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3 years ago

Pls can someone help with question 37 asap pls

Physics

1 answer:

adell [148]3 years ago

4 0

The answer should be 2times3.14 which equals 6.28

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R: Law of Conservation of Energy Pre-Write

Bad White [126]

Answer:

V at C is 3.6 m/s

Explanation:

At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J

By conservation of energy.

KE+PE=Constant

At C PE=0.6 J

the KE=2.943-0.6=2.343 J

KE=0.5*m*v^2

v=√[KE/(0.5*m)]=3.06 m/s

3 0

3 years ago

A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base

-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g= $426\times 10^{-3} kg$

1 kg=1000 g

Radius,r= $\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

$I=\frac{2}{2}mr^2$

a.By law of conservation of energy

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh$

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy= $\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J

6 0

3 years ago

When a boat is placed in liquid, two forces act on the boat. Gravity pulls the boat down with a force equal to the weight of the

wel

Answer:

the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

Explanation:

This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law

B- W = 0

B = W

the thrust force is equal to the weight of the liquid that is dislodged

B = ρ g V

we substitute

ρ g V = m g

V = m /ρ_fluid 1

we can write the mass of the pot as a function of its density

ρ_body = m / V_body

m = ρ_body V_body

V_fluid / V_body = ρ_body / ρ _fluid 2

Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

The effect appears the pot as if it had a lower apparent weight

3 0

4 years ago

Define projectile in your own .

lorasvet [3.4K]

Answer:

a body which was thrown in space ,moves under the influence of gravity only is defined as projectile.

5 0

3 years ago

Read 2 more answers

Based on the velocity-time graph given, the acceleration of the object is..

fredd [130]