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chubhunter [2.5K]

2 years ago

12

If the net force on a chair is 6 newtons to the right, what will occur? *

1 answer:

Tatiana [17]2 years ago

8 0

Answer:

The chair will accelerate to the right.

Explanation:

As we know by Newton's Law that when some net unbalanced force is exerted on the system then it will accelerate in the direction of force and the equation is given as

$F = ma$

now we know that 6 N force is exerted on the chair towards right

So here the chair will accelerate in the direction of net force

so correct answer is given as

The chair will accelerate to the right.

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he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

8 0

1 year ago

Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o

Mila [183]

Answer:

a) $x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) x = 4.47 cm

c) $x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

$x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}$

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

$x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3} ) }{m_{1}+m_{2}+m_{3} }$

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

$x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm$

c) The center of the mass of the beads realtive to the center of bead is:

$x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }$

d) $x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm$

6 0

2 years ago

Read 2 more answers

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

2 years ago

A car with a mass of 1.5 × 103 kilograms is traveling west at a velocity of 22 meters/second. It hits a stationary car with a ma

Licemer1 [7]

A. 14 meters/second to the west

3 0

2 years ago

Read 2 more answers

The Cenozoic Era is best described as the era in which

zhenek [66]

The era after the KT event occurred

6 0

3 years ago