When falling through the air, which of the following objects will hit the ground first?

The average kinetic energy of the particles in an object is directly proportional to its A) heat. B) volume. C) temperature. D)

yanalaym [24]

the average kinetic energy of the particles in an object is directly proportional to its TEMPERATURE.

5 0

3 years ago

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A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = $\frac{1}{2} mV^{2} + mgh$

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In our situation

$\frac{0,71}{sin90} =\frac{h}{sin32}$ → $h=0,376$

Therefore

$1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376$

→ $V=2,416$

7 0

3 years ago

Which type of force if due to the masses of objects.

statuscvo [17]

Answer: C

Explanation: weak nuclear

7 0

2 years ago

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A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

inn [45]

Answer:

The answer to the question is

The ladybug begins to slide

Explanation:

To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same

Where the frictional force equals $F_{Friction}$ = μ×N = m×g×μ

and the centripetal force is given by m·ω²·r

If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have

m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁

and for the gentleman bug we have

m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂

But r₁ = 2×r₂

Therefore substituting the values of r₁ =2×r₂ we have

g×μ = ω²·r₁ = g×μ = ω²·2·r₂

Therefore ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide

The ladybug begins to slide

6 0

3 years ago

A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ = (v² - u²)/2a

h₂ = h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ = 1.3 m + 0.54 m

h₂ = 1.84 m

7 0

2 years ago