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3 years ago

12

When evidence from many new experiments does not support an existing scientific theory like daltons, which of the following is m

ost likely to happen

1 answer:

iragen [17]3 years ago

8 0

I cant see the awnser but its probably replicate experiment to see if done right

You might be interested in

A stretched string is 2.11 m long and has a mass of 19.5 g. When the string oscillates at 440 Hz , which is the frequency of the

blsea [12.9K]

Answer:

The tension of the string is 41.876 N

Explanation:

Given;

length of the string, L = 2.11 m

mass of the string, m = 19.5 g = 0.0195 kg

frequency of the wave, f = 440 Hz

wavelength, λ = 15.3 cm = 0.153 m

The velocity of the wave is given by;

v = fλ

v = 440 x 0.153

v = 67.32 m/s

Also the velocity of the wave is given by

$v = \sqrt{\frac{T}{\mu} }$

where;

μ is mass per unit length = 0.0195 / 2.11 = 0.00924 kg/m

T is the tension of the string

T = v²μ

T = (67.32)²(0.00924)

T = 41.876 N

Therefore, the tension of the string is 41.876 N

5 0

4 years ago

A forklift lifts 5 boxes from the ground to a height of 2 meters (m). The boxes push down with a force of 1000 newtons (N). How

Nata [24]

Hello!

Answer:

2000 J

Explanation

Work equation is expressed as:

$W=F.d.Cos \alpha$

Where:

F: Applied force
d: traveled distance
α: Angle between the direction of the force and the direction of the movement. (in this case, both of the direction are the same, so the angle is 0°)

By substituting:

$F=1000N.2m.Cos(0)=2000N.m=2000 J$

Have a nice day!

8 0

3 years ago

On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth’s atmosphere over Chelyabinsk, Russia, and expl

topjm [15]

Explanation:

A.

Using the wave equation:

v = d/t

Where,

v = velocity

d = distance

= 23.5 km

= 23500 m

t = time

= 2m 30s = 150 s

v = 23500/150

= 156.7 m/s

B.

From the calculated value above, it can be seen that the velocity of the heat blast is >>> than the velocity of sound above sea level.

3 0

3 years ago

At noon, ship A is 110 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is

Katyanochek1 [597]

Answer:

$4.47\ \text{km/h}$

Explanation:

$\dfrac{da}{dt}$ = Rate at which the distance between A and starting point of B is changing = -20 km/h

$\dfrac{db}{dt}$ = Rate at which the distance of B is changing = 15 km/h

$\dfrac{dc}{dt}$ = Rate at which the distance between A and B is changing

Time after which the rate at which the distance between A and B is changing is 4 hours

Distance covered by A in 4 hours = $20\times 4=80\ \text{km}$

a = Distance remaining to the start point of B = $110-80=30\ \text{km}$

b = Distance covered by B in 4 hours = $15\times 4=60\ \text{km}$

Distance between A and B after 4 hours

$c=\sqrt{a^2+b^2}\\\Rightarrow c=\sqrt{30^2+60^2}\\\Rightarrow c=67.08\ \text{km}$

$c^2=a^2+b^2$

Differentiating with respect to time we get

$c\dfrac{dc}{dt}=a\dfrac{da}{dt}+b\dfrac{db}{dt}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{a\dfrac{da}{dt}+b\dfrac{db}{dt}}{c}\\\Rightarrow \dfrac{dc}{dt}=\dfrac{30\times -20+60\times 15}{67.08}\\\Rightarrow \dfrac{dc}{dt}=4.47\ \text{km/h}$

The rate at which the distance between the ships is changing at 4 PM is $4.47\ \text{km/h}$ .

7 0

3 years ago

A box of spherical jawbreaker candies is 23 g and its volume is 32.3 cm3. If the average mass of a single jawbreaker is 0.94 g,

Alex777 [14]

The radius of each jawbreaker is approximately 0.68 cm.

<h3>Volume of a sphere;</h3>

v = 4 /3 πr³

where

r = radius

Therefore,

23 g = 32.3 cm³

0.94 g = ?

cross multiply

volume of a single jawbreaker = 32.3 × 0.94 / 23 = 30.362 / 23 = 1.32 cm³

Therefore,

volume of each jawbreaker = 4 /3 πr³

1.32 = 4 / 3 × 3.14 × r³

r³ = 1.32 /4.18666666667

r³ = 0.31533683707

r = ∛0.31533683707

r = 0.680651651 = 0.68

Therefore, the radius of each jawbreaker is approximately 0.68 cm.

learn more on radius here: brainly.com/question/19172427

5 0

3 years ago