Will give brainliest! A team of engineers is asked to evaluate several different

madam [21]

1) hypothesis
2) data
3) method

I think these are correct.

7 0

3 years ago

A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitr

kirill115 [55]

Answer:

The heat capacity of a sample is 37.7 J/K.

Explanation:

Given that,

Submerged temperature of tissue sample = 275 K

Mass of liquid nitrogen= 2 kg

Temperature = 70 K

Final temperature = 75 K

We need to calculate the heat

Using formula of heat

$Q=mc(T_{f}-T_{i})$

Put the value into the formula

$Q=2\times1.039\times10^{3}\times(75-70)$

$Q=10390\ J$

We need to calculate the heat capacity of a sample

Using formula of heat capacity

$\Delta S=\dfrac{Q}{T}$

Put the value into the formula

$\Delta S=\dfrac{10390}{275}$

$\Delta S=37.7\ J/K$

Hence, The heat capacity of a sample is 37.7 J/K.

7 0

3 years ago

Read 2 more answers

a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th

Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

3 0

3 years ago

An object is attached to a hanging unstretched, ideal and massless spring and slowly lowered to its equilibrium position, a dist

Ad libitum [116K]

Answer:

10.6cm

Explanation:

We are given 5.3cm below the starting point (spring extension).

Therefore, to find static vertical equilibrium, we use the equation:

kx = mg

Where:

k = spring constant =

=mg/5.3 kg/s²

We are told the object was dropped from rest.

Therefore:

loss in potential energy = gain in spring p.e

Let's use the expression:

mgx = ½kx²

We are asked to find the stretch at maximum elongation x.

To find x, we make x subject of the formula.

Therefore, we have:

x = 2mg/k (after rearranging the equation above)

x = (2mg) / (mg/5.3)

x = 10.6cm

3 0

2 years ago

A tungsten wire has resistance R at 20°C. A second tungsten wire at 20°C has twice the length and half the cross-sectional area

Colt1911 [192]

Answer:

Resistance will become 4 times of first wire resistance.

Explanation:

At the temperature of both the the tungsten wire is same so we can apply ohm's law

Let the length of first wire is $l_1$ and cross sectional area is $A_1$

Resistance of first wire $R=\frac{\rho l_1}{A_1}$ ......1

Now length of second wire is twice the length of first wire

$l_2=2l_1$ and cross sectional area $A_2=\frac{A_1}{2}$ .......2

Resistance of wire 2 $R_2=\frac{\rho l_2}{A_2}$ ........2

Dividing equation 1 by equation 1

$\frac{R}{R_2}=\frac{\rho l_1}{A_1}\times \frac{0.5A_1}{\rho 2l_1}$

$R_2=4R$

Therefore resistance will become 4 times.

8 0

3 years ago