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ahrayia [7]
3 years ago
13

When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le

ngth of the shorter stick in feet?
Physics
1 answer:
77julia77 [94]3 years ago
7 0

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

x+2.93

Both cover a magnitude of 8.32 ft, therefore

x +(x+2.97) = 8.32

Now solving for x we have,

x + (x + 2.93) = 8.32

2x + 2.93 = 8.32

2x = 8.32 - 2.93

x = \frac{ 8.32 - 2.93}{2}

x = 2.695 ft

Therefore the shorter stick is 2.695ft long.

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1. A man travels 45 m in 35 seconds. How fast is that?<br> What’s the answer
Natalija [7]

Answer:

1.28

Explanation:

If you want to find the m/s you would divide distance by time, so

45 divided by 35 would equal 1.28571429 and so on.

you can just write the three first numbers.

8 0
3 years ago
Explain why secondary succession usually occurs more rapidly than primary succession.
kiruha [24]

Answer:

Explanation:

As it mostly occurs on a surface where an ecosystem has previously existed. Primary succession only occurs on a surface where no ecosystem existed ever before!

6 0
3 years ago
Question 2
Delvig [45]

Answer:

Approximately 73\; {\rm N}, assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, a, is constant during the entire descent.

Let x denote the displacement of this ball and let t denote the duration of the descent. The SUVAT equation x = (1/2)\, a\, t^{2} would apply.

Rearrange this equation to find an expression for the acceleration, a, of this ball:

\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}.

Note that x = 11\; {\rm m} and t = 1.5\; {\rm s} in this question. Thus:

\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}.

Let m denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is a, the net external force on this ball would be m\, a.

Since m = 7.5\; {\rm kg} and a \approx 9.78\; {\rm m\cdot s^{-2}}, the net external force on this ball would be:

\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}.

4 0
2 years ago
Diffusion of particles through protein channels is called
mamaluj [8]

Answer:

Facilitated diffusion.

Explanation:

Facilitated diffusion is the movement of molecules across a biological cell membrane by the use of a membrane protein.

It is a spontaneous passive movement and it can also be reffered to as FACILITATED TRANSPORT or PASSIVE-MEDIATED TRANSPORT.

4 0
3 years ago
A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va
ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

U=\dfrac{1}{2}CV^2

C=\dfrac{\varepsilon _oA}{d}

U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2

If energy become double U' = 2 U then d'

U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0
3 years ago
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