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3 years ago

10

Magnet A and B are of equal magnetic strength and which position will magnets A&B have the greatest attraction for toward ea

ch other

2 answers:

Oksanka [162]3 years ago

8 0

Its D.

(sorry the picture got cut off)

77julia77 [94]3 years ago

3 0

When they are facing each other, almost touching

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Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh

coldgirl [10]

Answer:

You cant draw on brainly

Explanation:

6 0

2 years ago

Read 2 more answers

The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler

guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

$v^{2}=u^{2}+ 2as$

$620^{2}=0^{2}+ 2\times a \times 0.89$

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.

6 0

3 years ago

What is meant by infinite slew rate

Dennis_Churaev [7]

Am infinite slew rate means that the changes in the output voltage occur immensely when the input voltage changes.
Slew rate is measurement of the response of an operational amplifier. For an ideal operational amplifier, time delay is negligible. Hence it has an infinite slew rate.
In simpler terms it means that it can provide output voltage simultaneously with the input voltage changes.

Hope this helps :)

3 0

2 years ago

Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 1.80 cm from the axis to equal

Arturiano [62]

Answer:

The required angular speed ω of an ultra-centrifuge is:

ω = 18074 rad/sec

Explanation:

Given that:

Radius = r = 1.8 cm

Acceleration due to g = a = 6.0 x 10⁵ g

Sol:

We know that

Angular Acceleration = Angular Radius x Speed²

a = r x ω ²

Putting the values

6 x 10⁵ g = 1.8 cm x ω ²

Converting 1.8 cm to 0.018 m, also g = 9.8 ms⁻²

6 x 10⁵ x 9.8 = 0.018 x ω ²

ω ² = (6 x 10⁵ x 9.8) / 0.018

ω ² = 5880000 / 0.018

ω ² = 326666667

ω = 18074 rad/sec

7 0

3 years ago

The quantity of charge through a conductor is modeled as Q = (3.00 mC/s4)t4 − (2.00 mC/s)t + 9.00 mC. What is the current (in A)

Mumz [18]

Answer:

The current at time t = 4.00 s is 0.766 A.

Explanation:

Given that,

The quantity of charge through a conductor is modeled as :

$Q=(3t^4-2t+9)\ mC$

We need to find the current (in A) at time t = 4.00 s. We know that the rate of change of electric charge is called electric current. It is given by :

$I=\dfrac{dQ}{dt}\\\\I=\dfrac{d(3t^4-2t+9)}{dt}\\\\I=12t^3-2$

At t = 4 s

$I=12(4)^3-2=766\ mC/s\\\\I=0.766\ C/s=0.766\ A$

So, the current at time t = 4.00 s is 0.766 A. Hence, this is the required solution.

7 0

3 years ago