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Dafna1 [17]
3 years ago
9

A drag racer starts from rest and accelerates at 7.4 m/s2. How far will he travel in 2.0 seconds?

Physics
1 answer:
Leto [7]3 years ago
3 0

Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2.  First we must determine the final velocity:

v_{final}=v_{initial}+\frac{1}{2}at\\\\v_{final}=0+(7.4m/s^2)(2s)=34.8m/s

Now we will determine the distance traveled:

v_{final}^2=v_{initial}^2+2a \Delta x\\\\\Delta x = \frac{v_{final}^2}{2a} =\frac{(34.8)^2}{(2)(7.4)}=81.83 m

Therefore, the drag racer traveled 81.83 meters in 2 seconds.

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What is the kinetic energy of an object that has a mass of 12 kg and moves with a velocity of 10 m/s
WITCHER [35]

Answer:

600 J

Explanation:

The formula to find the kinetic energy of an object is \frac{1}{2}mV^2

  • m = mass in kg
  • V = velocity in m/s
  • KE is measured in Joules just as all other forms of energy.

Now, let's plug in the variables we're given and simplify.

  • \frac{1}{2}(12)(10)^2
  • \frac{1}{2}(12)(100)
  • (6)(100)
  • (600)

Thus, the answer is 600 Joules.

6 0
2 years ago
Read 2 more answers
Explain how an object moving at a constant speed can have a nonzero acceleration
SOVA2 [1]
Its very simple if a body is moving in circle the magnitude of its velocity remain constant  but its direction changes because velocity is directed towards tangent and at every point in a cirlce its direction will be different (along tangent) so velocity is not uniform .As acceleration is the rate  change of velocity  so it will be non zero because velocity is changing due to its direction.
8 0
3 years ago
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The wavelength of the visible line in the hydrogen spectrum that corresponds to m = 5 in the Balmer equation is: A. 656 nm. B. 4
BaLLatris [955]

Answer:

The wavelength of the visible line in the hydrogen spectrum is 434 nm.

Explanation:

It is given that, the wavelength of the visible line in the hydrogen spectrum that corresponds to n₂ = 5 in the Balmer equation.

For Balmer series, the wave number is given by :

\dfrac{1}{\lambda}=R(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})

R is the Rydberg's constant

For Balmer series, n₁ = 2. So,

\dfrac{1}{\lambda}=1.097\times 10^7\times (\dfrac{1}{2^2}-\dfrac{1}{5^2})

\lambda=4.34\times 10^{-7}\ m

or

\lambda=434\ nm

So, the wavelength of the visible line in the hydrogen spectrum is 434 nm. Hence, this is the required solution.

6 0
3 years ago
A scientist decides to replicate an experiment completed by another scientist. Which statement describes something that would no
tatyana61 [14]

The statement describes something that would not affect the results of the replicated experiment is option A. (The two experiments were completed by two different people)

<h3>What is an experiment?</h3>

An experiment can be defined as a procedure carried out to support or refute a hypothesis, or determine the efficacy or likelihood of something previously untried.

Experiments helps in the provision of insight into the cause-and-effect by demonstrating what outcome occurs when a particular factor is manipulated.

If two experiments were completed by two different people, it would not affect the results if the experiment is replicated.

The purpose of any experiment is to test or verify a hypothesis

Learn more about experiments at: brainly.com/question/26117248

#SPJ1

6 0
1 year ago
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed
Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

\Delta x \Delta p\geq \frac{h}{4\pi}      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}

The minimum kinetic energy is calculated by using the following formula:

k=\frac{(\Delta p)^2}{2m}       (2)

m: mass of the proton = 1.67*10^{-27}kg

k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J

in eV you have:

2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV

The kinetic energy of the proton is 1.29eV

7 0
3 years ago
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