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3 years ago

Please help will give the brainliest!please answer correctly.Urgent!!!!!!!!!!!!

Physics

1 answer:

ZanzabumX [31]3 years ago

6 0

a) Live wire

b) It has gained extra electrons.

c) Part of circuit can be turned off while other parts remain on.

d) $R=\frac{V}{I}=60 \Omega$

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If we know the total energy in a system is 30 J, and we know the PE is 20 J. What is the KE?

svetoff [14.1K]

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3 years ago

What is time and speed and acceleration

Andre45 [30]

Answer:

Acceleration is the rate of change of velocity with time.Acceleration occurs anytime an object's speed increases or decreases, or it changes direction. Much like velocity, there are two kinds of acceleration: average and instantaneous. Average acceleration is determined over a "long" time interval.

Explanation:

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3 years ago

A 26-g steel-jacketed bullet is fired with a velocity of 630 m/s toward a steel plate and ricochets along path CD with a velocit

Alecsey [184]

Answer:

F = - 3.53 10⁵ N

Explanation:

This problem must be solved using the relationship between momentum and the amount of movement.

I = F t = Δp

To find the time we use that the average speed in the contact is constant (v = 600m / s), let's use the uniform movement ratio

v = d / t

t = d / v

Reduce SI system

m = 26 g ( 1 kg/1000g) = 26 10⁻³ kg

d = 50 mm ( 1m/ 1000 mm) = 50 10⁻³ m

Let's calculate

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

With this value we use the momentum and momentum relationship

F t = m v - m v₀

As the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26 10⁻³ (-500 - 630) / 8.33 10⁻⁵

F = - 3.53 10⁵ N

The negative sign indicates that the force is exerted against the bullet

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3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

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3 years ago

What Is The frequency of an X Ray That had A Wavelength of 1.5* 10^-9

musickatia [10]

Here I come and we wanna go home!!!

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3 years ago

Please help will give the brainliest!please answer correctly.Urgent!!!!!!!!!!!!​

Please help will give the brainliest!please answer correctly.Urgent!!!!!!!!!!!!