1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Anarel [89]
3 years ago
7

Pls send the answer..pls...​

Physics
1 answer:
Sophie [7]3 years ago
4 0

Answer:

The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

Explanation:

Given;

initial temperature of the liquid, t₁ = 76.3  +/-  0.4⁰C

final temperature of the liquid, t₂ = 67.7  +/-  0.3⁰C

The change in temperature of the liquid is calculated as;

Δt = t₂  -  t₁

Δt = (67.7 - 76.3)  +/-  (0.3 - 0.4)

Δt = (-8.6)  +/-  (-0.1)

Δt = 8.6 +/- 0.1 ⁰C

Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C

You might be interested in
A brick falls from a wall to the ground 3.42 m below. How much time will it take to
ddd [48]

Answer:

The time taken by the brick to hit the ground, t = 0.84 s

Explanation:

Given that,

A brick falls from a height, h = 3.42 m

The initial velocity of the brick is zero.

Since the brick is under free-falling. The time equation of a free-falling body when the displacement is given is

                                     t = \sqrt{2h/g}

where,

                                 h - height from surface in meters

                                 g - acceleration due to gravity

on substituting the values in the above equation,

                                 t =  \sqrt{2X3.14/9.8}

                                   = 0.84 s

Hence, time taken by the brick to hit the ground is t = 0.84 s

6 0
3 years ago
Road rage is an aggressive driving incident where the driver has:
WARRIOR [948]
I would say that it would be A.) Been insulted. This is because you are angered by someone else. It does not necessarily mean that something has de-escelated, you've lost control of your car, or you've nothing to lose. Hopefully that helps. :)

5 0
3 years ago
If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com
Lelu [443]

Answer:

<em> B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.</em>

<em></em>

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = 100*10^{-6} R

for dry skin, voltage will be

V = IR = 100*10^{-6}*200R = 0.02R

Comparing both voltages

0.02R ÷  100*10^{-6} R  = 200

<em>this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.</em>

4 0
3 years ago
Which is an example of an unstructured activity that promotes resistance training?
aleksandrvk [35]
One possible unstructured activity that promotes resistance training would be climbing playground equimpent - A. 

This is by nature a unstructured ctivity. Furthermore, it promotes resistance training because you're forced to move and pull and push yourself. 
3 0
3 years ago
This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
2 years ago
Other questions:
  • What conditions are required for resonance to occur?
    10·2 answers
  • You can see lightning before you can hear the thunder. Which is the best explanation for this?
    11·2 answers
  • Two identical balls balls P and Q, each of mass 0.20 g, carry identical charges and are suspended by two threads of equal length
    15·1 answer
  • I'm very confused on this home work question How did magnetism 'drive' the theory of plate tectonics?
    10·1 answer
  • Water is pumped from a reservoir to a tank suspended 12 m above the reservoir. The water flows through the same size pipe throug
    5·1 answer
  • Which of the following describes work?
    5·2 answers
  • "A window washer pulls herself upward using the bucket-pulley apparatus. The mass of the person plus the bucket is 65 kg.
    7·2 answers
  • If a golf ball is dropped from the thirteenth floor of a building, ignoring air resistance, after falling for 7.00 seconds the s
    13·1 answer
  • Tyler and his family drove from Atlanta to Baltimore. They traveled a total of 678 miles over 2 days.
    10·2 answers
  • the velocity of a body of mass 60kg reaches 15m/s from 0m/s in 12 second. calculate the kinetic energy and power of the body.​
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!