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3 years ago

Explain on any two modifications that you would expect to use for the thermos flask which would enable it to be handier/helpful.

Physics

2 answers:

LuckyWell [14K]3 years ago

7 0

Answer:

The two modifications in the Thermos flask which would enable it to be more helpful:

1) To have an internal temperature indicator that indicates temperature.

2) A wireless rechargeable battery (USB connector)

Pls mark me brainliest.

quester [9]3 years ago

5 0

Here, we are required to explain any two modifications to use for the thermos flask which would enable it to be handier/ helpful.

The presence of a built-in temperature monitoring system would enable it to be handier/helpful.
The presence of a built-in solar panel on the thermos flask cover for harvesting solar energy which may further be converted to electric energy and dispensed through USB ports.

Value addition is important in improving existing products such that they become handier/helpful.

Therefore, the 2 modifications stated above can be used for the Thermos flask to enable it to be handier/handler.

Read more:

brainly.com/question/24019227

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The three branches of science are ____, earth, and physical

Rom4ik [11]

The three branches of science are life, earth, and physical science.
Life science deals with living organisms, so, biology is the most important of these sciences, although there are many.
Earth science deals with Planet Earth, studying its atmosphere, lithosphere, etc.
Physical science studies the inorganic world.

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4 years ago

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

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3 years ago

If a rock climber accidentally drops a 52.5-g piton from a height of 325 meters, what would its speed be just before striking th

alex41 [277]

Ignoring air resistance, the Kinetic energy before hitting the ground will be equal to the potential energy of the Piton at the top of the rock.
So we have 1/2 MV^2 = MGH
V^2 = 2GH
V = âš2GH
V = âš( 2 * 9.8 * 325)
V = âš 6370
V = 79.81 m/s

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3 years ago

Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determin

tia_tia [17]

Answer with Explanation:

We are given that

A=3i-3j m

B=i-4 j m

C=-2i+5j m

a. $D=A+B+C$

$D=3i-3j+i-4j-2i+5j$

$D=2i-2j$

Compare with the vector r=xi+yj

We get x=2 and y=-2

Magnitude= $\mid D\mid=\sqrt{x^2+y^2}=\sqrt{(2)^2+(-2)^2}=2\sqrt 2$ units

By using the formula $\mid r\mid=\sqrt{x^2+y^2}$

Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

$E=-3i+3j-i+4j-2i+5j$

$E=-6i+12j$

$\mid E\mid=\sqrt{(-6)^2+(12)^2}=13.4$ units

Direction of E= $\theta=tan^{-1}(\frac{12}{-6}=tan^{-1}(-2)=-63.4^{\circ}$

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3 years ago

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coldgirl [10]

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3 years ago