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3 years ago

How much would your 20.0-kg dog weigh on Neptune?

Physics

2 answers:

Helga [31]3 years ago

7 0

Answer:

22.73 kg

Explanation:

Lesechka [4]3 years ago

3 0

22.73 kg ish yeh :D Yee

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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k

Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s$

Now, we will apply the law of conservation of momentum:

$m_1v_1 = m_2v_2$

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

$(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s$

Now, we again use the third equation of motion for the upward motion of the ball:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

$(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\$

6 0

3 years ago

An open diving chamber rests on the ocean floor at a water depth of 60 meter. Find the air pressure (gage pressure relative to t

Fofino [41]

Answer:

Gauge Pressure required = 606.258 kPa

Explanation:

Water will not enter the chamber if the pressure of air in it equals that of the water which tries to enter it.

Thus at a depth of 60m we have pressure of water equals

$P(z)=P_{0}+\rho _wgh$

Now the gauge pressure is given by

$P(z)-P_{0}=\rho _wgh$

Applying values we get

$P(z)-P_{0}=\rho _wgh\\\\P_{gauge}=1.03\times 1000\times 9.81\times 60Pa\\\\P_{gauge}=606258Pascals\\\\P_{gauge}=606.258kPa$

8 0

4 years ago

If you have 10.0 g of a substance that decays with a half-life of 14 days, then how much will you have after 42 days?

Nat2105 [25]

The formula for the mass that remains:
$m=m_0 \times (\frac{1}{2})^\frac{t}{T}$
m₀ - the initial mass, t - time, T - the half-life

$m_0=10 \ g \\
T=14 \ d \\
t=42 \ d \\ \\
m=10 \times (\frac{1}{2})^\frac{42}{14}=10 \times (\frac{1}{2})^3=10 \times \frac{1}{8}=10 \times 0.125=1.25$

The answer is c. 1.25 g.

6 0

3 years ago

Three monkeys A, B, and C weighing 20, 26, and 25 lb, respectively, are climbing up and down the rope suspended from D. At the i

Marina86 [1]

Answer: $2235.2lb-ft/s^2$