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3 years ago

WILL GET BRAINLIEST!!

Physics

2 answers:

Ket [755]3 years ago

8 0

It’s definitely the 3rd one

Yakvenalex [24]3 years ago

6 0

Answer:

It's the third one because the second is under water the first is fossil so not all areas of geologic and it's not the last its number 3

Explanation:

Using relative and radiometric dating methods, geologists are able to answer the question: how old is this fossil?

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Joey throws a football 30 meters with a force of 10 N. How much work does Joey do?

stich3 [128]

Answer:

300 Joules

Explanation:

Just use the Work formula:

W = F . D

W = 10 . 30

W = 300 Joules

8 0

3 years ago

A 0.5 kg yellow bouncy ball rolls to the right at a speed of 8 m/s while a 0.25 kg blue bouncy ball rolls

inessss [21]

Explanation:

Let yellow ball be m1 = 0.5kg with u1 = 8 m/s and blue ball be m2 = 0.25 kg with u2 = - 4 m /s respectively.

After collision, blue ball travels 12 m/s.

Using conservation of Linear Momentum :

m1u1 + m2u2 = m1v1 + m2v2

0.5* 8 + 0.25 * - 4 = 0.5 * v1 + 0.25 * 12

v1 = 0 m/sec i.e. Yellow ball comes to rest.

3 0

3 years ago

A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts

Viefleur [7K]

Answer:

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

F = 0.140(37+ 28.9) /1.85x10^-3

F = 9.226/1.85x10^-3

F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

8 0

3 years ago

A 1000-kg car traveling at 70 m/s takes 3 m to stop under full braking. the same car under similar road conditions, traveling at

azamat

We assume $a=const$ (acceleration is constant. We apply the equation
$v^2=v0^2+2as$ where s is the distance to stop $v=0(m/s)$ . We find the acceleration from this equation
$a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)
$
We know the acceleration, thus we find the distance necesssary to stop when initial speed is $v=140 (m/s)$
$s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)$

5 0

3 years ago

A student uses a spring to launch a marble vertically in the air. The mass of the marble is

Fed [463]

Answer:

the maximum height reached by the marble is 11.84 m.

Explanation:

Given;

mass of the marble, m = 0.02 kg

extension of the string, x = 0.08 m

force applied to the string, F = 58 N

Apply the principle of conservation of energy;

elastic potential energy = gravitational potential energy

¹/₂fx = mgh

$h = \frac{Fx}{2mg} \\\\h = \frac{58 \ \times \ 0.08}{2 \ \times \ 0.02 \ \times \ 9.8} \\\\h = 11.84 \ m$

Therefore, the maximum height reached by the marble is 11.84 m.

5 0

3 years ago