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3 years ago

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A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. the balls under

go a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle umax = 50. what was v0 ?

1 answer:

Novay_Z [31]3 years ago

7 0

Sorry but i dont know i was doing this for some point sorry

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What is one advantage of doing a field experiment instead of a laboratory experiment

Flauer [41]

It mimics the real world accurately

Explanation:

Experiments conducted in the field clearly presents the real world at it is to the scientist. Hardly can any part be controlled precisely and this gives a near to perfect scenario.

In the laboratory, for example, an organism is isolated from its environment and might not fully display its natural instinct and physiological capabilities.
Most laboratory set up are driven towards a model instead of real life settings.
The laboratory is more controlled and less varied and might truly represent the real world. It will only portray a part of the real world and series of further tests might have to be carried out to have a better model.

Learn more:

Experiment brainly.com/question/5096428

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3 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

3 years ago

Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.

valentina_108 [34]

The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics

5 0

3 years ago

What is the mass of an object that has an acceleration of 2.63 m/s^2 when an unbalanced force of 112 N is applied to it?

shusha [124]

112/2.63= 42.586
42.586 is your answer I need 20 characters

6 0

3 years ago

Two small nonconducting spheres have a total charge of 93.0 μC . Part A

Travka [436]

Answer:

charge on each

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

Q1Q2 = 1.49 × $10^{-9}$ C²

and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

5 0

3 years ago