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3 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

Physics

1 answer:

pav-90 [236]3 years ago

7 0

Answer:

θ = 33°

Explanation:

Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:

$T = \frac{2\ u\ Sin\theta}{g} \\\\Sin\theta = \frac{Tg}{2u}$

where,

θ = launch angle = ?

T = Total time of flight = 0.6 s

g = acceleration due to gravity = 9.81 m/s²

u = launch speed = 5.4 m/s

Therefore,

$Sin\theta = \frac{(0.6\ s)(9.81\ m/s^2)}{(2)(5.4\ m/s)}\\\\\theta = Sin^{-1}(0.545)$

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Answer:

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Explanation:

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3 0

3 years ago

Which of the following choices is an example of an allele?

VikaD [51]

Answer:

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7 0

3 years ago

HELP!!

QveST [7]

Answer:

Total displacement 155 m

Explanation:

From the question, the formula to apply will be that of cosine rule where:

$c=\sqrt{a^2+b^2-2ab*cosC}$

where

a= 43 m , b=130m and c =? , <C=28° + 90° =118°

Substituting values in the equation as;

$c=\sqrt{43^2+130^2-2*43*130*cos118}$

$c=\sqrt{1849+16900-11180*-0.46947} \\\\c=\sqrt{1849+16900+5249} \\\\c=\sqrt{23998} \\\\\\c=155m$

6 0

3 years ago

Please help: I don't know how to do these problems

antiseptic1488 [7]

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3 years ago

The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

$\frac{0.60}{R_A^2}=\frac{1}{R_B^2}$

So the ratio between the radii of the two planets is

$\frac{R_A}{R_B}=\sqrt{0.60}=0.775$

6 0

3 years ago