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Genrish500 [490]

3 years ago

12

A football is kicked from ground level with an initial velocity of 23.6 m/s at angle of 57.5° above the horizontal. How long, in

seconds, is the football in the air before it hits the ground? Ignore air resistance.

1 answer:

aniked [119]3 years ago

3 0

Answer:

6 seconds is the best answer

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What is Resistance? How is it measured?

ser-zykov [4K]

Answer:

Resistance is a measure of the opposition to current flow in an electrical circuit.

Explanation:

6 0

2 years ago

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(URGENT 75 points)You are given two same party balloons (one filled with helium and the

skad [1K]

$\frac{100}{0.214+-0.006}$ = 454.55 g/cm3

I'm not too sure since the graduated cylinder was missing and I really don't know how to do it then. But give this a shot. Are you sure it wasn't a graduated cylinder, because I have no idea what that means

5 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

A camera lens with focal length f = 50 mm and maximum aperture f>2

Brut [27]

Answer:

The minimum distance between two points on the object that are barely resolved is 0.26 mm

The corresponding distance between the image points = 0.0015 m

Explanation:

Given

focal length f = 50 mm and maximum aperture f>2

s = 9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D

Substituting the given values, we get –

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5

Y’ = 0.0015 m

8 0

3 years ago

Dòng điện dịch là:

yarga [219]

Answer: the answer is a

Explanation:

8 0

3 years ago