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alina1380 [7]
3 years ago
14

A cannon fires a 0.652 kg shell with initial

Physics
2 answers:
nataly862011 [7]3 years ago
8 0

Answer:

height= 6.429m

Explanation:

h is equall to Max height traveled by trajectory body.

Max height= (U²Sin²tita)/2g

Max height = (12²(sin61)²)/(2*9.8)

Max height =( 144*0.875)/19.6

Max height= 6.429 m

laila [671]3 years ago
6 0
 Mass have no effect for the projectile motion and  u want to know the  height "h"   
first,
        find the vertical and horizontal components of velocity 
 vertical component of velocity = 12 sin 61                  
horizontal component of velocity = 12 cos 61
now for the vertical motion ;               
             S = ut + (1/2) at^2
where
 s = h 
u = initial vertical component of velocity 
t = 0.473 s 
a = gravitational deceleration (-g) = -9.8 m/s^2    
       
         h=[12×sin 610×0.473]+[−9.8×(0.473)2] 

u can simplify this and u will get the answer

h=.5Gt2 

H=1.09m
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3 years ago
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What is the density of a 150cm 3 block with a mass of 700 grams?
Vinvika [58]

Answer:

The Density of the block is 4.667g/mL

Explanation:

Given the following data;

Mass of block = 700g

Volume of block = 150cm³

Density = ?

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Simply stated, density is mass per unit volume of an object.

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3 0
3 years ago
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
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