Vector a cross b whole square<br><br>

a 2.0 kg ball is dropped from a height of 20 m onto a soft surface and rebounds to a height of 5.0 m . what is the magnitude of

Yuki888 [10]

Based on the data provided, the impulse of the floor on the ball is 59.4 Ns.

<h3>What is the impulse of the floor on the ball?</h3>

Using the equation of motion to determine the velocity at the end of the fall

v^2 = u^2 + 2gh

Where v is velocity at the end of fall

u is initial velocity = 0

g is acceleration due to gravity = 9.81 m/s^2

h is height = 20

Taking downward velocity as negative and up as positive

v^2 = 0 + 2 (9.81)(20)

v^2 = 392.4

v = - 19.8 m/s

The velocity, v after bouncing is calculated also:

u = 0

g = 9.81 m/s^2

h = 5.0 m

v^2 = 0 + 2(9.81)(5)

v^2 = 98.1

v = 9.904 m/s

Impulse = change in momentum
Impulse = m(v- u)

Impulse = 2.0 × (9.9 -(-19.8)

Impulse = 59.4 Ns

Therefore, the impulse of the floor on the ball is 59.4 Ns.

Learn more about impulse at: brainly.com/question/904448

5 0

2 years ago

Can a 20 N force and 40 N force ever produce a resultant with magnitude of 27 N?

SVEN [57.7K]

Sure ,Let's find angle between forces

Vectors be A and B and resultant be R

$\\ \sf\longmapsto R^2=A^2+B^2+2ABcos\theta$

$\\ \sf\longmapsto 27^2=20^2+40^2+2(20)(40)cos\theta$

$\\ \sf\longmapsto 729=400+1600+1600cos\theta$

$\\ \sf\longmapsto 729=2000+1600cos\theta$

$\\ \sf\longmapsto 1600cos\theta=-271$

$\\ \sf\longmapsto cos\theta=-0.169$

$\\ \sf\longmapsto \theta=cos^{-1}(-0.169)$

$\\ \sf\longmapsto \theta=80.2°$

7 0

3 years ago

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI

OverLord2011 [107]

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

<u>Explanation: </u>

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } \times \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} \times 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^{-2}$

So, the work done can be expressed in $k g m s^{-2}$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2}$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^{2} / \mathrm{s}^{2}$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^{2} \mathrm{s}^{-2} \times \frac{1000 \mathrm{g}}{1 \mathrm{kg}} \times \frac{100 * 100 \mathrm{cm}^{2}}{m^{2}}=3200 \mathrm{g} \mathrm{cm}^{2} \mathrm{s}^{-2}$

Thus, work done in ergs is 3200 ergs.

6 0

3 years ago

A moving sidewalk in an airport terminal moves at 1 m/s and is 35 m long. If a person steps on at one end of the sidewalk and wa

Serhud [2]

Answer:

a) (1.4 + 1.9) = 3.3m/sec.

(35/3.3) = 10.6 secs.

b) (1.9 - 1.4) = 0.5m/sec.

(35/0.5) = 70 secs.

Explanation:

4 0

3 years ago

How long does it take venus to orbit the sun

kodGreya [7K]

The answer is 225 days

5 0

3 years ago

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Vector a cross b whole square​

Vector a cross b whole square