All categories

3 years ago

Please help me because it is due 12 am tonight will give brainlist

Physics

1 answer:

8_murik_8 [283]3 years ago

8 0

Answer:

The one to the right. His diagram was earth with moon, sun, mercury, mars, Venus, the fixed stars, and Jupiter and Saturn.

You might be interested in

A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate

exis [7]

Answer:

The velocity of water at the bottom, $v_{b} = 28.63 m/s$

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, $P_{gauge} = 2.90 atm$

Solution:

Now,

Atmospheric pressue, $P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa$

At the top, the absolute pressure, $P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa$

Now, the pressure at the bottom will be equal to the atmopheric pressure, $P_{b} = 1 atm = 1.01\times 10^{5} Pa$

The velocity at the top, $v_{top} = 0 m/s$ , l;et the bottom velocity, be $v_{b}$ .

Now, by Bernoulli's eqn:

$P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}$

where

$h_{t} - h_{b} = 12.8 m$

Density of sea water, $\rho = 1030 kg/m^{3}$

$\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} = v_{b}$

$\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} = v_{b}$

$v_{b} = 28.63 m/s$

5 0

3 years ago

Which equation would you use to calculate how much work is done in pushing a rock to the edge of the cliff?

Natalka [10]

Answer:

Explanation:

d because a b c are energy equation not work

4 0

2 years ago

Brutus, a champion weight lifter, raises 220 kg a distance of 3.10 m. (a) how much work is done by brutus lifting the weights? j

Nana76 [90]

A) work = force * distance
mass is not a force, weight is, so we have to find the weight of the block.
Weight = mg
Weight = (220kg)(9.8)
Weight = 2156N
Work = 2156N * 3.10m
work = 6683.6J
b) Since he is holding the weights, it's not moving, therefore, he doesn't do any work
c) The answer is still the same amount of work when he lifted them.
d) The answer is no since when he let go the weight, he doesn't apply any force to the weight.
e) P = work/time
P = 6683.6J / 2.1s
P = 3182.67 watts

3 0

3 years ago

Can someone make a list of 10 simple things people can do to maintain fitness ? Please I literally need help.

ivanzaharov [21]

Answer:

1. drink lots of water

2. eat healthy breakfast

3. plan your meals

4. set a deadline

5. exercise at least 30 minutes a day even 10 minutes 3 times a day is good

6. eat fruits and vegetables

7. eat whole grains instead of white

8. get a good night's sleep

9. stay motivated

10. take time for mental health

6 0

3 years ago

Read 2 more answers

A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

3 years ago