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3 years ago

10. A frog leaps from its resting position at the

Physics

1 answer:

lana [24]3 years ago

5 0

Answer:

The answer is C

Explanation:

F = m × a

m = 0.5kg

a = 3m/s²

F = 0.5 × 3

= 1.5N

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3 years ago

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PLZ HELP

Nadya [2.5K]

1) The mass of the continent is $3.3\cdot 10^{21}kg$

2) The kinetic energy of the continent is 1683 J

3) The speed of the jogger must be 6.57 m/s

Explanation:

The continent can be represented as a slab of size

$d=5850 km = 5.85\cdot 10^6 m$

and depth

$t = 35 km = 3.5\cdot 10^4 m$

So its volume is

$V=d^2 t = (5.85\cdot 10^6)^2(3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that the density of the continent is

$\rho = 2750 kg/m^3$

Therefore, we can calculate its mass as:

$m=\rho V=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

The kinetic energy of the continent is given by

$K=\frac{1}{2}mv^2$

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

The mass is

$m=3.3\cdot 10^{21} kg$

So, the kinetic energy of the continent is

$K=\frac{1}{2}(3.3\cdot 10^{21})(1.01\cdot 10^{-9})^2=1683 J$

Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

$K=\frac{1}{2}mv^2$

where

m = 78 kg is the mass of the jogger

v is his speed

We can therefore re-arrange the equation to find the speed of the man, and we get:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

2 years ago

How much work is done lifting a 12-m chain that is initially coiled on the ground and has a density 2 kg/m so that its top end i

Misha Larkins [42]

Answer:

2587.2 J.

Explanation:

From potential energy,

The work done to lift the chain = potential energy of the chain.

W = mgh............... Equation 1

Where W = work done to lift the chain, m = mass of the chain, g = acceleration due to gravity of the chain, h = height of the chain.

But,

m = m'd............... Equation 2

Where m' = density of the chain, d = length of the chain.

Substitute equation 2 into equation 1

W = m'dgh................ Equation 3

Given: m' = 2 kg/m, d = 12 m, h = 11 m, g = 9.8 m/s²

Substitute into equation 3

W = 2(12)(11)(9.8)

W = 2587.2 J.

3 0

3 years ago