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jeyben [28]
3 years ago
12

A bicyclist steadily speeds up from rest to 9.00m/s during a 7.20s time interval. Determine all unknowns and answer the followin

g questions. What is the magnitude of the bicyclist's acceleration? unit How far did the bicyclist travel during this time? unit​
Physics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. Since elapsed time is

Δ

t

=

t

f

−

t

0

, taking

t

0

=

0

 means that

Δ

t

=

t

f

, the final time on the stopwatch. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity. That is,

x

0

is the initial position and

v

0

is the initial velocity. We put no subscripts on the final values. That is,

t

is the final time,

x

is the final position, and

v

is the final velocity. This gives a simpler expression for elapsed time—now,

Δ

t

=

t

. It also simplifies the expression for displacement, which is now

Δ

x

=

x

−

x

0

. Also, it simplifies the expression for change in velocity, which is now

Δ

v

=

v

−

v

0

. To summarize, using the simplified notation, with the initial time taken to be zero,

Δ

t

=

t

Δ

x

=

x

−

x

0

Δ

v

=

v

−

v

0

 

}

Explanation:

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A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the t
zhannawk [14.2K]

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

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8 0
3 years ago
A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given t
Mademuasel [1]

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

8 0
2 years ago
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