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3 years ago

What does an atomic nucleus give off a particle?

Physics

1 answer:

____ [38]3 years ago

5 0

The correct answer is answer choice B.

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An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi

Amanda [17]

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

8 0

3 years ago

What is the term to describe the rate of flow of electricity?

SVEN [57.7K]

The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.

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3 years ago

What can happen when two plates bordering each other suddenly move and release energy

Luda [366]

A. Earth crust shakes
Sudden movements in the fault line, ex: San Andreas fault line, can cause sudden earthquakes,

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3 years ago

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contractor will use a package emulsion having a specific gravity of 1.25 and relative bulk strength of 130 to open an excavation

tester [92]

Answer:

The burden distance is 7 ft

Solution:

As per the question:

Specific gravity of package emulsion, $SG_{E} = 1.25$

Specific gravity of diabase rock, $SG_{R} = 2.76$

Diameter of the packaged sticks, d = 3 in

Now,

To calculate the first trail shot burden distance, B:

$B = [\frac{2SG_{E}}{SG_{R}} + 1.5]\times d$

$B = [\frac{2\times 1.25}{2.76} + 1.5]\times 3 = 7.22$

B = 7 ft

5 0

3 years ago

A 25kg mass is suspended at the end of a horizontal, massless rope that extends from a wall on the left and from the end of a se

patriot [66]

<h2>Answer:20.97g N,32.63g N</h2>

Explanation:

We consider the forces at the knot.

The vertical forces are

$T_{2}Sin(50^{0}})$ is the vertical component of tension $T_{2}$ at the knot.

$-25g$ is the weight of the mass $25Kg$ acting downwards.

The horizontal forces are

$-T_{1}$ is the tension in the rope acting left.

$T_{2}Cos(50^{0})$ is the horizontal component of tension $T_{2}$ acting towards right.

Since the knot has no mass,it is always in equilibrium.

So,the sum of forces acting on it will be zero.

Balancing vertical forces gives,

$T_{2}Sin(50^{0}})$ $-25g=0$

$T_{2}$ = $32.63gN$

Balancing horizontal forces gives,

$T_{2}Cos(50^{0})$ $-T_{1}=0$

$T_{1}=32.63g\times Cos(50^{0})=20.97gN$

7 0

3 years ago

Read 2 more answers