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igor_vitrenko [27]

3 years ago

11

The Falcon 9 rocket has a mass of 549,000 kg and in 70 seconds into the launch, the rocket reaches a speed of 343.2 m/s. What is

the rocket's momentum?

1 answer:

yaroslaw [1]3 years ago

8 0

Explanation:

=5.49*10^5*343.2

=1.8846*10^8

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Physics help please!

Oliga [24]

Answer:

student attach a save block to a horizontal spring so that the block spring system will oscillator with the block spring system released from rest horizontal position that is not the systems equilibrium position well this question regards about the energy used the answer may be 0.73 Joel ok you just try it ok verified

Explanation:

apply applied the potential energy value mean the formula MGH write it means what mass into gravitation in to height

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3 years ago

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

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1 year ago

The OBD-II catalytic converter monitor uses _______ sensor(s) to monitor catalytic converter efficiency.

oksano4ka [1.4K]

C comment for an explanation

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3 years ago

In a vacuum, two particles have charges of q1 and q2, where q1 = +3.11 μC. They are separated by a distance of 0.241 m, and part

lesya [120]

Answer:

Charge of particle 2, $q_2=-7.13\ \mu C$

Explanation:

Given that,

Charge 1, $q_1=3.11\ \mu C=3.11\times 10^{-6}\ C$

The distance between charges, r = 0.241 m

Force experienced by particle 1, F = 3.44 N

We need to find the magnitude of electric charge 2. It can be calculated using formula of electrostatic force. It is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$q_2=\dfrac{Fr^2}{kq_1}$

$q_2=\dfrac{3.44\times (0.241)^2}{9\times 10^9\times 3.11\times 10^{-6}}$

$q_2=7.13\times 10^{-6}\ C$

or

$q_2=7.13\ \mu C$

So, the magnitude of electric charge 2 is $q_2=7.13\ \mu C$ . Since, the force is attractive then the magnitude of charge 2 must be negative.

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3 years ago

How much energy is released for alpha decay of 236u?

lapo4ka [179]

<span>
E = mc^2 = 8.152*10^-30 *(3.00 *10^8)^2 = 7.336 *10^-13 J. </span>

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3 years ago