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3 years ago

A bicycle accelerates from rest to 6 m/s in 2 s. What is the bicycle's acceleration?

Physics

2 answers:

USPshnik [31]3 years ago

5 0

So the right answer is 3m/s^2.

Look at the attached picture

hope it will help you

Good luck on your assignment

Angelina_Jolie [31]3 years ago

4 0

Answer:

Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me...

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Noah drops a rock with a density of 1.73 g/cm3 into a pond. Will the rock float or sink? Explain your answer

stealth61 [152]

Answer:

Sink

Explanation:

Water has a density of 1 g/cm³. Since the density of the rock is greater than the density of water, the rock will sink.

4 0

3 years ago

Read 2 more answers

Red laser light passes through a single slit to form a diffraction pattern. If the width of the slit is increased by a factor of

SCORPION-xisa [38]

If the width of the slit is increased by a factor of two , then the central maxima will get decreased by half.

since , width of the central maxima is inversely proportional to the width of the slit . Which implies if width of the slit get increased then width of the central maxima will get decreased and vice versa.

central maxima = (2 * wavelength * D) / width of the slit

= (2 * lambda * D) / d

if initially width of the slit = d

then, after increasing it by a factor of two it become = 2d

central maxima = (2 * lambda * D) / 2d

= (lambda * D) / d

If the width of the slit is increased by a factor of two , then the central maxima will get decreased by half.

Learn more about central maxima :

brainly.com/question/14279157?referrer=searchResults

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5 0

2 years ago

for a freely falling object dropped from rest, what is the acceleration at the end of the fifth second of fall?

lara31 [8.8K]

acceleration due to gravity is always 9.8 m/s/s (on earth)

3 0

3 years ago

In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t

USPshnik [31]

Answer:

$W_x = 0.9156\ c$

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

$W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}$

u_x = 0.543 c

v_x = - 0.741 c

$W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}$

$W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}$

$W_x = \dfrac{1.284c}{1+0.402363}$

$W_x = 0.9156\ c$

Relative velocity of the particle is $W_x = 0.9156\ c$

5 0

3 years ago

A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

Papessa [141]

Answer:

The acceleration is $a = 3.45*10^{3} m/s^2$

Explanation:

From the question we are told that

The radius is $d = 6.5 cm = \frac{6.5}{100} = 0.065 m$

The magnitude of the magnetic field is $B = 5.5 T$

The rate at which it decreases is $\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s$

The distance from the center of field is $r = 1.5 cm = \frac{1.5}{100} = 0.015m$

According to Faraday's law

$\epsilon = - \frac{d \o}{dt}$

and $\epsilon = \int\limits {E} \, dl$

Where the magnetic flux $\o = B* A$

E is the electric field

dl is a unit length

$\int\limits {E} \, dl = - \frac{d}{dt} (B*A)$

${E} l = - \frac{d}{dt} (B*A)$

Now $l$ is the circumference of the circular loop formed by the magnetic field and it mathematically represented as $l = 2\pi r$

A is the area of the circular loop formed by the magnetic field and it mathematically represented as $A= \pi r^2$

${E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} [ - \frac{db}{dt} ]$

Substituting values

$E = \frac{0.015}{2} (24*10^{-4})$

$E = 3.6*10^{-5} V/m$

The negative signify the negative which is counterclockwise

The force acting on the proton is mathematically represented as

$F_p = ma$

Also $F_p = q E$

$ma = qE$

Where m is the mass of the the proton which has a value of $m = 1.67 *10^{-27} kg$

$q = 1.602 *10^{-19} C$

$a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}$

$a = 3.45*10^{3} m/s^2$

8 0

3 years ago