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3 years ago

8

Help please correct answer i will mark brainliest

1 answer:

murzikaleks [220]3 years ago

6 0

Answer:

19.21ms-¹

Explanation:

that is the solution above

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4 years ago

According to Newton's third law, which describes what happens when a 4.5 kg pumpkin is launched from a catapult with a force of

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3 years ago

Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How hig

Salsk061 [2.6K]

Answer:

So height will be 47.387 m

Explanation:

We have given gauge pressure $P=4.63\times 10^5N/m^2$

Density of water $\rho =997kg/m^3$

Acceleration due to gravity $g=9.8m/sec^2$

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So height will be 47.387 m

4 0

3 years ago

The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration. How long will

Setler [38]

Answer:

Explanation:

Given

half life $(t_{\frac{1}{2})=4.5$ billion year

10 % to decay i.e. 90 % remaining

And $\ln (\frac{C}{C_0})=-kt$

where k= constant

t=time

and $k=\frac{\ln (2)}{t_{\frac{1}{2}}}=\frac{\ln (2)}{4.5}$

so $\frac{C}{C_0}=0.9$

$\ln (0.9)=-\frac{\ln (2)}{4.5}\times t$

t=0.684 billion year

(b) $C_0=1.5\times 10^{18}$

t=13.8 billion year

$\ln (\frac{C}{C_0})=-0.15403\times 13.8$

$\ln (\frac{C}{C_0})=-2.125$

$C=C_0e^{-2.125}$

$C=1.5\times 10^{18}\times 0.1194=0.179\times 10^{18}$ atoms

6 0

4 years ago

A bungee jumper of mass m jumps off a bridge. Assume that the bungee cord behaves like am ideal spring of spring constant k. Whe

DanielleElmas [232]

Answer:

b) √[(kx²/m) - 2gx]

Explanation:

The energy at the lowest point is equal to:

$E_{elas}=\frac{1}{2} *k*x^{2}$

where:

Eelas = elastic energy [J]

k = spring constant [N/m]

x = extension of the spring [m]

We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.

$E_{elas}=E_{pot}+E_{kine}\$

$\frac{1}{2}*k*x^{2}=m*g*x +\frac{1}{2} *m*v^{2} \\v^{2} = \frac{k*x^{2} }{m}-2*g*x\\ v=\sqrt{\frac{k*x^{2} }{m}-2*g*x}$

8 0

3 years ago