A 0.54 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the bullfrog?

Physics

1 answer:

RUDIKE [14]3 years ago

7 0

Answer:

Explanation:

Step one:

given data

mass of bullfrog= 0.54kg

Step two:

Required

The force F in Newton

From newton first law

F=mg

The acceleration due to gravity acting on the frog is 9.81m/s^2

hence

F=0.54*9.81

F=5.3N

The normal force is 5.3N

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An object of mass 700700 kg is released from rest 20002000 m above the ground and allowed to fall under the influence of gravity

qwelly [4]

Answer:

59.503987 seconds

Explanation:

b = Proportionality constant = 50 Ns/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object = 700 kg

We have the equation of velocity

$v(t)=\dfrac{mg}{b}+\left(V_0-\dfrac{mg}{b}\right)e^{\dfrac{bt}{m}}$

The equation of motion

$x(t)=\dfrac{mg}{b}+\dfrac{m}{b}\left(V_0-\dfrac{mg}{b}\right)(1-e^{\dfrac{bt}{m}})$

$x(t)=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})$

when x(t)=2000

$2000=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})\\\Rightarrow 2000\times \:50=\frac{700\times \:9.81}{50}\times \:50+\frac{9.81}{50}\left(0-\frac{700\times \:9.81}{50}\right)\left(1-e^{\frac{50t}{700}}\right)\times \:50\\\Rightarrow 6867-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)=100000\\\Rightarrow 50\left(-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)\right)=93133\times \:50\\\Rightarrow \frac{-67365.27\left(1-e^{\frac{50t}{700}}\right)}{-67365.27}=\frac{4656650}{-67365.27}\\\Rightarrow 1-e^{\frac{50t}{700}}=-69.12538\dots\\\Rightarrow -e^{\frac{50t}{700}}=-70.12538\dots\\\Rightarrow t=14\ln \left(70.12538\dots \right)\\\Rightarrow t=59.50398\ s$