2. is a variable that gives location relative to an origi

The vine has a fixed end, so Tarzan's path is circular. So the vine must not only support Tarzan's weight (or some component of

Minchanka [31]

Answer: hello your question lacks some data attached below is the missing data

answer : T - mg = ma $_{c}$

Explanation:

Given that the vine has a fixed end and Tarzan's path is circular

At Tarzans lowest point the point can be expressed as shown below.

It can be expressed as : T - mg = ma $_{c}$

8 0

3 years ago

At a time when mining asteroids has become feasible, astronauts have connected a line between their 3740-kg space tug and a 5690

Gwar [14]

Answer:

61.4 s

Explanation:

The distance d₁ traveled by the asteroid:

$d_1=\frac{1}{2}a_1t^2$

The distance d₂ traveled by the space ship:

$d_2=\frac{1}{2}a_2t^2$

The total distance d:

$d=d_1+d_2=\frac{1}{2}(a_1+a_2)t^2$

Solving for time t:

$t=\sqrt{\frac{2d}{a_1+a_2}}=\sqrt{\frac{2d}{F(\frac{1}{m_1}+\frac{1}{m_2})}}=\sqrt{\frac{2dm_1m_2}{F(m_1+m_2)}}$

3 0

3 years ago

Read 2 more answers

archer shoots his arrow towards a target at a distance of 90 m, and hits ‘bullseye’ . Calculate the acceleration and time taken

mario62 [17]

Answer:

a =45 m/s2

t = 2 seconds

Explanation:

Hi, to answer this question we have to apply the next formula:

v^2 = u^2 +2 a d

Where:

v = final velocity = 90 m/s

u = initial velocity = 0 m/s (shots from rest)

a = acceleration (m/s2)

d = distance = 90m

90^2 = 0^2 + 2a(90)

Solving for a:

8,100= 180 a

8,100/180 = a

a = 45 m/s^2

For time:

v = u + at

90 = 0 + 45t

90/45=t

t =2 seconds

6 0

3 years ago

Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a ho

antiseptic1488 [7]

To solve this problem it is necessary to apply the concepts related to Malus' law. Malus' law indicates that the intensity of a linearly polarized ray of light that passes through a perfect analyzer with a vertical optical axis is equivalent to:

$I = I_0 cos^2(\theta)$

$I_0 =$ Indicates the intensity of the light before passing through the Polarizer,

I = The resulting intensity, and

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

There is 3 polarizer, then

For the exit of the first polarizer we have that the intensity is,

$I_2 = I_0cos^2(45)$

For the third polarizer then we have,

$I_3 = I_2 cos^{2}(45)$

Replacing with the first equation,

$I_3 = I_0cos^2(45)cos^{2}(45)$

$I_3 = I_0 (\frac{1}{2})(\frac{1}{2})$

$I_3 = I_0 \frac{1}{4}$

Therefore the transmitted intensity now is $\frac{1}{4}$ of the initial intensity.

5 0

4 years ago

What type of image can be larger or smaller than the object?

Stolb23 [73]

It’s D. An enlargement (hope this helps!)

4 0

3 years ago