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GenaCL600 [577]
3 years ago
11

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

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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...

You might be interested in
what is the energy (in eV units) carried by one photon violet light that has a wavelength of 4.5e-7?
DaniilM [7]
The energy of a photon is given by
E=hf
where h is the Planck constant and f is the photon frequency.

We can find the photon's frequency by using the following relationship:
f= \frac{c}{\lambda}
where c is the speed of light and \lambda is the photon's wavelength. By plugging numbers into the equation, we find
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.67 \cdot 10^{14}Hz

And so now we can find the photon energy
E=hf=(6.6 \cdot 10^{-34} Js)(6.67 \cdot 10^{14}Hz )=4.4 \cdot 10^{-19} J

We know that 1 Joule corresponds to
1 J = 1.6 \cdot 10^{-19} eV
So we can convert the photon's energy into electronvolts:
E= \frac{4.4 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=2.75 eV
4 0
4 years ago
A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be
nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

W=\mu_kmgd

W=0.2\times 5\times 9.8\times 2

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

5 0
3 years ago
Read 2 more answers
500km is equal to how many millimeters
Sindrei [870]

Answer:

500000000

if you can give me brainliest that would be great

7 0
4 years ago
if a person can jump 2m in earth surface how high can he jump in the moon (g of moon = 1.66m/s, g of earth = 9.8 m/s) [hint: use
julia-pushkina [17]

Answer:

h_{moon} = 11.8\ m

Explanation:

Since the work done is same everywhere in the universe. Hence, the work done in jumping will be same for the person on moon and earth:

W_{moon} = W_{earth}\\\\P.E_{moon} = P.E_{earth}\\\\mg_{moon}h_{moon} = mg_{earth}h_{earth}\\\\g_{moon}h_{moon} = g_{earth}h_{earth}\\\\(1.66\ m/s^2)h_{moon} = (9.8\ m/s^2)(2\ m)\\\\h_{moon} = \frac{(9.8\ m/s^2)(2\ m)}{1.66\ m/s^2}\\\\h_{moon} =  11.8\ m

5 0
3 years ago
Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the
Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

m=180 kg the mass of the raft

F=75 N the force applied by Sawyer

V=2 m/s the raft's final speed

V_{o}=0 m/s the raft's initial speed (assuming it starts from rest)

We have to find the time t

Well, according to Newton's second law of motion we have:

F=m.a (1)

Where a is the acceleration, which can be expressed as:

a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}} (2)

Substituting (2) in (1):

F=m\frac{V-V_{o}}{t-t_{o}} (3)

Where t_{o}=0

Isolating t from (3):

t=\frac{m(V-V_{o})}{F} (4)

t=\frac{180 kg(2 m/s-0 m/s)}{75 N}

Finally:

t=4.8 s

6 0
3 years ago
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