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wlad13 [49]
3 years ago
8

I'm doing something for science but I'm not sure how to answer these questions.

Physics
1 answer:
Tomtit [17]3 years ago
5 0

1,  yeti hair is thick and strong

2, a warm blanket and a change of clothes

3, to circulate the heat around the house/room

You might be interested in
Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did
Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

5 0
3 years ago
If y^2= 3.249 x 10^-11, y = ?
Leni [432]

Answer:

Scientific Notation: 3.45 x 10^5

E Notation: 3.45e5

5 0
3 years ago
Read 2 more answers
A student places his hand in front of a plane mirror as shown in the diagram. Which terms correctly describe the image that the
ANTONII [103]
Either A or D but maybe D
4 0
3 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32 .1 kj/mol. de
atroni [7]
Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
5 0
3 years ago
A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
Lemur [1.5K]

Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

                  = 5 kJ

4 0
3 years ago
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