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3 years ago

A bike accelerates uniformly(from rest to a speed of 7 m/s over a distance of 40 m. Determine

Physics

1 answer:

marshall27 [118]3 years ago

3 0

Answer:

$0.61 m/s^2$

Explanation:

The bike's acceleration can be found by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the bike

u is the initial velocity

a is the acceleration

s is the distance covered

For the bike in the problem,

u = 0

v = 7 m/s

d = 40 m

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{7^2-0}{2(40)}=0.61 m/s^2$

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In the diagram, q1= -2.60*10^-9 C and

Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

$E_{net}=90.37\: N/c$

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

$|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}$

Where:

k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
q1 is the first charge
d1 is the distance from q1 to P

$|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}$

$|E_{1}|=80.84\: N/C$

And E2 will be:

$|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}$

$|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}$

$|E_{2}|=40.39\: N/C$

Finally, we need to use the Pythagoras theorem to find the magnitude of the net electric field.

$E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}$

$E_{net}=\sqrt{80.84^{2}+40.39^{2}}$

$E_{net}=90.37\: N/c$

I hope it helps you!

7 0

3 years ago

When was the copernican treatise published

LiRa [457]

They were published in 1542.

5 0

4 years ago

A nucleus with a mass number of 64 has a mean radius of about: _________.

dangina [55]

Answer:

The correct option is A

Explanation:

From the question we are told that

The mass number is $A= 64$

Generally the mean radius is mathematically evaluated as

$R = R_o A^{\frac{1}{3} }$

Here $R_o$ is a constant with a value $R_o =1.2*10^{-15}$

$R = 1.2*10^{-15} * 64^{\frac{1}{3} }$

$R = 4.8 *10^{-15}$

$R = 4.8\ fm$

5 0

3 years ago

When you stand on tiptoes on a bathroom scale, there is an increase in

pychu [463]

Answer:

B) Pressure on the scale, not registered as weight.

Explanation:

This is because energy (derived from weight) becomes compiled on the tips of your toes, and therefore does not increase your weight, but simply the pressure at a smaller point

3 0

3 years ago

A 60cm long string of an ordinary guitar is tuned to produce the note a4 (frequency 440hz) when vibrating in its fundamental mod

pishuonlain [190]

Answer:

0.78 m

Explanation:

The relationship between wavelength and frequency of a wave is given by

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the sound wave in this problem, we have

$f=440 Hz$ is the frequency

v = 344 m/s is the speed of sound in air

Substituting into the equation and re-arranging it, we find the wavelength:

$\lambda=\frac{v}{f}=\frac{344 m/s}{440 Hz}=0.78 m$

8 0

3 years ago

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