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pav-90 [236]
2 years ago
9

IF YOU ANSWER THESE 2 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 15 POINTS!!!

Physics
1 answer:
sergey [27]2 years ago
3 0

Answer:

Explanation:

Ahhhhhhhhhhhhhhhhjhhhhhhhhhhhhhhhhh

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A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam
ahrayia [7]

Answer:

  1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

  (1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

8 0
3 years ago
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A 25 n object requires a 5.0 n to start moving over a horizontal surface. what is the coefficient of static friction?
olganol [36]
Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N =  0.2
Hence 0.2 is the coefficient of static friction
3 0
3 years ago
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What causes water to move from the liquid part of the hyrdrosphere to the cyrosphere?
bezimeni [28]

Answer:

When the liquid moves through the hydrosphere, the water collects into a cloud. When it falls to the earth, turning into snow and sleet collecting in rivers and lakes.

Explanation:

Hope that helps

7 0
3 years ago
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A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i
Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]          (1)

n = 4

Now,

We know the standard eqn is given by:

y = AsinKxcos\omega t           (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = 14.4 m^{- 1}

\omega = 166\ rad/s

where

A = Amplitude

K = Propagation constant

\omega = angular velocity

Now, to calculate the string's wavelength,

(a) K = \frac{2\pi}{\lambda}

where

K = propagation vector

\lambda = \frac{2\pi}{K}

\lambda = \frac{2\pi}{14.4} = 0.436\ m

(b) The length of the string is given by:

l = \frac{n\lambda}{2}

l = \frac{4\times 0.436}{2} = 0.872\ m

(c)  Now, we first find the frequency of the wave:

\omega = 2\pi f

f = \frac{\omega}{2\pi}

f = \frac{2\pi}{166} = 26.42\ Hz

Now,

Speed of the wave is given by:

v = f\lambda

v = 26.419\times 0.436 = 11.518\ m/s

4 0
3 years ago
Which conclusion is best supported by the data?
sattari [20]

Answer: b

Explanation: the chart

8 0
3 years ago
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