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3 years ago

A 5 kg rock is dropped down a vertical mine shaft. How long does it take to reach the bottom, 79 meters below?

Physics

1 answer:

likoan [24]3 years ago

7 0

Answer:

The time for the rock to reach the bottom is 4.02 seconds.

Explanation:

Given;

mass of the rock, m = 5 kg

height of the rock fall, h = 79 meters

The time to drop to the given height is given by;

$t = \sqrt{\frac{2h}{g} }$

where;

t is the time to fall to the bottom

g is acceleration due to gravity = 9.8 m/s²

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*79}{9.8} }\\\\t = 4.02 \ s$

Therefore, the time for the rock to reach the bottom is 4.02 seconds.

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How to calculate the mass of a body whose weight is 200 N

puteri [66]

Answer:

20.4081633 kilograms.

Explanation:

To find an object's mass using its weight, the formula is Mass equals Weight divided by the Acceleration of Gravity (M = W ÷ G). Convert the weight measured in pounds to the equivalent in Newtons.

Divide the weight in Newtons by the acceleration of gravity to determine the mass of an object measured in Kilograms. On Earth, gravity accelerates at 9.8 meters per second squared (9.8 m/s2). For example, to determine the mass of an object weighing 667 Newtons, calculate as follows:

200 Newtons ÷ 9.8 m/s2 = 20.4081633 kilograms.

3 0

3 years ago

The objective lens in a microscope with a 17.0 cm long tube has a magnification of -50.0 and the eyepiece has a magnification of

Bas_tet [7]

Answer:

(a) 0.34 cm

(b) 1.32 cm

Explanation:

Given that, $m_{o}=-50$

$m_{e}=20 cm$

And the length of objective lens in a microscope

$L=17$

And we have d of the eyepiece is 25 cm.

(a) Focal length of the objective is,

$f_{o}=-\frac{L}{m_{o} }\\ f_{o}=-\frac{17}{-50} \\f_{o}=0.34 cm$

Therefore, objective lens's focal length is 0.34 cm.

(b) Focal length of an eyepiece,

$m_{e}=(1+\frac{d}{f_{e} })\\ 20=(1+\frac{25}{f_{e} }\\ f_{e}=\frac{25}{19}cm\\ f_{e}=1.32cm$

Therefore, eyepiece's focal length is 1.32 cm.

5 0

3 years ago

Convert gravitational potential energy to kinetic energy.

Digiron [165]

Answer:

1.7 J

Explanation:

The energy carried by a single photon is given by

E=hf

where h is the Planck's constant and f is the frequency of the photon.

The photon of our exercise has a frequency of f=1.7 \cdot 10^{17} Hz, therefore its energy is

E=hf=(6.63 \cdot 10^{-34}Js)(1.7 \cdot 10^{17} Hz)=1.1 \cdot 10^{-16} J

4 0

3 years ago

Will mark as brainliest!

Papessa [141]

A. Between planet A and planet B i believe is the correct answer

3 0

3 years ago

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago