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natima [27]
3 years ago
15

a ball was projected in such a way that it attained the maximum horizontal distance with a velocity of 20m/s calculate the verti

cal distance reached​
Physics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

80m

Explanation:

u=20,R=?,sin theta=1,g=10

R=u²sin2theta/g

R=20²x2/10

R=400x2=800/10

R=80m

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A small object carrying a charge of -4.00 nC is acted upon by a downward force of 24.0 nN when placed at a certain point in an e
PIT_PIT [208]

Answer:

E  = -6 \  N/C

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

Explanation:

From the question we are told that

   The charge on the small object is Q = -4.00 \ nC =  -4.00 *10^{-9} \  C

   The force is  F  =  24  \ nN  =  24 *10^{-9} \  N

    Generally the magnitude of the electric  field is mathematically represented as

       E  =  \frac{F}{Q}

=>    E  =  \frac{ 24 *10^{-9}} {-4 *10^{-9 }}

=>     E  = -6 \  N/C

Generally given that the electric field is negative it mean that its direction is opposite to that of the force    

 

6 0
3 years ago
Need help ??? Please
Yanka [14]
Is there information in the previous question which relates to this one?
8 0
3 years ago
Challenge! A marshmallow is dropped from a 5-meter high pedestrian bridge and 0.83 seconds later, it lands right on the head of
WINSTONCH [101]

a₀).  You know ...
         -- the object is dropped from 5 meters
             above the pavement;
         --  it falls for 0.83 second.

a₁).  Without being told, you assume ...
         -- there is no air anyplace where the marshmallow travels,
             so it free-falls, with no air resistance;
         -- the event is happening on Earth,
            where the acceleration of gravity is  9.81 m/s² .

b).  You need to find how much LESS than 5 meters
       the marshmallow falls in 0.83 second.
    
c).  You can use whatever equations you like.
       I'm going to use the equation for the distance an object falls in
       ' T ' seconds, in a place where the acceleration of gravity is ' G '.

d).  To see how this all goes together for the solution, keep reading:


The distance that an object falls in ' T ' seconds
when it's dropped from rest is

                                 (1/2 G) x (T²) .

On Earth, ' G ' is roughly  9.81 m/s², so in 0.83 seconds,
such an object would fall

                               (9.81 / 2) x (0.83)² = 3.38 meters .

It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was

                         (5.00 - 3.38)  =  1.62 meters

above the pavement.  That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.

7 0
3 years ago
I don’t know where each one goes
Usimov [2.4K]
I classified them by numbers 1 to 8
Force: #s 2,4,5,67
Friction: #s 1 3 8
That’s it what I would say its the answer
3 0
2 years ago
Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
olga_2 [115]

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

6 0
3 years ago
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