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3 years ago

Write the ordered pair that represents vec yz .Then find the magnitude of vec yz Y(- 4, 12), (1, 19)

Physics

1 answer:

sladkih [1.3K]3 years ago

7 0

Answer with Explanation:

We are given that

Y=(-4,12)

Z=(1,19)

We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.

Vector YZ=Z-Y=<1,19>-<-4,12>

Vector YZ=<5,7>

Magnitude of vector r

$r=xi+yj$

$\mid r\mid=\sqrt{x^2+y^2}$

Using the formula

$\vec{YZ}=\sqrt{5^2+7^2}$

$\vec{YZ}=\sqrt{25+49}=\sqrt{74} units$

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You are making pesto for your pasta and have a cylindrical measuring cup 12.0 cmcm high made of ordinary glass [β=2.7×10−5(C∘)−1

katrin [286]

Answer:

81.8°C

Explanation:

βglass = 2.7×10^(−5) (C∘)−1

βoil = 6.8 × 10^(−4) (C∘)−1

T1 = 22°C

Total cup height = 12cm = 0.12m

Height below top of cup= 4.5mm or 0.0045m

From the question, we can see that the coefficient volume of expansion of oil is greater than that of glass and therefore, the heat absorbed by oil will be greater.

Thus;

δVoil =δVglass + 0.0045A

So, 0.0045A = δVoil - δVglass

0.0045A = δV(oil) - δVglass

Now,

δV(oil) is expressed fully as;

δT x V(oil) xβ(oil)

Likewise, δV(glass) is expressed fully as;

δT x V(glass) xβ(glass). Thus;

0.0045A = [δT x V(oil) xβ(oil)] - [δT x V(glass) xβ(glass)]

So,

0.0045 = δT[V(oil) xβ(oil)] - [V(glass) xβ(glass)]

So;

δT = 0.0045/[V(oil) xβ(oil)] - [V(glass) xβ(glass)]

Plugging in the relevant values to obtain ;

δT = 0.0045/[V(oil) xβ(oil)] - [V(glass) xβ(glass)]

V(oil) = Area of cup x height = A(0.12 - 0.0045) = 0.1155A

V(glass) = 0.12A

So,δT = 0.0045/[(0.1155A) x 6.8 × 10^(−4) ] - [0.12A x 2.7×10^(−5))]

= 0.0045/[0.7854 -x 10^(−4) - 0.0324 x 10^(−4)] = 0.0045/(0.753 x 10^(-4) = 59.76°C

Now, δT is the change in temperature and it's ;

δT = T2 - T1

Thus, T2 = δT + T1

T2 = 59.76 + 22 = 81.76°C

Approximating to 3 significant figures, T2 = 81.8°C

5 0

3 years ago

What is a transverse and longitudinal wave ?

kodGreya [7K]

Transverse waves travel perpendicular to neighboring wave motion like a slinky while longitudinal waves depend on the displacement of a medium which include sound waves.

8 0

3 years ago

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon

WINSTONCH [101]

1) velocity: 98 m/s downward

Explanation:

The marble moves by uniformly accelerated motion, with constant acceleration a=g=-9.8 m/s^2 directed towards the ground. Therefore, its velocity at time t is given by:

$v(t)=v_0 +at$