Find V V=25m. ____

A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavel

sattari [20]

Answer:

The wavelength is $\lambda = 1.2 * 10^8 nm$

Explanation:

From the question we are told that

The frequency of operation of the microwave is $f = 2.50 GHz = 2.50 *10^{9} \ Hz$

Generally the wavelength is mathematically represented as

$\lambda = \frac{c}{f}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$\lambda = \frac{3.0 *10^{8}}{ 2.50 *10^{9}}$

=> $\lambda = 0.12 \ m$

converting to nanometer

$\lambda = 1.2 * 10^8 nm$

6 0

3 years ago

An arrow is projected by a bow vertically up with a velocity of 40 m/s, and reaches a target in 3 s. What is the velocity of the

Fittoniya [83]

Answer:

Explanation:

Step one:

given data

initial velocity u= 40m/s

time taken t=3seconds

final velocity v=?

Step two:

applying the first equation of motion

v=u-gt--- (the -ve sign implies that the arrow is against gravity)

assume g=9.81m/s^2

v=40-9.81*3

v=40-29.43

v=10.57m/s

Step three:

how high the target is located

applying

s=ut-1/2gt^2

s=40*3-1/2(9.81)*3^2

s=120-88.29/2

s=120-44.145

s=75.86m

6 0

3 years ago

Which nucleus completes the following equation? A. 299 Np B. 20Pa C. 2 Pa D. - Np

blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

Collect like terms

92 + 1 = x

93 = x

x = 93

239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0

3 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

3 years ago

beth walked to the school in 80 seconds from a position 100 m west of the school. what was her velocity?

Ugo [173]

1.8 meters a second

100/80

4 0

3 years ago

Find V V=25m. _____=?? 3.5s.