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3 years ago

According to the "Future Potential Source #2: Wind" article, what natural resource does wind power free up

1 answer:

6 0

In short, the key value added of CDR data over census or survey approaches is the potential to access current and comprehensive evidence on population size, density, and dynamics, information that is fundamentally necessary for managing any humanitarian emergency or disease-related disaster but which is often

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Calculate the amount of work done if you use a 100N force to push a 50kg box 5m across the kitchen floor.

irinina [24]

So if the formula for work is force times displacement times cosine(theta), you'd plug in the numbers

100x5 (since there's no angle in the problem, cosine(theta) isn't used

100x5 = 500

So the answer would be B.

Hope that helps!

6 0

2 years ago

Read 2 more answers

A car is initially traveling at 12 m/s when the driver sees a yellow light ahead. He accelerates at a constant 7 m/s^2 for 6 s i

Vitek1552 [10]

Answer:

v = 54m/s

Explanation:

$a = \frac{v - u}{t}$

a = 7m/s²

u = 12m/s

t = 6s

7 = (v-12)/6

v - 12 = 42

v = 54m/s

(Correct me if i am wrong)

3 0

3 years ago

Can someone help me?!!!!!

SOVA2 [1]

Answer:magnitude -5; angle 160°

Explanation:

Vector A is described as having magnitude 5 and angle -20°.

To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.

5 @ -20° = 5 @ 340°

5 @ -20° = -5 @ 160°

The third one is the answer.

8 0

3 years ago

ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po

Levart [38]

Answer:

$6.046N$

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

$F_n_e_t=F_g+F_t$

$F_t$ is the tension force. $F_g=9.8N/kg$ is the force of gravity.

$F_n_e_t=ma_c=mv^2/r\\$

where $r$ is the rope's radius from the fixed point.

From the net force equation above:

$F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N$

Hence the tension force is 6.046N

8 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

2 years ago