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2 years ago

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A potato and raisins salad has been warmed up to a temperature of 80∘C80∘C and let it stand for three minutes. Then one tries a

bite.
Would the potatoes and raisins be equally warm? Potatoes have a specific heat of 3430 J/(kg⋅∘C)3430 J/(kg⋅∘C). Raisins have a specific heat of 1630 J/(kg⋅∘C)1630 J/(kg⋅∘C).

a) No. Potatoes will be warmer.

b) No. Raisins will be warmer.

c) Yes

1 answer:

barxatty [35]2 years ago

6 0

Answer : The correct option is, (a) No. Potatoes will be warmer.

Explanation :

Formula used :

$Q=m\times c\times \Delta T$

where,

Q = heat taken

m = mass of substance

c = specific heat of substance = ?

$\Delta T$ = change in temperature

As we are given that the potatoes have a specific heat of $3430J/(kg^oC)$ and raisins have a specific heat of $1630J/(kg^oC)$ . That means higher the value of specific heat the more heat taken by the substance (more warmer will be the substance). So, the potatoes will be more warmer as compared to raisins.

Hence, the correct option is, (a) No. Potatoes will be warmer.

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The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

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A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

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