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3 years ago

If a net horizontal force of 175 N is applied to a bike whose mass is 50 kg what acceleration is produced?

Physics

1 answer:

Crazy boy [7]3 years ago

7 0

The acceleration of the bike is $3.5 m/s^2$

Explanation:

We can solve this problem by using Newton's second law of motion, which states that:

$F=ma$

where

F is the net force on an object

m is the mass of the object

a is its acceleration

For the bike in this problem, we have

F = 175 N is the net force

m = 50 kg is the mass

And solving for a, we find the acceleration:

$a=\frac{F}{m}=\frac{175}{50}=3.5 m/s^2$

Learn more about Newton's second law and acceleration:

brainly.com/question/3820012

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Answer:

The valves prevent the backward flow of blood. These valves are actual flaps that are located on each end of the two ventricles (lower chambers of the heart). They act as one-way inlets of blood on one side of a ventricle and one-way outlets of blood on the other side of a ventricle.

Explanation:

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3 years ago

The frequency of the second harmonic of a certain musical instrument is 100 Hz. What is the fundamental frequency of the instrum

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3 years ago

Initial velocity 10 m/s accelerates at 5 m/s for 2 seconds whats the final velocity

stiks02 [169]

Answer:

The final velocity is 20 m/s.

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

$v_f=v_o+at$

The provided data is: vo=10 m/s, $a=5\ m/s^2$ , t=2 s. The final velocity is:

$v_f=10~m/s+5\ m/s^2\cdot 2\ s$

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The final velocity is 20 m/s.

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3 years ago

A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a

kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

$\lambda = 720 \ nm$

or,

$= 720\times 10^{-9} \ m$

$D=0.85 \ m$

$d = 0.22 \ mm$

or,

$=0.22 \times 10^{-3} \ m$

As we know,

Fringe width is:

⇒ $\beta=\frac{\lambda D}{d}$

hence,

Separation between second and third bright fringes will be:

⇒ $\theta=\beta=\frac{\lambda D}{d}$

$=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}$

$=2.78\times 10^{-3} \ m$

or,

$=2.78 \ mm$

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3 years ago

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