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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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Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen
m_a_m_a [10]

Answer:

F=2.25\times 10^{-5}\ N

Explanation:

According to question,

Charge 1 and charge 2 are 10^{-7}\ C

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N

So, the electric force of repulsion is 2.25\times 10^{-5}\ N.

8 0
3 years ago
Can you help and explain these for me?
ollegr [7]
Yes I can do you want me to
8 0
3 years ago
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Airida [17]

Answer:

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Explanation:

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7 0
4 years ago
Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot
m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force F_1=430 N

other Pulls it with a rope of rope F_2=360 N

mass of crate m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

f_r=F_1+F_2

where f_r is the friction force

f_r=430+360

f_r=790 N

f_r=\mu N

where \muis the coefficient of static friction

N=mg

790=\mu 29\cdot 9.8

\mu =0.27

6 0
4 years ago
You want to move a 4- kg bookcase to a different place in the living room. If u push with a force of 65 n and the bookcase accel
IrinaK [193]

Answer:

1.65

Explanation:

The equation of the forces along the horizontal direction is:

F-F_f = ma (1)

where

F = 65 N is the force applied with the push

F_f is the frictional force

m = 4 kg is the mass

a=0.12 m/s^2 is the acceleration

The force of friction can be written as F_f = \mu R (2), where

\mu is the coefficient of kinetic friction

R is the normal force exerted by the floor

The equation of forces along the vertical direction is

R-mg=0 (3)

since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find

F-\mu mg = ma

And solving for \mu,

\mu = \frac{F-ma}{mg}=\frac{65-(4)(0.12)}{4(9.8)}=1.65

7 0
3 years ago
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