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Dmitrij [34]
2 years ago
6

Please help: 2.Le pluriel Questions

Physics
1 answer:
sp2606 [1]2 years ago
8 0

1. bleus

2. rouges

3. petites

4. actifs

5. gross

Just add an -s to make it plural

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Flauer [41]
     Using the Definition of Kinetic Energy, we have:

E_{c}= \frac{mv^2}{2}  \\ E_{c}= \frac{8*4^2}{2}  \\ \boxed {E_{c}=64J}
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Suppose that Lisa walks her dog around the block for a little exercise. The block is 1 mile around. If she walks around the bloc
natita [175]
Displacement only measure how far between the starting and ending point. In this case, Lisa walks around the block as a circle so the starting point is the same as the ending point. Thus, displacement is 0mile.
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A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?
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8 0
3 years ago
Read 2 more answers
You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body
stellarik [79]

A) The change in internal chemical energy is 1.15\cdot 10^7 J

B) The time needed is 1 minute

Explanation:

First of all, we start by calculating the power output of you and the bike, given by:

P=Fv

where

F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed

v = 8.0 m/s is the velocity

Substituting,

P=(80)(8.0)=640 W

The energy output is related to the power by the equation

P=\frac{E}{t}

where:

P = 640 W is the power output

E is the energy output

t = 30 min \cdot 60 = 1800 s is the time elapsed

Solving for E,

E=Pt=(640)(1800)=1.15\cdot 10^6 J

Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by

\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J

B)

From the previous part, we found that in a time of

t = 30 min

the amount of internal chemical energy converted is

E=1.15\cdot 10^7 J

Here we want to find the time t' needed to convert an amount of chemical energy of

E'=3.8\cdot 10^5 J

So we can setup the following proportion:

\frac{t}{E}=\frac{t'}{E'}

And solving for t',

t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min

Learn more about power and energy:

brainly.com/question/7956557

#LearnwithBrainly

3 0
3 years ago
A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
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