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3 years ago

A particle moves along a straight line and its position at time t is given by

1 answer:

3 0

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

if a(t) >0, it means 12t-48>0, and t >4, I = ]4, infinity[

the particle is speeding down

if a(t) < 0, it means 12t-48>0, and t < 4, I = ] -infinity, 4[

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

moving backward if v(t)< 0

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

so after taking care the sign of 6t² - 48 t +90,

v(t)> o when t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward

v(t) < o when t is in I = ]1, 4[,the particle is moving backward

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A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55

SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point. At the rest, block doesn't have kinetic energy and since it is on the reference point(as we decided) it also has no potential energy.

Under the force block gains;

W = F.x → $W=2,55.0,71=1,8105\frac{N}{m}$

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = $\frac{1}{2} mV^{2} + mgh$

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In our situation

$\frac{0,71}{sin90} =\frac{h}{sin32}$ → $h=0,376$

Therefore

$1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376$

→ $V=2,416$

7 0

3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago

Unbalanced contact forces between several object would result in ?

Svetlanka [38]

In a reaction with a directing vendor of the sum forces towards some object or nothing

3 0

3 years ago

A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicul

gladu [14]

Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

$\theta$ is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

$\theta=90^{\circ}$

and the formula reduces to

$F=qvB$

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

So, the path of the particle will be a circle.

4 0

4 years ago

Which of the following objects would have the smallest wavelength at the same

Liono4ka [1.6K]

Answer:

Planet Earth

Explanation:

7 0

3 years ago