Question 1 of 10

The formula $F = \frac{9}{5} C + 32$ can be used to convert temperatures between degrees Fahrenheit ($F$) and degrees Celsius ($

slamgirl [31]

Answer:

-30° C

Explanation:

Data provided in the problem:

The formula for conversion as:

F = (9/5)C + 32

Now,

for the values of F = -22 , C = ?

Substituting the value of F in the above formula, we get

-22 = (9/5)C + 32

or

-22 - 32 = (9/5)C

or

(9/5)C = - 54

or

C = - 54 × (5/9)

or

C = - 30 °

Hence, -22 Fahrenheit equals to -30°C

7 0

2 years ago

What must happen to an atom of magnesium in order to become a magnesium ion

Zielflug [23.3K]

I would assume that the magnesium would have to come into contact with an ion in order to become a magnesium ion.

6 0

2 years ago

A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn

vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

$f_{m} = \mu_{s}N$

where

N = Normal reaction = mg

Thus

$f_{m} = \mu_{s}mg$

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

$f\leq f_{m}$

$f \leq \mu_{s}mg$

$109 \leq \mu_{s}\times 31.2\times 9.8$

$\mu_{s}\geq 0.356$

7 0

2 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

Provide two ways that objects can become charged

kap26 [50]

Answer:

friction, conduction and induction

Explanation:

Had it in class I had it correct hope this helps.

8 0

2 years ago