A 20kg bike accelerates at 10 m/s2. With what force was the person padeling?

I’ll give u BRAINLIEST PLEASE!! HURRY

zalisa [80]

Answer:

C

Explanation:

6 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is

Sonja [21]

Answer:

0.8895m

Explanation:

Cable diameter = 0.0125m

Mass of elevator = 6450kg

Young Modulus(E) = 2.11*10¹¹N/m

∇l (change in length) =

L = 362m

A = Πr², but r = d / 2 = 0.0125 / 2 = 0.00625m

A = 3.142 * (0.00625)² = 1.227*10^-4m²

Young Modulus (E) = Tensile stress / Tensile strain

E = (F / A) / ∇l / L

F = mg = 6450 * 9.8 = 63210N

2.11*10¹¹ = (63210 / 1.22*10^-4) / (∇l / 362)

2.11*10¹¹ = 5.18*10⁸ / (∇l / 362)

2.11*10¹¹ = (5.18*10⁸ * 362) / ∇l

2.11*10¹¹ = 1.875*10¹¹ / ∇l

∇l = 1.875*10¹¹ / 2.11*10¹¹

∇l = 0.8895m

The change in length is 0.8895m

8 0

3 years ago

What determines who is the most important person

qwelly [4]

Answer:

Putting your partner first means his or her needs, feelings, and wellbeing take priority over other people or things. A “sense of we” forms as you maintain this priority on purpose each day. You protect your relationship from being destroyed or damaged. You tend to your connection so it feels good to you b

5 0

2 years ago

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0

2 years ago