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3 years ago

11

Herbie's mother sent him to the store and told him 12 items to get. He didn't write them down and when he got to the store he ha

d some retrieval problems. Herbie is most likely to remember the items _____________ of the list.

1 answer:

Eva8 [605]3 years ago

7 0

I would say important or last items on the list

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In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer spce. From stationary positions, they push agains

Hoochie [10]

Answer:

$\frac{m_B}{m_E}=1.18$

Explanation:

$m_B$ = Mass of Bonzo

$m_E$ = Mass of Ender

$u_B$ = Initial Velocity of Bonzo = 2.2 m/s

$u_E$ = Initial Velocity of Ender = -2.6 m/s

From conservation of linear momentum

$m_Bu_B=m_Eu_E\\\Rightarrow \frac{m_B}{m_E}=\frac{u_E}{u_B}\\\Rightarrow \frac{m_B}{m_E}=\frac{2.6}{2.2}\\\Rightarrow \frac{m_B}{m_E}=1.18$

$\therefore \frac{m_B}{m_E}=1.18$

8 0

3 years ago

Assume the earth to be a nonrotating sphere with mass MEME and radius RERE. If an astronaut weights WW on the ground, what is hi

Solnce55 [7]

Answer:

The weight at a distance 2 RE from surface of earth is <em>W/9</em>

Explanation:

For the value of acceleration due to gravity (g), we have a formula, that is:

g = (G)(ME)/(RE)² ----- equation (1)

where,

G = Gravitational Constant

ME = Mass of Earth

RE = Radius of Earth

g = Acceleration due to gravity on surface of earth = 9.8 ms²

When the person goes 2RE, distance above earth's surface. Then the total distance from center of earth becomes: 2RE + RE = 3RE.

Therefore, equation (1) becomes:

gh = (G)(ME)/(3RE)²

where,

gh = acceleration due to gravity at height

gh = (G)(ME)/(RE)²9

using equation (1), we get:

gh = g/9

Now, he weight is given by formula:

W = mg ------- equation (2)

At height 2RE

Wh = (m)(gh)

where,

Wh = Weight at height = ?

m = mass of astronaut

Therefore, using vale of gh, we get:

Wh = mg/9

Using equation (2), we get:

<u>Wh = W/9</u>

6 0

3 years ago

Carbohydrates are made of hydrogen, oxygen, and which of the following? A. nitrogen B. carbon C. helium D.argon

k0ka [10]

The answer would be carbon

4 0

3 years ago

I will GIVE BRAINLYEST AND POINTS PLEAS HELP!!!

Elenna [48]

Answer:

1. Sports education-uses seasons and teams to educate

2. Adventure education-uses challenging recreational activities

3. Outdoor education-may use camps to teach activities

4. Dance education-uses cultural movement activities

5. fitness education-teaches concepts and self-management skills

Explanation:

I figured that these answers made the most sense so ya.

3 0

3 years ago

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at

GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0

2 years ago