Find the direction of this vector

In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text

Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second , change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

= 4.859 x 10⁻² oscillations .

5 0

2 years ago

A shaving or makeup mirror is designed to magnify your face by a factor of 1.40 when your face is placed 20.0cm in front of it

Dennis_Churaev [7]

Answer:

(a) convex mirror

(b) virtual and magnified

(c) 23.3 cm

Explanation:

The having mirror is convex mirror.

distance of object, u = - 20 cm

magnification, m = 1.4

(a) As the image is magnified and virtual , so the mirror is convex in nature.

(b) The image is virtual and magnified.

(c) Let the distance of image is v.

Use the formula of magnification.

$m =-\frac{v}{u}\\1.4=-\frac{v}{-20}\\v =28 cm$

Use the mirror equation, let the focal length is f.

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\frac{1}{f}=\frac{1}{28}+\frac{1}{20}\\\frac{1}{f}=\frac{28+20}{560}\\f=11.67cm$

Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm

5 0

3 years ago

A rock is launched at angle theta=53.2∘ above the horizontal from an altitude of ℎ=182 km with an initial speed ????0=1.61 km/s.

Mariulka [41]

Answer:

The rock's final speed at the required altitude will be 42.24 m/s.

Explanation:

Let's start by finding the initial vertical speed.

Vertical Speed = 1.61 * Sin (53.2°)

Vertical Speed = 0.8 m/s

We want to know the speed of the rock when it is at an altitude of 91 km.

The total displacement of the rock from its starting position will thus be equal to -91 km

We can use this in the following equation:

$s=u*t+\frac{1}{2} (a*t^2)$

$-91=0.8*t+\frac{1}{2} (-9.8*t^2)$

t = 4.3918 seconds

Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:

$V=U+at$

$V=0.8+(-9.8)*(4.3918)$

$V = -42.24$

Thus the rock's final speed at the required altitude will be 42.24 m/s.

8 0

3 years ago

Read 2 more answers

What should we do to be a computer engineer ‍ ‍?

ycow [4]

Answer:

to be an eginere u would have to go to college and study hard

Explanation:

6 0

2 years ago

Read 2 more answers

A student sits at rest on a piano stool that can rotate without friction. The moment of inertia of the student-stool system is 4

irina [24]

Here We can use principle of angular momentum conservation

Here as we know boy + projected mass system has no external torque

Since there is no torque so we can say the angular momentum is conserved

$mvL = (I + mL^2)\omega$

now we know that

m = 2 kg

v = 2.5 m/s

L = 0.35 m

I = 4.5 kg-m^2

now plug in all values in above equation

$2\times 2.5 \times 0.35 = (4.5 + (2\times 0.35^2))\omega$

$1.75 = [4.5 + 0.245]\omega$

$1.75 = 4.745\omega$

$\omega = 0.37 rad/s$

so the final angular speed will be 0.37 rad/s

4 0

3 years ago