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2 years ago

Convert 90 oF to Celcius, (write your answer up to 1st decimal point)

Physics

2 answers:

Fudgin [204]2 years ago

6 0

Answer:

32.2

HOPE THIS HELPS :)

Mice21 [21]2 years ago

5 0

Answer:

The answer is 32.2

Explanation:

Hope this helps

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A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro

Mrac [35]

Answer:

The work done to get you safely away from the test is 2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope = 2 lb/ft x 70 ft

= 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs

= 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s² x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.

4 0

2 years ago

What effect does the sun have on Earth's tides? A. The pull of the sun cancels the pull of the moon when they are on opposite si

Kisachek [45]

Answer:

Explanation:

because two vectors which align in the same line adds one to another

6 0

2 years ago

How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr

Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

$Q =mc\Delta T$

where,

m = mass

c = specific heat

$\Delta T$ = Change in temperature

Therefore the total heat exchange is given as

$\Delta Q = Q_w+Q_v$

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

Our values are given as,

Total mass is $M_T$ = 200lb ,however the mass of solid vegetable and water is given as,

$m_v= 0.4*200lb = 80lb$

$m_w=0.6*200lb=120lb$

$T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF$

Replacing at our equation we have,

$\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)$

$\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)$

$\Delta Q = 22411.2Btu$

Therefore the heat removed is 22411.2 Btu

6 0

2 years ago

Organisms whose whole population was at risk of dying out are called???

natali 33 [55]

A nearly extinct species is an endangered species

8 0

3 years ago

Read 2 more answers

A sample of a monatomic ideal gas initially at temperature T, undergoes a process during which the pressure of the gas triples a

Allisa [31]

Answer:

B) 8 T.

Explanation:

For an ideal gas , the gas law that it follow is as follows

P₁V₁ / T₁ = P₂ V₂ /T₂

This law is followed by all the gases whether mono atomic or diatomic.

In this case following information are given

P₂ / P₁ = 3

V₂ / V₁ = 3

T₁ = T

T₂ = ?

From the gas law formula given , we have

$T_2 =\frac{P_2\times V_2\times T_1}{P_1\times V_1}$

$V_2 = \frac{P_2}{P_1} \times\frac{V_2}{V_1}\times T$

= 3 X 3 T

= 9 T

Change in temperature = 9T - T = 8 T.

7 0

2 years ago