A boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?

DESCRIBE THE REQUIREMENTS OF AN INTERNET CONNECTION?<br>please tell me the answer

AlekseyPX

Answer: The basic requirements for connecting to the Internet are a computer device, a working Internet line, and the right modem for that Internet line. In addition, software programs such as Internet browsers, email clients, Usenet clients, and other special applications are needed in order to access the Internet.

Explanation: brainleist pls :)

4 0

3 years ago

A bullet of 10g strikes a sand bag at a speed of 100 m/s and gets embedded after travelling

Tasya [4]

Answer:

Solving for time :

(There are 4 formulas from linear motion. These formulas are very helpful as it allows us to prevent complicated calculations. Choose among the four that has : 1. The most constants known

2. The unknown constant that we want to solve)

s = (1/2)(u+v)t <--- one of the formulas

from linear motion

s (distance) = 0.05m

u (initial velocity) = 100m/s

v (final velocity) = 0 m/s (it stops)

t (time taken for change in velocity) = to be found

0.05 = (1/2)(100+0)t

t = 0.001 seconds

Solving for the resistant force :

Since the bullet hits the bag with an impulsive force and stops, the force that stops the bullet is the resistant force.

When the bullet stops :

F net = 0

F r = F imp

F r = (mu -mv)/t

F r = (0.01x100-0.01x0)/0.001

F r = 1/0.001

F r = 1000N

4 0

3 years ago

Given a force of 88n and an acceleration of 4m/s2 what is the mass?

Hitman42 [59]

F=ma
m=F/a = 88/4 = 22 kg

5 0

3 years ago

Read 2 more answers

One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d

stira [4]

Answer:

a) Diffusion coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

$^{2} = 2Dt$ ..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

$36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr$

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

$^{2} = 2Dt$

$0.1^{2} = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds$

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0

3 years ago

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting a

IRISSAK [1]

A. $4.64\cdot 10^{11}m$

The orbital speed of the clumps of matter around the black hole is equal to the ratio between the circumference of the orbit and the period of revolution:

$v=\frac{2\pi r}{T}$