Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run?

Maya and Kenzie are discussing oil spills and how they impact the environment. How can humans help reduce the impact of oil spil

madam [21]

The correct answer is A.

6 0

3 years ago

Read 2 more answers

How are cactus adapted to survive in deserts?

andreev551 [17]

1. They have evolved their leaves into spikes for minimum water loss through transpiration.

2. They have a waxy layer for minimum water loss.

3. They have thick walls for minimum water loss.

4. They can take water from atmosphere.

5. They change the photo energy from Sun into an intermediate stage and store it, so that they can make food even in night.

7 0

2 years ago

Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo

Naily [24]

Answer:

t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

y = y₀ + $v_{oy}$ t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

0 = y₀ – ½ g t²

t= √ (2 y₀/g)

calculemos

t= √ ( 2 12 / 9,8)

t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida

x = v₀ₓ t

x = 80 1,56

x= 124,8 m

La velocidad de impacto cuando toca el suelo

vx = v₀ₓ = 80 ms

$v_{y}$ = v_{oy} – gt

v_{y} = - 9,8 1,56

v_{y} = - 15,288 m/s

la velocidad es

v = (80 i^ - 15,288 j ) m/s

Traducttion

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( $v_{oy}$ = 0)

0 =y₀I - ½ g t²

t = √ (2y₀ / g)

let's calculate

t = √ (2 12 / 9.8)

t = 1.56 s

Projectile range is the horizontal distance traveled

x = v₀ₓ t

x = 80 1.56

x = 124.8 m

Impact speed when it hits the ground

vₓ = v₀ₓ = 80 ms

v_{y} = v_{oy} - gt

v_{y} = - 9.8 1.56

v_{y} = - 15,288 m / s

the speed is

v = (80 i ^ - 15,288 j) m / s

7 0

3 years ago

Which term describes an image that is in the opposite orientation than the object from which it was formed? virtual erect invert

DiKsa [7]

Answer:

lateral

...............

7 0

3 years ago

An electron with speed 2.45 x 10^7 m/s is traveling parallel to a uniform electric field of magnitude 1.18 x 10^4N/C . How much

cupoosta [38]

Answer:

time will elapse before it return to its staring point is 23.6 ns

Explanation:

given data

speed u = 2.45 × $10^{7}$ m/s

uniform electric field E = 1.18 × $10^{4}$ N/C

to find out

How much time will elapse before it returns to its starting point

solution

we find acceleration first by electrostatic force that is

F = Eq

here

F = ma by newton law

so

ma = Eq

here m is mass , a is acceleration and E is uniform electric field and q is charge of electron

so

put here all value

9.11 × $10^{-31}$ kg ×a = 1.18 × $10^{4}$ × 1.602 × $10^{-19}$

a = 20.75 × $10^{14}$ m/s²

so acceleration is 20.75 × $10^{14}$ m/s²

and

time required by electron before come rest is

use equation of motion

v = u + at

here v is zero and u is speed given and t is time so put all value

2.45 × $10^{7}$ = 0 + 20.75 × $10^{14}$ (t)

t = 11.80 × $10^{-9}$ s

so time will elapse before it return to its staring point is

time = 2t

time = 2 ×11.80 × $10^{-9}$

time is 23.6 × $10^{-9}$ s

time will elapse before it return to its staring point is 23.6 ns

7 0

3 years ago