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Sever21 [200]
3 years ago
13

An object is thrown upward from the edge of a tall building as with a velocity of 10m/s. where will the object be 3s after it is

thrown? take g=10s^2
Physics
1 answer:
KonstantinChe [14]3 years ago
6 0
20 m is the answer, hope that helps
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A body is at aheight of 81m and is ascending upwards with a velocity of 12m/s .a body of 2 kg weight is dropped from it.if g=10m
butalik [34]
First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s
7 0
2 years ago
Need help will mark you the Brainliest
Hatshy [7]
The last one is correct (D)
8 0
2 years ago
Convert to the fractional equivalent and reduce 21.12
nignag [31]

The decimal point is placed after two digits starting from the end. For each decimal place, we can write the number divided by 100.

21.12 can be written as \frac{2112}{100}.

Divide the numerator and denominator by 2:

\frac{2112}{100}= \frac{156}{50}

The numerator and denominator can be divided by 2 again:

\frac{78}{25}

There is no other common factor between numerator and denominator other than 1. Hence, it is the reduced form.




3 0
2 years ago
Read 2 more answers
For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas
inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

F= ma

And we know as well that

a= \frac{v}{t}

Replacing the aceleration value in the equation force we have,

F= \frac{mv}{t}

Substituting our values we have,

F= \frac{(0.060)(55)}{4*10^{-3}}

F=825N

The weight of the person is then,

W = mg

W = (60)(9.8) = 558N

<em>We can conclude that force on the ball is large to lift the ball</em>

6 0
2 years ago
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

    The mass suspended in mid-air is m_a =  5.0 \mu g =  5.0 *10^{-6} \ g =  5.0 *10^{-9} \  kg

   

Generally the electric field on the carpet is mathematically represented as

           E =  \frac{q}{ 2 *  A  *  \epsilon _o}

Where \epsilon _o is the permittivity of free space with value \epsilon_o = 8.85*10^{-12}  \ \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

           E =  \frac{-10*10^{-6}}{ 2 *  (2 * 4 )  *  8.85*10^{-12}}

           E = -70621.5 \  N/C

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

=>     q_d  = 6.94 *10^{-13} \  C

4 0
3 years ago
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