All categories

2 years ago

Where is the acceleration zero?

Physics

1 answer:

ladessa [460]2 years ago

3 0

Answer:

When acceleration is zero (that is, a = dv/dt = 0), rate of change of velocity is zero. That is, acceleration is zero when the velocity of the object is constant. so probably D

Explanation:

You might be interested in

What do a control group and an experiment group have in common ?

Vlad [161]

Whatever the focus of the experiment is, plus any others factors that might influence the outcome of the experiment. If you are testing a new cancer drug, the experimental group and the control group must both be people with the same type of cancer, and both be a representative distribution of the population, all races, genders, ages, etc. You want the only difference in the two groups to be what you are studying, i.e. the effects of the drug.

7 0

2 years ago

Read 2 more answers

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

3 years ago

who brought astronomy out of the dark ages by declaring: "finally we shall place the sun himself at the center of the universe.

solong [7]

Nicolaus Copernicus was the one who brought astronomy out of the dark ages.

In one of his books "On the Revolutions of the Heavenly Bodies", he published the renowned line which declares his theory.

Nicolaus Copernicus, A polish astronomer put forward the theory that the is th Sun one that rests in the middle or center of the Universe and the planet Earth revolves around it on its axis every day.which was considered to be called the Heliocentric system.

To know more about the Heliocentric system refer to the link brainly.com/question/3491738?referrer=searchResults.

To know about Sir Nicolaus Copernicus' work refer to the link brainly.com/question/6699117?referrer=searchResults.

#SPJ4

6 0

2 years ago

A physics student swings a pail of water in a vertical circle 1.0 m in radius at a constant speed. If the water is NOT to spill

love history [14]

Answer:

(A) 3.1 m/s

(B) 2.0 s

Explanation:

At the minimum speed, the force of gravity equals the centripetal force.

mg = m v² / r

v = √(gr)

v = √(9.8 m/s² × 1.0 m)

v = 3.1 m/s

The time is the circumference divided by the speed.

t = (2π × 1.0 m) / (3.1 m/s)

t = 2.0 s

7 0

2 years ago

A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of

Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

$\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}$

Where, $\beta=\dfrac{c}{v}$

$\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}$

$\lambda_{o}=115\ nm$

Hence, The observer detects light of wavelength is 115 nm.

8 0

3 years ago