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murzikaleks [220]

3 years ago

14

Newton's third law states that for every action force there is an equal and opposite reaction force. An idiot in your class says

, "Wow that means everything cancels and nothing ever moves, it is all an illluussion! Wowwwwww" What statement best proves to him he is an idiot.
a) The equal and opposite forces act on different objects
b) If there is even a slight imbalance in the third law there will be a net force causing acceleration
C) he's right, all forces cancel, any motion I have ever seen is wrong

2 answers:

Otrada [13]3 years ago

4 0

A. Because the third laws say that for every action force the is an equal and opposite reaction force

Veronika [31]3 years ago

4 0

$\huge \mathfrak{Answer.... }$

The Correct Answer is :

B. if there is even a slight imbalance in third law there will be a net force causing acceleration.

A slight difference in the forces can result in acceleration of an object.

$\mathrm{✌TeeNForeveR✌}$

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How does potential and kinetic energy change in to each other and back again?

PolarNik [594]

They can change because the law of conservation of energy allows it to happen, for example when you are sitting, your body is at a potential energy state, meaning you are inert, you are not moving, but when you get up and suddenly start walking or running, that energy is converted to kinetic energy, meaning that you are moving and can be changed back into potential energy if all of a sudden you stop running or walking to rest or sit down. This is just an example of how energy can are transferred multiple ways

6 0

3 years ago

A block of mass 0.510 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. T

maria [59]

Answer:

x=0.46m, speed=7.9m/s

Explanation:

Using the concept of conservation of energy:

1. kinetic energy of mass m and velocity v: $E_k=\frac{1}{2}mv^2$

2. gravitational potential energy of mass m, grav. acc. g and height h: $E_g=mgh$

3. potential energy in a spring with spring constant k and displacement from equilibrium x: $E_s=\frac{1}{2}kx^2$

Calculating x:

$\frac{1}{2}mv_a^2=\frac{1}{2}kx^2$

$x=\sqrt{\frac{m}{k}}v_a$

Calculating the speed:

$\frac{1}{2}mv_a^2 +mgh_a=\frac{1}{2}mv_b^2+mgh_b + W_{friction}$

$h_a=0, h_b=2R,W_{friction}=F_{friction}\times distance=7\pi R$

$\frac{1}{2}mv_a^2=\frac{1}{2}mv_b^2+2mgR+7\pi R$

Solving for $v_b$ :

$v_b=\sqrt{v_a^2-4gR-14\pi\frac{R}{m}}$

7 0

3 years ago

The human body is 65% oxygen by mass. If a person has a mass of 75.0 kg, what is the mass oxygen in his body?

Gekata [30.6K]

65% x 70 = 45.5
The mass of oxygen in his body is 45.5 kg

4 0

3 years ago

5. A cable is attached 32.0 m from the base of a flagpole that is about to

soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, $F_v$ = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

$F_v$ + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0

2 years ago

In Trial II, the same spring is used as in Trial I. Let us use this information to find the suspended mass in Trial II. Use 0.51

Masteriza [31]

Answer:

$M_2=0.79kg$

Explanation:

From the question we are told that:

Period $T=0.517s$

Trial 1

Spring constant $\mu=117N/m$

Period $T_1=0.37$

Mass $m=0.400kg$

Trial 2

Period $T_2=0.52$

Generally the equation for Spring Constant is mathematically given by

\mu=\frac{4 \pi^2 M}{T^2}

Since

$\mu _1=\mu_2$

Therefore

$\frac{4 \pi^2 M_1}{T_1^2}=\frac{4 \pi^2 M_2}{T_2^2}$

$M_2=M_1*(\frac{T_2}{T_1})^2$

$M_2=0.400*(\frac{0.52}{0.37}})^2$

$M_2=0.79kg$

5 0

3 years ago