What did it mean if something was in blue glass?

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

2 years ago

The rate of spin of a disk is decreasing. A point on the disk that is near the rim of the disk will have...

NNADVOKAT [17]

At a point near the rim of the disk, it will have a non-zero radial acceleration and a zero tangential acceleration. Also known as centripetal acceleration, radial acceleration takes place along the radius of the disk. On the other hand, the tangential acceleration is along the path of disk's motion.

8 0

2 years ago

Help with these questions please , 15 PTS & brainliest

myrzilka [38]

For these two questions, first you need to know that the voltage across each branch of a parallel circuit is the same.

So, for Q5, we can first find out the voltage across R₂ by V=IR.

Voltage across R₂ = 2.5 × 8 = 20V

Since R₂ and R₃ are in parallel circuit, their voltage should be the same. Thus, voltage across R₃ is 20V.

So, by V=IR,

current of R₃ = $\frac{20}{4}$ = 5A

Q6. voltage across R₁ = 2 × 4 = 8V

∴voltage across R₂ = 8V

current of R₂ = $\frac{8}{8}$ = 1A

<h3>Alternative method</h3>

From these two examples, you can find out that the current of each branch of the parallel circuit is inversely proportional to the resistance of the branch.

ie. for Q5,

$\frac{R2}{R3}$ = $\frac{I3}{I2}$

$\frac{8}{4}$ = $\frac{I3}{2.5}$

I₃ = 5A

Q6. $\frac{R1}{R2}$ = $\frac{I2}{I1}$

$\frac{4}{8}$ = $\frac{I2}{2}$

I₂ = 1A

6 0

2 years ago

A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin

blsea [12.9K]

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

8 0

2 years ago

at the extact same time, youj drop a piece of notebook paper and your science out the window. the science book hit first? why?

jeyben [28]

The piece of paper has less mass and will glide down the window, whereas the textbook will go straight to the ground. Since the textbook has more mass and less ways of it being able to 'glide' the textbook will hit the ground first.

8 0

2 years ago