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2 years ago

As best you can, describe how we can tell the age of the Earth and rocks found on it.

Physics

1 answer:

Viktor [21]2 years ago

4 0

Answer:

We can determine the age of the Earth and the rocks found on Earth with techniques such as measuring ice cores and digging and analysing fossils.

Explanation:

Hopefully this helped!

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I need help!!! 60 POINTS!!!!

Scrat [10]

Correct one is b
Good luck

3 0

3 years ago

Read 2 more answers

The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes it 0.90 s to reach the floor. Wha

Anna007 [38]

Answer:

The mass of the another block is 60 kg.

Explanation:

Given that,

Mass of block M= 100 kg

Height = 1.0 m

Time = 0.90 s

Let the mass of the other block is m.

We need to calculate the acceleration of each block

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$1.0=0+\dfrac{1}{2}\times a\times(0.90)^2$

$a=\dfrac{2\times1.0}{(0.90)^2}$

$a=2.46\ m/s^2$

We need to calculate the mass of the other block

Using newton's second law

The net force of the block M

$Ma=Mg-T$

$T=Mg-Ma$ ....(I)

The net force of the block m

$ma=T-mg$

Put the value of T from equation (I)

$ma=Mg-Ma-mg$

$m(a+g)=M(g-a)$

$m=\dfrac{M(g-a)}{(a+g)}$

Put the value into the formula

$m=\dfrac{100(9.8-2.46)}{2.46+9.8}$

$m=59.8\ \approx60\ kg$

Hence, The mass of the another block is 60 kg.

8 0

2 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

Please help I literally don’t understand

Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0

2 years ago

An 800 N man climbs 5 m up a ladder. How much gravitational potential energy does he gain?

Artemon [7]

Answer:

4000J

Explanation:

Given parameters:

Weight of the man = 800N

Height of ladder = 5m

Unknown:

Gravitational potential energy gained = ?

Solution:

The gravitational potential energy is due to the position of a body.

Gravitational potential energy = weight x height

Now insert the parameters;

Gravitational potential energy = 800 x 5 = 4000J

5 0

3 years ago