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salantis [7]
3 years ago
7

A box is 5 cm high, 4 cm wide, and 9 cm long. What is the

Physics
1 answer:
FromTheMoon [43]3 years ago
6 0

Answer:

18

Explanation:

  1. 5+4=9+9=<em>18</em><em> </em><em>IS</em><em> </em><em>THE</em><em> </em><em>VOLUME</em><em> </em><em>OF</em><em> </em><em>CUBIC</em><em> </em><em>CENTIMETRES</em><em> </em>
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Liz has designed a sailboat. She wants to build a prototype of the boat. Liz has $12,000 to spend on the boat. $7,000 of that mo
Naddika [18.5K]

Answer:

A. Marine plywood only.

Explanation:

3 0
3 years ago
Read 2 more answers
Help with this question.....​
mariarad [96]

Answer:

g_{moon}=1.67 [m/s^{2} ]

Explanation:

The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.

w =m*g\\

where:

w = weight [N]

m = mass = 0.06 [kg]

g = gravity acceleration = 10 [N/kg]

Therefore:

w=0.06*10\\w=0.6[N]

By Hooke's law we know that the force in a spring can be calculated by means of the following expression.

F=W\\F = k*x

where:

k = spring constant [N/m]

x = deformed distance = 6 [cm] = 0.06 [m]

We can find the spring constant.

k= F/x\\k=0.6/0.06\\k=10 [N/m]

Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.

F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]

Since this force is equal to the weight, we can now determine the gravitational acceleration.

F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]

6 0
3 years ago
An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
marissa [1.9K]
The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
where 
k is the spring constant
x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
4 0
3 years ago
The hydraulic lift in a car repair shop has an output diameter of 30 cm and is to lift cars up to 2000 kg. determine the fluid g
leonid [27]
Given Data: Diameter 'd' = 30 cm = 0.3 m Lifting Weight 'W' = mg = 2000*9.81 N = 19,620 N Calculations: Area of the lift 'A' = <span>pi\over4*d^2=pi\over4*0.3^2=0.07 m^2

Thank you for posting your question here at brainly. I hope the answer helps. </span>
4 0
3 years ago
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An engine extracts 452.8kJ of heat from the burning of fuel each cycle, but rejects 266.7 kJ of heat (exhaust, friction,etc) dur
Svetlanka [38]

Answer: The thermal efficiency of the engine is 41.09 %.

Explanation:

Efficiency is the ratio of the useful work performed to the total energy expended or heat taken in.

Formula for thermal efficiency of engine is

\eta=1- \frac{Q_2}{Q_1}\times 100

\eta = efficiency

{Q_2} = heat rejected = 266.7 kJ

{Q_1} = heat extracted = 452.8 kJ

Putting in the values we get:

\eta=1- \frac{266.7 kJ}{452.8 kJ }\times 100

\eta=0.41\times 100

\eta =41.09\%

The thermal efficiency of the engine is 41.09 %.

7 0
4 years ago
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