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Fofino [41]
2 years ago
14

Newton believed that time and space are absolute, while Einstein believed that the ______ is absolute.

Physics
2 answers:
nadezda [96]2 years ago
8 0

Curve of space time

pickupchik [31]2 years ago
8 0

Newton believed that time and space are absolute, while Einstein believed that the <em>speed of light</em> is absolute.

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Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

___

9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
3 years ago
Perform the following calculations and give your answer with the correct number of significant figures
love history [14]

Answer:

see below

Explanation:

a. 0.1886 x 12 =2.2632

This has 2 sig figures so the answer can only have 2 sig figures

2.3

b. 2.995 - 0.16685 =2.82815

The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place

2.828

c. 910 x 0.18945=172.3995

The  least number of significant figures is 3 so the answer can only have 3 significant figures

172

3 0
2 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
WHICH ONE IS CORRECT???
sveta [45]

Answer:

A

Explanation:

5 0
2 years ago
Read 2 more answers
A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the
mash [69]

Answer:

\Delta x = v_0 t + \frac{1}{2}at^2

Explanation:

To solve this problem, we can use the following suvat equation:

\Delta x = v_0 t + \frac{1}{2}at^2

where

\Delta x is the vertical displacement of the frog

v_0 is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from v_0, all the other quantities are known. In fact:

\Delta x =0.1 m is the vertical displacement

t = 2 s is the total time of flight

a=g=-9.8 m/s^2 is the acceleration due to gravity (negative because it is downward)

Therefore, solving for v_0, we find the initial velocity of the frog:

v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s

4 0
3 years ago
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