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Veseljchak [2.6K]

3 years ago

6

I have my finals exams and I have an activate exam paper, I wanna find it on the internet for free, does anyone know a website t

hat I can get it for free oj

1 answer:

soldi70 [24.7K]3 years ago

8 0

So if you wanna 4155162622

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4. How does nearness to water affect the climate zone of an area?

Afina-wow [57]

Answer:

Choice A.

Nearness to a body of water causes an increase in humidity, due to the higher rate of evaporation.

6 0

2 years ago

Read 2 more answers

a 65 kg skater at rest on a frictionless rink throws a 2 kg ball, giving the ball a velocity of 7 m/s. What is the velocity of t

gayaneshka [121]

The answer to your question is 33

8 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

3 years ago

A point moves on the x-axis in such a way that its velocity at time t (t > 0) is given by v=ln t/t . At what value of t does

Olenka [21]

Answer:

Explanation:

Given

Velocity of point is given by $v=\frac{\ln t}{t}$

To get maximum or minimum velocity differentiate v w.r.t t

$\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\frac{1}{t}\times t-1\times \ln(t)}{t^2}$

so $1-\ln (t)$ should be equal to zero

$\ln (t)=1$

$t=e$

i.e. $t=2.718\ s$

5 0

4 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago