$F = 1386.62N$

Therefore the force that the water exert on the man is 1386.62

3 0

4 years ago

We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

3 years ago

A 25 kg child stands 2.5 m from the center of a frictionless merry‐go‐round, which has a 200 kg*m^2 moment of inertia and is spi

vodka [1.7K]

Answer:

Explanation:

a ) Time period T = 2 s

Angular velocity ω = 2π / T

= 2π / 2 = 3.14 rad /s

Initial moment of inertia I₁ = 200 + mr²

= 200 + 25 x 2.5²

=356.25

Final moment of inertia

I₂ = 200 + 25 X 1.5 X 1.5

= 256.25

b ) We apply law of conservation of momentum

I₁ X ω₁ = I₂ X ω₂

ω₂ = I₁ X ω₁ / I₂

Putting the values

$w_2=\frac{356.25\times3.14}{256.25}$

ω₂ = 4.365 rad s⁻¹

c ) Increase in rotational kinetic energy

=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²

.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²

= 684.95 J

This energy comes from work done against the centripetal pseudo -force.

7 0

3 years ago

Does the air exert a buoyant force on all objects in air or only on objects such as balloons that are very light for their size?

Citrus2011 [14]

Answer:

See explanation

Explanation:

Solution:-

Buoyancy is the force that causes objects to float. It is the force exerted on an object that is partly or wholly immersed in a fluid. Buoyancy is caused by the differences in pressure acting on opposite sides of an object immersed in a static fluid. It is also known as the buoyant force. Buoyancy is the phenomena due to Buoyant Force.

It is as an upward force exerted by a fluid that opposes the weight of an object immersed in a fluid. As we know, the pressure in a fluid column increases with depth. Thus, the pressure at the bottom of an object submerged in the fluid is greater than that at the top. The difference in this pressure results in a net upward force on the object which we define as buoyancy.

- The formula for buoyant force (Fb) is given:

Fb = ρ*g*V

- The force acts on all objects. However, it depends on the fluid density and amount of volume displaced.

- The Buoyant force exerted by air with density = 1.225 kg/m^3 on an object with volume (V) is:

Fb = ρ*g*V = 1.225*9.81*V = 12.02*V

- For the similar object with mass (m), the downward weight would be:

W = m*g

- For the object to float the buoyant force (Fb) must be greater than weight of the object:

Fb > W

12.02*V > m*9.81

V / m > 0.816

- The ratio of V / m must be at-least = 0.816.

- Assuming the object is fully immersed in air, then the volume displaced V = ρ_material*V

ρ_material < 1 / 0.816

ρ_material < 1.225 or ( ρ_air )

- So the for an object to float in air its material density must always be less than that of air. That why in balloons lighter gas is used which have density less than that of air like Helium.

4 0

4 years ago