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Helen [10]
3 years ago
8

As scientists research, they often find information that does not fit with current theories. What happens when new, contradictor

y information is found?
A) new information is ignored
B) old theories are completely discarded
C) old theories are adjusted to incorporate all old and new information
D) new theories are made solely from new information and old information is ignored
Physics
2 answers:
timofeeve [1]3 years ago
8 0
The correct answer is "C". 'Old theories are adjusted to incorporate all old new information.' This makes the most sense, regarded the old and new information should be taken into consideration.

I hope this helped you!

Brainliest answer is always appreciated!
wel3 years ago
8 0

The correct answer is: C

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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Bess [88]

Answer:

Explanation:

given,

initial velocity of the ball = 20 m/s

angle of ramp = 22°

ball travel at a distance  = 5 m

a) for friction less  

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh

             v^2 = u^2 - 2gh

             v = \sqrt{u^2- 2 g h cos 22^0}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0 }

                    v = 17.58 m/s

b) considering the friction

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh-\mu_kmgl

             v^2 = u^2 - 2gh-2\mu_kmgl

             v = \sqrt{u^2- 2 g h cos 22^0-2\mu_kgl}

             v = \sqrt{20^2- 2\times 9.8 \times 5 cos 22^0-2\times 0.15\times 9.8 \times 5 }

                   v = 17.16 m/s

7 0
3 years ago
If the weight of 1kg is 10 N in water what is the density of the stone<br>​
vekshin1

Answer:

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Explanation:

5 0
3 years ago
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup
KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx

Where

x_{o}, x_{f} - Initial and final position, respectively, measured in meters.

F(x) - Force as a function of position, measured in newtons.

Given that F = k\cdot x and the fact that F = 25\,N when x = 0.3\,m - 0.2\,m, the spring constant (k), measured in newtons per meter, is:

k = \frac{F}{x}

k = \frac{25\,N}{0.3\,m-0.2\,m}

k = 250\,\frac{N}{m}

Now, the work function is obtained:

W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx

W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]

W = 0.313\,J

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be r(\theta) = 2\cdot \sin 5\theta. The area of the region enclosed by one loop of the curve is given by the following integral:

A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta

A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta

By using trigonometrical identities, the integral is further simplified:

A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta

A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta

A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta

A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)

A = 4\pi

The area of the region enclosed by one loop of the curve r(\theta) = 2\cdot \sin 5\theta is 4\pi.

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3 years ago
What is an example of kinetic energy
Olenka [21]

An airplane has a large amount of kinetic energy in flight due to its large mass and fast velocity.

6 0
3 years ago
Which illustration represents low accuracy but high precision?​
Misha Larkins [42]

Answer:

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Explanation:

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3 years ago
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