How do you solve for frequency?

You accelerate from 2m/s to 6m/s while traveling a distance of 2m; what was your acceleration?

nata0808 [166]

Answer:

The acceleration is 8 m/s²

Explanation:

The given parameters are;

The initial velocity, u = 2 m/s

The final velocity, v = 6 m/s

The distance the acceleration took place, s = 2 m

The acceleration, a, can be found from the following kinematic equation;

v² = u² + 2·a·s

By substituting the values, we have;

6² = 2² + 2 × a × 2

6² - 2² = 2 × a × 2

32 = 4·a

a = 32/4 = 8 m/s²

The acceleration, a, of the given motion = 8 m/s².

7 0

3 years ago

the force of friction between a 1000 kg car and the road is 10000 N, what is the fastest acceleration the car can achieve?

nordsb [41]

Answer:

1000

Explanation:

7 0

3 years ago

Read 2 more answers

A wave with 2.0 m amplitude has a frequency of 500 Hz is travelling at a speed of 200 m/s. What is the wavelength?

Serggg [28]

Answer: 0.4m

Explanation:

Given that:

Amplitude of wave = 2.0 m

Wavelength (λ)= ?

Frequency F = 500Hz

Speed V = 200 m/s

The wavelength is measured in metres, and represented by the symbol λ.

So, apply the formula:

Wavespeed V= Frequency F xwavelength λ

200m/s = 500Hz x λ

λ = 200m/s / 500Hz

λ = 0.4m

Thus, the wavelength is 0.4 metres

3 0

3 years ago

You are moving into an apartment and take the elevator to the 6th floor. Suppose your weight is 660 N and that of your belonging

Ivan

Answer:

Explanation:

Total weight

My weight+weight of belongings

660+1100=1760N.

a. Work done by the elevator to travel a total height of 15.2m

Using newton law of motion

ΣF = ma

There are only two forces acting upward, the weight and the reaction by the elevator

Also note it is moving at constant velocity then, a=0

N - W=0

Then, N=W

N=1760N

So, workdone is given as

Wordone, =force × distance

Work done=1760×15.2

W=26,752J

W=26.752KJ

b. Work done on me alone is still need to go through the same process but will remove the weight of the belonging

Therefore,

Weight now = 660N

And using the same equation of motion

ΣF = ma

Comstant velocity, a=0

N - W=0

N=W

N=660N

Then, workdone

W=F×d

W=660×15.2

W=10,032J

W=10.032KJ

6 0

4 years ago

Read 2 more answers

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)

From equation 2 and 3, we get

$M_{a}$ *a = $M_{b}$ $\sqrt{a^{2} +g^{2} }$

Squaring both sides we get

$M_{a} ^{2}$ * $a^{2}$ = $M_{b} ^{2}$ * ( $a^{2}$ + $g^{2}$ )

$M_{a} ^{2}$ * $a^{2}$ = ( $M_{b} ^{2}$ * $a^{2}$ )+ ( $M_{b} ^{2}$ * $g^{2}$ )

( $M_{a} ^{2}$ - $M_{b} ^{2}$ ) * $a^{2}$ = $M_{b} ^{2}$ * $g^{2}$

$a^{2}$ = $M_{b} ^{2}$ * $g^{2}$ /( $M_{a} ^{2}$ - $M_{b} ^{2}$ )

Taking square root on both sides, we get acceleration a

a = $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ )

Hence substituting the value of a in equation 1, we get

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

3 0

3 years ago