CAN SOME PLZ HELP I REALY NEED THIS PLZ!!!!

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;

<em>t is the time of motion </em>
<em /> $v_0_y$ <em> is the initial vertical velocity of the stone = 0</em>

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

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.

"<em>Your question is not complete, it seems be missing the following information;"</em>

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

What kind of habitat do we live in

Aleksandr-060686 [28]

the worst because its not enough habitat and when the cutting down a whole lot of rainforest at a time which is killing animals escpecialy rainforest animals

8 0

3 years ago

A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and

irina [24]

The eccentricity of its orbit is $U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where $, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$$

$M=1.99 \times 10^{30} \mathrm{~kg}=$ mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

To learn more about mass, refer to:

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4 0

2 years ago

Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a

weeeeeb [17]

Answer:

force becomes one - ninth

Explanation:

According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.

Here both the charges remains same but the distance is variable.

So, we can say that

$F \alpha \frac{1}{d^{2}}$ .... (1)

Where d be the distance between the tow charges

As the distance between two charges increases by factor of three, let the new force be F'.

$F' \alpha \frac{1}{9d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

$\frac{F'}{F}=\frac{d^{2}}{9d^{2}}$

$F'=\frac{F}{9}$

Thus, the force becomes one - ninth times the initial force.

6 0

3 years ago

A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with

Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

½ m v₀² + m g h = ½ m v²

v² = v₀² + 2 g h

Let's calculate

v = √(1.23² + 2 9.8 1.69)

v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

W = ΔK

-fr x = 0- ½ m v²

Newton's second law

N- W = 0

The equation for the friction is

fr = μ N

fr = μ m g

We replace

μ m g x = ½ m v²

x = v² / 2μ g

Let's calculate

x = 5.89² / (2 0.255 9.8)

x = 6.94 m

6 0

3 years ago

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