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2 years ago

3. Check that the dimensions of both side of the equations below are agreeing to each other:

Physics

1 answer:

Reptile [31]2 years ago

4 0

Answer:

b is the answer of that question answer sapai

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Which of the following never cause a change in the motion of an object? A. Net forces B. Unbalanced forces O C. Balanced forces

Karolina [17]

Answer:

balanced force never changes its motion

5 0

2 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

What is a tool that can be used to measure the size of a force

Travka [436]

The answer to this question would be: a spring scale.

The spring scale that you use to determine your body weight is actually a device that measures your body gravitational force. The force itself influenced by your body weight, that is why it can determine your body weight.
More weight means more force, more force will shrink the spring more.

7 0

3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

vampirchik [111]

Answer:

$\large \boxed{\text{30 rev/s}}$

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

I₁ = 3.5 kg·m²s⁻¹

ω₁ = 6.0 rev·s⁻¹

I₂ = 0.70 kg·m²s⁻¹

Calculation:

$\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}$

8 0

3 years ago

A van has a weight of 4000 lb and center of gravity at Gv. It carries a fixed 900 lb load which has a center of gravity at Gl. I

natulia [17]

Answer:

x = 25 / μ [ ft]

Explanation:

To solve this exercise we can use Newton's second law.

Let's set a reference system where the x axis is parallel to the road

Y axis

N_B + N_A - W_van - W_load = 0

N_B + N_A = W_van + W_load

X axis

fr = ma

a = fr / m

the total mass is

m = (W_van + W_load) / g

the friction force has the expression

fr = μ N_{total}

fr = μy (W_van + W_load)

we substitute

a = μ (W_van + W_load) $\frac{g}{W_van + W_load}$

a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

v² = v₀² - 2 a x

0 = v₀² -2a x

x = $\frac{v_o^2}{2a}$

x = $\frac{v_o^2}{2 \mu g}$

x = $\frac{40^2}{2 \ 32 \ \mu}$

x = 25 / μ [ ft]

5 0

2 years ago