A 37 N object is lifted to a height of 3 meters, What is the potential energy of this object?

Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000

babunello [35]

Answer:

6000 lines/cm

Explanation:

From the question we are told that:

Grating 1=4000 lines/cm

Grating 2=6000 lines/cm

Generally The Spread of fringes is Larger when the Grating are closer to each other

Therefore

Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm

7 0

3 years ago

If Zack has a mass of 80 kg calculate his weight on Earth. Please show work, thank you!

Alina [70]

784 Newton ( mass x gravity = 80 x 9.8)

4 0

3 years ago

a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s

Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

$h = v_0_yt + \frac{1}{2} gt^2$

where;

t is the time of motion
$v_0_y$ is the initial vertical velocity of the stone = 0

$h = \frac{1}{2} gt^2$

The time taken by the first stone to hit the ground is calculated as;

$t_1 = \sqrt{\frac{2h}{g} }$

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

$t_2 = \sqrt{\frac{2h}{g} } + 1$

$t_2 = t_1 + 1$

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"Your question is not complete, it seems be missing the following information;"

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0

3 years ago

If the archerfish spits its water 30 degrees from the horizontal aiming at an insect 1.2 m above the surface of the water, how f

Burka [1]

Answer:

The speed is $v = 9.8 \ m/s$

Explanation:

From the question w are told that

The angle made is $\theta = 30^o$

The distance above the surface of the water is $h_{max} = 1.2 \ m$

The value of $g = 10 \ m/s^2$

The maximum height attained by the fish is mathematically evaluate as

$h_{max} = \frac{v^2 sin ^2 \theta }{2g }$

Making v which is the speed of the fish the subject of the formula

$v = \sqrt{ \frac{2gh_{max}}{ sin^2 \theta } }$

substituting values

$v = \sqrt{ \frac{2*10 *1.2 }{ [sin (30)]^2 } }$

$v = 9.8 \ m/s$

3 0

3 years ago

For horizontally-launched projectiles, which of the following describes acceleration in both directions with a = 0 and a = -9.8m

natulia [17]

Answer:

4 is the best option for it as there are some stuff in grade to ye

5 0

3 years ago