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3 years ago

6. What is the resultant force on each car below? Remember TWO pieces of

Physics

1 answer:

tamaranim1 [39]3 years ago

4 0

Explanation:

In first case, the forces on LHS and on RHS is the same i.e. 3 N. The force acting on the car is balanced force. As a result, the car will not move at all.

In second case,

Force on RHS = 2000 N

Force on LHS = -6000 N

Net force acting on it is given by :

F = 2000+(-6000)

= -4000 N

Hence, this is the required solution.

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olga2289 [7]

4.6 billion years agony

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3 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

$d = 2 - 0 = 2cm$

so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

$a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2$

at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

At t = 2 we have

$y = 2 cm$

$a = -\frac{\pi^2}{16}*2$

$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

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3 years ago

A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser sp

uranmaximum [27]

Answer:

a) 351351.35m/s

b) 1.044*10^{-8}kg/C

Explanation:

a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.

$F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}$

b) the mass-charge ratio is given by:

$\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}$

hope this helps!!

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3 years ago

Lenny loves physics and math. In which Energy career pathway would these interests be the most helpful?

vazorg [7]

Answer:

it is D

Explanation:

cause i took the test

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4 years ago

Read 2 more answers

A 2500 N force accelerates a car at a rate of 3.0 m/s^2. What is the car’s mass? 250 kg

Ronch [10]

Apply Newton's second law to the car's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

F = 2500N, a = 3.0m/s²

Plug in and solve for m:

2500 = m(3.0)

m = 830kg

5 0

4 years ago