All categories

1 year ago

A stone is dropped from a high cliff vertically. After 6

Physics

1 answer:

bonufazy [111]1 year ago

4 0

Answer:

Ay=v₂t + ½ gt² is the best choice to solve this problem.

The height of the cliff is = 176.4 meters.

Explanation:

Given,

initial velocity = 0m/s as it is dropped vertically

acceleration = g = 9.8 m/s²

time = 6s

Height =?

Using the equation Ay=v₂t + ½ gt², we can calculate Ay, which is the height.

So,

Ay=v₂t + ½ gt²

by substituting v, t, and g, we get:

Ay= 0 × 6 + ½ × 9.8 × 6²

Ay = 0 + 4.9 × 36

Ay = 176.4 m

You might be interested in

An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary

brilliants [131]

Answer:

Explanation:

When the craft was stationary , weight will be balanced by tension

T = mg

T = 7000 N

when the craft was being lowered to the seafloor

drag force will act in upper direction , so

T₁ + 1800 = mg

T₁ = mg - 1800

= 7000 - 1800

= 5200 N

52 X 10² N

when the craft was being raised from the seafloor , Tension will act in downward direction

T₂ = mg+ 1800

T₂ = 7000 - 1800

= 8800N

3 0

3 years ago

Two resistors of 2 ohms each are connected in parallel, another resistor of 1 ohm is connected in series with the parallel combi

insens350 [35]

Answer:

The power dissipated in either one of the parallel resistors is 2 V

Explanation:

Given;

two parallel resistors, R₁ and R₂ = 2 ohms

The total resistance of the Two resistors of 2 ohms connected in parallel is;

$R_T = \frac{R_1R_2}{R_1+R_2} = \frac{2*2}{2+2} = \frac{4}{4} = 1 \ ohm$

when connected to another resistor of 1 ohm in series, the total resistance becomes;

Rt = R₁ + R₂

Rt = 1 + 1 = 2 ohms

Current in the circuit, I = voltage / total resistance

= 2 /2 = 1 A

the overall circuit has been resolved to series connection, and current flow in series circuit is constant.

Power = I²R

Thus, power dissipated in either one of the parallel 2 ohms resistors is;

Power = I²R = (1)² x 2 = 2 V

3 0

3 years ago

Two magnets are held apart. Once released, the South Pole of one magnet moves toward the North Pole of another magnet until the

tresset_1 [31]

Answer:

Kinetic energy and velocity are at zero when the magnets are held apart, and both increase rapidly when they are released and move together. Energy stored in the magnetic field decreases.

Explanation:

easy

6 0

3 years ago

A vertical beam of power intensity P (in Watts/m2 ) passes downward through a particular substance. The rate at which P decrease

Mashcka [7]

Answer:

x = 1162.5 W/m²

Explanation:

Since, the power decrease is proportional to the depth of the beam. Therefore, interpolation can be used to find the intensity of power at a depth of 1.25 m. First we calculate the slope from know points:

$Slope = \frac{\Delta y}{\Delta x} \\\\Slope = \frac{(1500 - 3000)\ W/m^{2}}{(2 - 0)\ m} \\\\Slope = -750\ W/m$

Now, we can find the unknown value by using this slope:

$Slope = -750\ W/m = \frac{(600 - x)\ W/m^{2}}{(2 - 1.25)\ m}\\\\(-750\ W/m)(0.75\ m) = (600 - x)\ W/m^{2}\\x = 600\ W/m^{2} + 562.5\ W/m^{2}\\$

<u>x = 1162.5 W/m² (Power intensity at depth of 1.25 m)</u>

5 0

3 years ago

A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the f

topjm [15]

Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s

8 0

3 years ago