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3 years ago

If a person climbed Mt. Everest has a mass of 105 kg and a weight of 625 N what would be the acceleration due to gravity?

Physics

1 answer:

DanielleElmas [232]3 years ago

3 0

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

Force = mass × acceleration.

From the parameters given:

Mass = 105 kg
Force = 625 N

By replacing the given values into the above equation, we can determine the acceleration.

∴

625 N = 105 kg × acceleration.

$\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}$

acceleration = 5.95 N/kg

Since 1 N/kg = 1 m/²

acceleration = 5.95 m/s²

Learn more about acceleration(a) here:

brainly.com/question/14344386

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Between which two points did they travel fastest?

marusya05 [52]

Answer:

During the section CD , the speed is fastest.

Explanation:

The rate of change of distance is called speed.

Speed = distance / time

Its SI unit ism/s. It is a scalar quantity.

The slope of the distance time graph is given by the speed of the object.

Here, the speed of AB is 30/3= 10 m/s .

The speed of BC is = 0 m/s

The speed of CD is (50 - 30)/(6 - 5) = 20 m/s

So, the speed is maximum during the section CD.

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3 years ago

A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

HACTEHA [7]

Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

So, the required magnetic field is equal to $2.91\ \mu T$ .

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3 years ago

Alan starts from his home and walks 1.3 km east to the library. He walks an additional 0.68 km east to a music store. From there

zepelin [54]

Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

Library to music store = 0.68km East

Music store to friend's house = 1.1km North

Friend's house to grocery store = 0.42 km North

Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

Fro = (0.68 + 1.3)km = 1.98 km East

Δ distance = (1.98 - 1.98) = 0

Northward direction:

To = (1.1 + 0.42)km = 1.52km north

Fro = (0.42 + 1.1)km = 1.52km North

Δ distance = (1.98 - 1.98) = 0

Hence final Displacement = 0

Total distance covered = 2 × (1.3 + 0.68 + 1.1 + 0.42) = 2 × 3.5

= 7km

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4 years ago

"Today you will use this chart to help you identify some common minerals," instructed Mr. Grant, Jimmy's teacher. Each group of

scoray [572]

The properties of minerals that might help Jimmy and his partners to identify the sample other than Hardness (scratch test) are:

1. Color

2. Crystalline Structure

3. Cleavage or Fracture

4. Transparency

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3 years ago

Read 2 more answers

Using this information...

Pepsi [2]

$19.2\:\text{m/s}$

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

$v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)$

$\:\:\:\:\:= 19.2\:\text{m/s}$

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3 years ago