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dsp73
3 years ago
9

If Styrofoam has a specific heat of 1131 J/kg°C, how much energy would it take to change a 0.5 kg Styrofoam cooler from 15°C to

2°C?
Physics
1 answer:
Y_Kistochka [10]3 years ago
8 0

Answer:

.

Explanation:

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(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

5 0
2 years ago
Which energy transformation converts energy from the sun’s core to light energy needed by plants?
Assoli18 [71]
Nuclear to electromagnetic
3 0
3 years ago
When the leaves are __________________, they fall straight down.
konstantin123 [22]

Answer:Done doing there job, in the winter they have a break

Explanation:

7 0
3 years ago
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou
Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

P=\frac{W}{t}

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

P=\frac{\frac{1}{2}mv^2}{t}=const.

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}

where:

t_1=1.40 s is the time the car needs to accelerates to v_1=28.0 mph

t_2 is the time the car needs to accelerate to v_2=57.0 mph

Therefore, solving for t_2,

t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

a=\frac{v-u}{t}

where:

u = 0 is the initial velocity

v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

a=\frac{12.5-0}{1.40}=8.9 m/s^2

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

F=ma

This means that the acceleration is also constant.

Now we want to find how long the car takes to accelerate to a final velocity of

v=57.0 mph \cdot \frac{1609}{3600}=25.5 m/s

From an initial velocity of

u = 0

Using again the same suvat equation, and using the acceleration we found previously, we find:

t=\frac{v-u}{a}=\frac{25.5-0}{8.9}=2.87 s

Learn more about accelerated motion:

brainly.com/question/9527152

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About power:

brainly.com/question/7956557

#LearnwithBrainly

6 0
3 years ago
1a) An object is thrown up with a velocity of 18m/s from a height of 520m.
xeze [42]

Answer:

Explanation:

An object is thrown up from a height of 520m

h_o = 520m

Initial velocity of thrown is 18m/s

u = 18m/s

What is the position of the object after 3seconds

t = 3s

Acceleration due to gravity

g = 9.81 m/s²

Let calculated the height the object will reached when thrown from the top.

Using equation of motion

h = ut + ½gt²

Since the body is thrown upward, it is acting against gravity then, gravity will be negative

Then,

h = ut — ½gt²

h = 18 × 3 — ½ × 9.81 × 3²

h = 54 — 44.145

h = 9.855 m

So, the body is above the top of the building at a distance of 9.855m

So, the total distance from the bottom is

Position = h + h_o

x = 9.855 + 520

x = 529.855m

x ≈ 530m,

The position of objects after 3seconds is 520m

3 0
3 years ago
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