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mariarad [96]
4 years ago
12

An asteroid has a radius of 110,000 m

Physics
1 answer:
nydimaria [60]4 years ago
7 0

Answer: M = 6.13 × 10^18 kg

Explanation:

g = GM/r2,

Where

The mass M of the asteroid = ?

The radius r = 110000 m

g = 0.0338 m/s^2

G is the gravitational constant.

SI units its value is approximately 6.674×10^−11m3⋅kg−1⋅s−2

Using the formula

g = GM/r2

Cross multiply

GM = gr^2

6.674×10^-11M = 0.0338 × 110000^2

M = 408×10^6/6.674×10^-11

M = 6.13 × 10^18 kg

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A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his
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Answer:

1 m/s²

Explanation:

From the question,

Using

a = (v-u)/t.................... Equation 1

Where a = accelartion of the bicycle, v = Final velocity, u = initial velocity, t = time.

Given: v = 15 m/s, u = 5 m/s, t = 10 s

Substitute these values into equation 1

a = (15-5)/10

a = 10/10

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Hence the acceleration of the bicycle is 1 m/s²

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3 years ago
An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t
scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

C = \frac{Q}{V}  = e_0\frac{A}{d}

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

V = \frac{Q}{e_0} *\frac{d}{A}

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V

So, if we double the distance between the plates, the potential difference will also be doubled.

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Mutations _____.
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3 years ago
We can compare these two interactions on the basis of impulse (see above), but sometimes, we are more interested in the forces (
kiruha [24]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 I =  476 \ N \cdot s

b

 I_1 =  14.21 \  N\cdot s

c

    F  = 20300 \  N

Explanation:

Considering the first question

From the question we are told that

   The force produced is F  =  3400 \ N

   The duration of the punch is  t =  0.14 \  s

Generally the impulse delivered is mathematically represented as

    I =  F  *  t

=>    I =  3400  *  0.14

=>    I =  476 \ N \cdot s

Considering the second  question

   The approaching velocity of the ball is  v_b  =  45 \ m/s

    The leaving  velocity of the ball is  v_l  =  -53 \ m/s

     The mass of the ball is  m_b  =  0.145 \  kg

Generally the magnitude of the impulse delivered is mathematically represented as

     I_1 =  m*  v_b  - m *  v_l

=>     I_1 =  [0.145 *  45]  - [0.145 * -53]

=>     I_1 =  14.21 \  N\cdot s

Considering the third  question

     The  duration of the impact of the bat is  t _1 =  0.7 \ ms  =  0.7 *10^{-3} \  s

      Generally the average force exerted by the bat is mathematically represented as  

       F  =  \frac{I_1}{t_1}

=>     F  =  \frac{14.21 }{0.7 *10^{-3}}

=>       F  = 20300 \  N

 

7 0
4 years ago
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