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3 years ago

zeros are always considered significant digits when they are to the left of the decimal point True or False

Physics

2 answers:

zlopas [31]3 years ago

8 0

Answer:

false

Explanation:

because they does'nt Work for anything, are like they are not there

Hunter-Best [27]3 years ago

4 0

Answer:

False

Explanation:

The rules for significant figures are:

Non-zero digits are always significant.
Zeros between significant digits are also significant.
Trailing zeros are significant only after a decimal point.

A zero before a decimal point is only significant if there's a non-zero digit before it.

For example, in the number 10.5, the zero is significant, because it is between significant digits.

But in the number 0.5, the zero is not significant.

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A projectile is launched into the air with the initial speed of vi = 40 m/s at a launch angle of 20 degrees above the horizontal

Sphinxa [80]

The range of the projectile is 188 m

Explanation:

The motion of the arrow in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

The path of a projectile is the combination of these two motions: see figure in attachment.

In order to find the horizontal range of the projectile, we just need to calculate the horizontal distance travelled.

We have:

t = 5.0 s (time of fligth of the projectile)

and the horizontal velocity is constant, and it is given by

$v_x = v_i cos \theta$

where

$v_i = 40 m/s$ is the initial velocity

$\theta=20^{\circ}$ is the angle of projection

Substituting,

$v_x = (40)(cos 20^{\circ})=37.6 m/s$

And therefore, the range of the projectile is:

$d=v_x t = (37.6)(5.0)=188 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac

xenn [34]

Answer:

$62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56) } \, dy = 7875 lb$

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

$\int\limits^a_b$ γhxdy

we know that γ=62.5 lb/ $ft^{3}$

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

$m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}$