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Svetlanka [38]
3 years ago
15

What is the force of gravity between two 40.0kg masses that are separated by 3.00m?

Physics
1 answer:
Slav-nsk [51]3 years ago
3 0

Answer:

f = g \times \frac{m1 \times m2}{ {d}^{2} }

f = 6.67 \times  {10}^{ - 11}  \times \frac{40 \times 40}{9}

<h2>F=1.2x 10^-8</h2>
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Plz help me make a nice C-E-R for science I will give brainly for the best C-E-R
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C-E-R stands for Chemistry Eastern Right. Rights has five words so add five to your question that’ll be 45 and boom you got it
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3 years ago
Electric field lines show the strength and _____ of an electric field.
Gwar [14]
So we want to know what do electric field lines show except strength of the electric field. Electric field lines are lines that show us the field strength and the DIRECTION of the electric field. If the lines point away from the charge that is producing the lines, the charge is positive, and if the lines point towards the charge that is producing them that charge is negative.
8 0
3 years ago
A body of mass 1kg is made to oscillate on a spring of force constant 16 n/m calculate 1 the angular frequency 2 the frequency o
Scilla [17]

Explanation:

Given that,

Mass of a body, m = 1 kg

Force constant, k = 16 N/m

We need to find the angular frequency and the frequency of oscillation.

(a) The angular frequency of a body is given by :

\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s

(b) The frequency of oscillation is given by :

f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz

Hence, this is the required solution.

7 0
3 years ago
a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha
defon

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 3 kg is the mass of the putty

u_1 = 10 m/s is the initial velocity of the putty (we take its direction as positive direction)

m_2 = 8 kg is the mass of the ball

u_2 = 0 m/s is the initial velocity of the ball (at rest)

v is the final combined velocity of the two putty+ball

Re-arranging the equation and substituting the values, we find the  final combined velocity:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s

And the positive sign indicates their final direction is the same as the initial direction of the putty.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

3 0
3 years ago
A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte
vaieri [72.5K]

Answer:

Part a)

H = 44.1 m

Part b)

y = 13.48 m

Part c)

d = 8.86 m

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

t = \frac{v sin\theta}{g}

now we have

3 = \frac{vsin\theta}{9.8}

v sin\theta = 29.4 m/s

Now maximum height above ground is given as

H = \frac{v^2sin^2\theta}{2g}

H = \frac{29.4^2}{2(9.8)}

H = 44.1 m

Part b)

Height of the fence is given as

y = (vsin\theta) t - \frac{1}{2}gt^2

y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)

y = 13.48 m

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

x = v_x t

97.5 = v_x (5.5)

v_x = 17.72 m/s

now total time of flight is given as

T = 3 + 3 = 6 s

so range is given as

R = v_x T

R = (17.72)(6)

R = 106.4 m

so the distance from the fence is given as

d = 106.4 - 97.5

d = 8.86 m

7 0
4 years ago
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