Are Styrofoam and wood the same temperature, if they are not what are each others temperature

HELP

dem82 [27]

Answer:

Explanation:

28 / 70 = 0.3857142... = 0.39 hr

280 / 100 = 2.8 hrs.

(100 - 0) / 10 = 10 m/s²

(60 - 20) / 4 = 10 m/s²

3 0

3 years ago

A book on a 2-meter high shelf has a mass of 0.4 kg. What is its potential energy?

poizon [28]

Answer:

$\boxed {\boxed {\sf 7.84 \ Joules}}$

Explanation:

The formula for potential energy is:

$PE=m*g*h$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.

$m=0.4 \ kg \\g=9.8 \ m/s^2 \\h=2\ m$

Substitute the values into the formula.

$PE=(0.4 \ kg)(9.8 \ m/s^2)(2 \ m)$

Multiply the first two numbers.

0.4 kg*9.8 m/s²= 3.92 kg*m/s²
If we convert the units now, the problem will be much easier later on.
1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N

$PE=(3.92 \ N )(2 \ m)$

Multiply.

3.92 N* 2 m=7.84 N*m
1 Newton meter is equal to 1 Joule (this is why we converted the units).
Our answer is equal to 7.84 Joules.

$PE=7.84 \ J$

6 0

3 years ago

Read 2 more answers

Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da

valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0

3 years ago

Please answer this question for me and explain why.

horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass $m_2$ and the 4kg mass $m_1$ . If the tension in the string is $T$ then for the mass $m_2$

(1). $T-m_2g=-m_2a$ (the negative sign on the right side indicates that acceleration is downwards)

And for the mass $m_1$

(2). $T-m_1g =m_1a$ (the acceleration is upwards, hence the positive sign)

Solving for $T$ in the 2nd equation we get:

$T =m_1a+m_1g$ ,

and putting this into the 1st equation we get:

$m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g$

$\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}$

Back to the question:

Using the formula for the acceleration we find

$a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g$

$a = \dfrac{g}{9},$

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0

3 years ago

Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant

Cerrena [4.2K]

Answer:

the force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

$F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81 + \frac{70 \times 120^2}{100}\\\\= 10766.7 N$

Hence, the force exerted by the seat on the pilot is 10766.7 N

4 0

3 years ago