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Julli [10]
3 years ago
14

How high does a rocket have to go above the Earth’s surface so that its weight is reduced to 58.8 % of its weight at the Earth’s

surface? The radius of the Earth is 6380 km and the universal gravitational constant is 6.67 × 10−11 N · m 2 /kg2 . Answer in units of km

Physics
1 answer:
julsineya [31]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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While standing on a bridge 40.0 m above the ground, you drop a stone from rest. when the stone has fallen 3.80 m, you throw a se
Norma-Jean [14]
Given: distance 1 d₁ = 40 m;  distance 2 d₂ = 3.8 m   g = -9.8 m/s²

 Initial Velocity Vi = 0  Final Velocity of stone 2 is unknown = ?

Total distance dₓ = d₁ - d₂  = 40 m - 3.8 m = 36.2 m

Formula: a = Vf² -  Vi²/2d   derive for Final Velocity Vf
 
acceleration is now due to gravity, therefore a = g

Vf = √2gd   Vf = √2(9.8 m/s²)(36.2 m)

Vf = 26.64 m/s  

Reason: The second stone will still start from rest.


3 0
3 years ago
Seasons are more noticeable in places that are____from to the equator
Liula [17]
1. farther
2. true

the equator is a weirdo
4 0
3 years ago
A ball is thrown off a cliff at a speed of 10 m/s in a horizontally direction. The ball reaches the ground 1.5 seconds. If the b
Tems11 [23]
I am pretty sure it is d
5 0
3 years ago
8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-
Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:  

A=A_{o}.2^{\frac{-t}{H}} (1)  

Where:  

A=0.375 g is the final amount of the material  

A_{o}=3 g is the initial amount of the material  

t=1 h is the time elapsed  

H is the half life of the material (the quantity we are asked to find)  

Knowing this, let's substitute the values and find h from (1):

0.375 g=(3 g)2^{\frac{-1h}{H}} (2)  

\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}} (3)  

Applying natural logarithm in both sides:

ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}}) (4)  

-2.079=-\frac{1 h}{H}ln(2) (5)  

Clearing H:

H=\frac{-1h}{-2.079}(0.693) (6)  

Finally:

h=0.333 h This is the half-life of the Bismuth-218 isotope

4 0
3 years ago
According to the law of reflection, the angle of incidence is equal to the angle of
Darina [25.2K]

Before going to answer this question first we have to understand reflection and laws of reflection.

Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .

Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.

Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]

The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]

There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.

There are two laws of reflection.

FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.

SECOND LAW- It states that that the angle of incidence is equal to the angle of reflection irrespective of the type of reflection.i.e i =r

Hence the correct answer will be angle of reflection.

                                         

3 0
3 years ago
Read 2 more answers
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