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2 years ago

List these things in order of size, starting with the smallest thing:

Physics

1 answer:

jek_recluse [69]2 years ago

4 0

Moon, earth, sun solar system, galaxy, universe

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You throw a bowling ball down an alley with speed v_0v 0 , and initially it slides without rolling. After a little while fri

telo118 [61]

Answer:

final speed after pure rolling is given as $R\omega = \frac{5}{7}v_0$

Explanation:

As we know that point of contact at ground is taken as reference then there is no external torque about this point on the ball

So we can use angular momentum conservation about this point

$L_i = L_f$

$L_i = mv_0R$

$L_f = \frac{7}{5}mR^2\omega$

so we have

$mv_0R = \frac{7}{5}mR^2\omega$

$R\omega = \frac{5}{7}v_0$

So final speed after pure rolling is given as $R\omega = \frac{5}{7}v_0$

4 0

2 years ago

A 2 kg block is pushed against a spring (k = 400 N/m), compressing it 0.3 m. When the block is released, it moves along a fricti

Kitty [74]

Answer:

$2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{s}}=0.4$

$4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{\mathrm{k}}=0.2$

Explanation:

Given values

Mass (m) = 2kg

K = 400 N/M

Compressing it 0.3 m

The law of conservation of energy:

$\frac{m v^{2}}{2}+\frac{k x^{2}}{2}=\text { constant }$

$\text { Where, } \frac{m v^{2}}{2} \text { is kinetic energy of the block. }$

$\frac{k \Delta l^{2}}{2}$ Energy of the spring deformation.

M mass of the block

x spring deformation

Therefore, if block left the spring (x = 0)

$\frac{m v^{2}}{2}+0=0+\frac{k \Delta l^{2}}{2}$

Where, Δl is initial spring deformation

$\frac{m v^{2}}{2}=\frac{k \Delta l^{2}}{2}$

$\mathrm{v}^{2}=\frac{k \Delta l^{2}}{m}$

$\mathrm{v}=\sqrt{\frac{k}{m} \times \Delta l^{2}}$

$v=\sqrt{\frac{400}{2} \times(0.3)^{2}}$

$\mathrm{v}=\sqrt{200 \times 0.09}$

The law of conservation of energy:

$\frac{m v^{2}}{2}+m g h=\text { constant }$

Where h is height

$\frac{m v^{2}}{2}+0=0+m g h$

$\frac{m v^{2}}{2}=m g h$

Cancel mass "m" each side

$\mathrm{h}=\frac{v^{2}}{2 g}$

Distance along incline equals

$\begin{array}{ll}{\text { For friction us }} & {\left(L=\frac{h}{u_{s}}\right)} \\ {\text { For friction } u_{k}} & {\left(L=\frac{h}{u_{k}}\right)}\end{array}$

$\begin{array}{l}{\mathrm{u}_{\mathrm{s}}=0.4} \\ {\mathrm{U}_{\mathrm{k}}=0.2} \\ {\text { For friction } \mathrm{u}_{\mathrm{s}}}\end{array}$

$\begin{array}{l}{\mathrm{h}=\frac{v^{2}}{2 g u_{s}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.4}} \\ {\mathrm{L}=\frac{17.9776}{784}}\end{array}$

$\begin{array}{l}{L=2.29 \mathrm{m}} \\ {2.29 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4} \\ {\text { For friction } \mathrm{u}_{\mathrm{k}}} \\ {\mathrm{L}=\frac{4.24^{2}}{2 \times 9.8 \times 0.2}}\end{array}$

$\begin{array}{l}{L=\frac{17.9776}{3.92}} \\ {L=4.58 \mathrm{m}} \\ {4.58 \mathrm{m} \text { the block slide if the } \mathrm{u}_{5}=0.4}\end{array}$

8 0

2 years ago

The potential energy of an apple is 6.00 joules. The apple is 3.00-meters high. What is the mass of the apple?

BaLLatris [955]

Potential energy = (mass) x (gravity) x (height)

Divide each side by (G x H) : Mass = PE / (G x H)

PE = 6 joules
Gravity = 9.8 m/s
Height = 3 m
 Mass = 6 / (9.8 x 3) =

 6 / (29.4) = 0.204 kg

(weighs about 2 newtons or 7.2 ounces on Earth)

8 0

2 years ago

What is it called when objects become positively charged when they lose electrons and negatively charged when they gain electron

Alexeev081 [22]

It is c: electric charge

4 0

2 years ago

Read 2 more answers

On the moon, the acceleration due to gravity is one-sixth that of earth. That is gmoon = gearth /6 = (9.8 m/s2 )/6 = 1.63 m/s2 .

MAVERICK [17]

The pendulum would differ from 300 inches

8 0

2 years ago