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3 years ago

Why do some nucleus release electrons?

Physics

1 answer:

fredd [130]3 years ago

5 0

Answer:

Electrons are not little balls that can fall into the nucleus under electrostatic attraction

Explanation:

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What is an example of a non contact force

olchik [2.2K]

Answer: Gravitational force

Explanation:

A non contact force can be described as a force applied to an object by another body that is not in direct contact with it.

For example, an object thrown upwards will return back due to the force of gravity acting on it. So, it means Gravitational force is acting on the body without necessarily being in contact with that body.

8 0

4 years ago

PLEASE SOLVE QUICKLY!!! Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

Explanation:

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0

3 years ago

Two points are on a disk turning at constant angular velocity. One point is on the rim and the other halfway between the rim and

DerKrebs [107]

Answer:

The point on the rim

Explanation:

All the points on the disk travels at the same angular speed $\omega$ , since they cover the same angular displacement in the same time. Instead, the tangential speed of a point on the disk is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the disk

As we can see, the tangential speed is directly proportional to the distance from the centre: so the point on the rim, having a larger r than the point halway between the rim and the axis, will have a larger tangential speed, and therefore will travel a greater distance in a given time.

3 0

4 years ago

Why does a compass give unreliable readings when used near electrical appliances

insens350 [35]

Answer:

Explanation:

Since the compass uses a magnetic field, if anything else magnetic is near it, the compass will start acting up. Making it unreliable so keep magnets away!

8 0

3 years ago

What is the minimum speed needed to fire a champagne cork a distance of 11m?

Crank

To start with solving this problem, let us assume a launch angle of 45 degrees since that gives out the maximum range for given initial speed. Also assuming that it was launched at ground level since no initial height was given. Using g = 9.8 m/s^2, the initial velocity is calculated using the formula:

(v sinθ)^2 = (v0 sinθ)^2 – 2 g d

where v is final velocity = 0 at the peak, v0 is the initial velocity, d is distance = 11 m

Rearranging to find for v0: 
v0 = sqrt (d * g/ sin(2 θ))

v0 = 10.383 m/s

8 0

4 years ago