If you exert a force of 51 n to walk 6 m up a flight of stairs in 9 seconds, how much power do you use? answer in units of w.

It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd tim

GaryK [48]

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

6 0

3 years ago

Problem 6: A bullet in a gun is accelerated from rest from the firing chamber to the end of the barrel at an average rate of 6.3

solong [7]

Answer:

v = 5.166 10² m / s

Explanation:

We can solve this exercise using the kinematics equations

v = v₀ + at

as the bullet starts from rest its initial velocity is zero

v = a t

let's calculate

v = 6.3 10⁵ 8.2 10⁻⁴

v = 5.166 10² m / s

3 0

2 years ago

Electromagnetic waves differ fundamentally from either water or sound waves because they

stepladder [879]

<span>Electromagnetic waves differ fundamentally from either water or sound waves because they does not require any medium, they can travel in free space (vacuum)

Hope this helps!</span>

8 0

3 years ago

Three objects of the same mass begin their motion at the same height. One object falls straight down, one slides down a low-fric

erik [133]

Answer:

D. Same

Explanation:

Because only gravity is doing the work on the objects, and gravity is constant for all the objects

4 0

3 years ago

The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when

mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, $a=-8\ m/s^2$ (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

$x_1=ut_1+\dfrac{1}{2}at^2$

$20=4u+\dfrac{1}{2}(-8)\times 4^2$

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

$v=u+at$

$0=21+(-8)t$

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

$v=21+(-8)\times 11$

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

$d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|$

$d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|$

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.

5 0

3 years ago