Ask question

Ask question

All categories

1 month ago

7

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

t time, y0 is the initial height (at t=0), v0 is the initial velocity, and g=9.8m/s2 (the acceleration due to gravity). If you stand at the edge of a cliff that is 75 m high and throw a rock directly up into the air with a velocity of 20 m/s, at what time will the rock hit the ground? (Note: The Quadratic Formula will give two answers, but only one of them is reasonable.)

1 answer:

Softa [3K]1 month ago

8 0

Answer:

$t=6.4534 s$

Explanation:

This problem requires applying concepts related to objects in free fall.

Our known variables consist of initial height, initial velocity, and the acceleration attributed to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the conclusion of the fall, the rock reaches the ground meaning the final height y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we analyze the equation:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This represents a standard Quadratic Formula, which we can resolve using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Given that time cannot be negative, the valid solution is

$t=6.4534s$

You might be interested in

Each shot of the laser gun most favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by th

Yuliya22 [3333]

U = 1794.005 × 10⁶ J. Explanation: Information provided indicates that the capacitance of the original capacitor is C = 1.27 F, and the potential difference applied to it is V = 59.9 kV, or 59.9 × 10³ V. The potential energy (U) for the capacitor is determined by the formula: U = (1/2) × C × V². Substituting the respective values, we find U = (1/2) × 1.27 × (59.9 × 10³)², resulting in U = 1794.005 × 10⁶ J.

7 0

18 days ago

Ability of the muscles to function effectively and efficiently without undue fatigue

Keith_Richards [3271]

Response:

Physical well-being

Clarification:

7 0

2 months ago

Why didn't the astronauts land on the moon 3.17 answers punchline?

kicyunya [3294]

Answer:

Responses to the 3.17 punchline varied among many individuals, with some suggesting that it was a "full" moon day which prevented the astronauts from landing.

Others claimed that the astronauts took off during daylight hours when the moon was not visible. There were also comments that indicated that 'astro' refers to stars rather than satellites, explaining why they did not land.

A few even noted that 'astro naut' sounds like 'naught,' meaning zero (0), as a possible reason for their failure to land.

3 0

1 month ago

Read 2 more answers

A loop of wire with cross-sectional area 1 m2 is inserted into a uniform magnetic field with initial strength 1 T. The field is

serg [3582]

Answer:

La magnitud del EMF es 0.00055 volts

Explanation:

El EMF inducido es proporcional al cambio en el flujo magnético según la ley de Faraday:

$emf\,=-\,N\, \frac{d\Phi}{dt}$

Como en nuestro caso hay solo un lazo de alambre, entonces N=1 y obtenemos:

$emf\,=-\,N\, \frac{d\Phi}{dt}$

Necesitamos expresar el flujo magnético dada la geometría del problema;

$\Phi=B\,\,A$ donde A es el área de la bobina que permanece constante con el tiempo, y B es el campo magnético que cambia con el tiempo. Por lo tanto, la ecuación para el EMF se convierte en:

$emf\,=-\,N\, \frac{d\Phi}{dt} = \frac{d\Phi}{dt} =-\frac{d\,(B\,A)}{dt} =-\,A\,\frac{d\,(B)}{dt}=- 1\,m^2(2\,\,T/h})= -2\,\,m^2\,T/(3600\,\,s)= -0.00055\,Volts$

8 0

16 days ago

Two charges of 15 pC and −40 pC are inside a cube with sides that are of 0.40-m length. Determine the net electric flux through

Keith_Richards [3271]

To address this issue, we will utilize the principles related to Gauss' law, which states that the electric flux across a surface corresponds to the object's charge divided by the permittivity of vacuum. In mathematical terms, this can be expressed as

$\phi = \frac{Q_{net}}{\epsilon_0}$

It's crucial to remember that the net charge equals the difference between the two specified charges, so upon substitution,

$\phi = \frac{(15-40)*10^{-12}C}{8.85*10^{-12}C^2/Nm^2}$

$\phi = 2.82WB$

The negative sign indicates that the flux is directed into the surface

4 0

1 month ago