2 years ago

The following nuclear reaction is balanced. True False

Physics

2 answers:

Dovator [93]2 years ago

8 0

False because one ways more than the other

elena-14-01-66 [18.8K]2 years ago

5 0

I think its true for tht one!!

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A(n) __________ refers to a substance or treatment that is often used in an experiment but does not have an actual medicinal or

sweet [91]

Answer:

placebo

Explanation:

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1 year ago

Read 2 more answers

The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will b

pogonyaev

Answer:

Lower

gsintheta (gsinθ)

Explanation:

The sum of forces resolved parallel to the inclined plane is given by;

F - mgsinθ = 0

ma - mgsinθ = 0

ma = mgsinθ

a = gsinθ

Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.

therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.

The acceleration along the ramp, is gsintheta (gsinθ)

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3 years ago

A body is accelerated continuously. What is the form of the graph

Andrej [43]

Answer:

If a body is constantly accelerating the graph would be one-quadrant

4 0

2 years ago

All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will

VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

$v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}$

since $v_y = 0$ at the top of its trajectory. Plugging in the numbers,

$t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}$

8 0

2 years ago

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

$v_i = 5.10 m/s$

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$d = -105 m$

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$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

7 0

3 years ago