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3 years ago

How can a small boy balance a big boyon asea- saw ? Show with a diagram

Physics

1 answer:

liq [111]3 years ago

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If an object undergoes a change in momentum of 10 kg m/s in 3 seconds what is the force?

kolbaska11 [484]

Answer:

The correct equation for calculating force is change in momentum divided by time. So 10 kg m/s divided by 3 s is 3.3 newtons (N

Explanation:

7 0

3 years ago

The coefficient of static friction between a book and the level surface it slides on is 0.65. If the mass of the book is 0.2 kg,

just olya [345]

Answer: 1.274 N minimum force required to slide the book across the surface

Explanation:

Mass of the book = m = 0.2 kg

Normal force acting on the book,N = mg

Coefficient of friction = $mu_s$ = 0.65

Minimum force required to slide the book across the surface= F

$F=\mu_s\times N$

$F=0.65\times 0.2 kg\times 9.8 m/s^2=1.274 N$

1.274 N minimum force required to slide the book across the surface

7 0

3 years ago

Read 2 more answers

Improvements in muscular strength will not affect muscular endurance.

olga2289 [7]

Answer:

the answer for your question is false ( F )

6 0

3 years ago

A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.

umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, $\lambda=0.0178\ h^{-1}$

(a) The half life of this nuclide is given by :

$t_{1/2}=\dfrac{ln(2)}{\lambda}$

$t_{1/2}=\dfrac{ln(2)}{0.0178}$

$t_{1/2}=38.94\ h$

(b) The decay equation of any radioactive nuclide is given by :

$N=N_oe^{-\lambda t}$

$\dfrac{N}{N_o}=e^{-\lambda t}$

Number of remaining sample in 4.44 half lives is :

$t_{1/2}=4.44\times 38.94$

$t_{1/2}=172.89\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 172.89}$

$\dfrac{N}{N_o}=0.046$

$t_{1/2}=14.6\times 24$

$t_{1/2}=350.4\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 350.4}$

$\dfrac{N}{N_o}=0.0019$

Hence, this is the required solution.

4 0

3 years ago

M1 is a spherical mass (25.0 kg) at the origin. M2 is also a spherical mass (10.6 kg) and is located on the x-axis at x = 94.8 m

Nata [24]

The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

<h3>Position of the third mass</h3>

m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2

a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;

Force on m3 due to m1 = Force on m3 due to m2

Gm1m3/d² = Gm2m3/r²

m1/d² = m2/r²

where;

d is the distance between m1 and m3 = x
r is the distance between m3 and m2 = 94.8 - x

m1/(x²) = m2/(94.8 - x)²

m1(94.8 - x)² = m2x²

(94.8 - x)² = (m2/m1)x²

(94.8 - x)² = (10.6/25)x²

(94.8 - x)² = 0.424x²

(94.8 - x)² = (0.651)²x²

94.8 - x = 0.651x

94.8 = 1.651x

x = 94.8/1.651

x = 57.42 m

Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

Learn more about gravitational force here: brainly.com/question/72250

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8 0

2 years ago

How can a small boy balance a big boyon asea- saw ? Show with a diagram​

How can a small boy balance a big boyon asea- saw ? Show with a diagram