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2 years ago

Answer this question please and you will can’t the brain list

Physics

1 answer:

Eduardwww [97]2 years ago

4 0

Answer:

We know that velocity equals to difference in distance divided by difference in time.

1st period: velocity is increasing. Where v = (8-0)/(3-0) = 8/3 m/s.

2nd one: velocity decreases. Where

v=(5-8)/(4-3) = (-3) m/s.

3rd one: velocity is constant because no change in distance, so v=0.

4th one: velocity increases again. Where v=(10-5)/(8-6)= 5/2 m/s.

5th one: velocity is constant again, so v=0.

6th one: velocity decreases, where

v=(6-10)/(15-12)= (-4)/3 m/s.

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What is the safety rules?

Black_prince [1.1K]

Answer:

Quick question do you mean what are some safety rules

Explanation:

Crosswalk, Stop sign,

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3 years ago

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What is the correct name for this formula BaCI 2?

KIM [24]

BaCI2 stands for Barium Chloride.

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2 years ago

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

7 0

2 years ago

The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the

pav-90 [236]

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A ten times.

So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:

$T=\frac{t}{n}$

t is the total time interval

n is the number of oscillations

$T=\frac{0.17}{10}$

10T = 0.17

T = 0.17/10 = 0.017s

8 0

2 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago