Does anyone know what these three are? I need help!

Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. As in the video, the mass of t

dimaraw [331]

Answer:

b) one-third as great.

Explanation:

As we know that same heat is supplied in this experiment

so we will have

$Q = ms\Delta T$

now we know that both are initially at same temperature

then their final temperatures are

$T_1 = 40 degree$

$T_2 = 80 degree$

now we have

$m_1s_1 (40 - 20) = m_2s_2(80 - 20)$

so we have

$m_2s_2 = \frac{20}{60} m_1s_1$

so heat capacity of mystery metal is 1/3 times that of water

6 0

3 years ago

Read 2 more answers

The displacement vector from your house to the library is 740 m long, pointing 40 ∘ north of east. Part A What are the x-compone

Ilia_Sergeevich [38]

The horizontal component of the displacement is 566.9 m

Explanation:

The horizontal (x-) and vertical (y-) components of a vector on the Cartesian plane can be found as follows:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v is the magnitude of the vector

$\theta$ is the angle representing the direction of the vector, measured as above the x-axis.

In this problem, we have:

v = 740 m (magnitude of the vector)

$\theta=40^{\circ}$ (direction of the vector)

Therefore, the two components are

$v_x = (740)(cos 40)=566.9 m$

$v_y = (740)(sin 40)=475.7 m$

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

4 0

3 years ago

A car moving at 95 km/h passes a 1.00-km-long train traveling in the same direction on a track that isparallel to the road. If t

victus00 [196]

Answer:

Time to pass the train=0.05 h

How far the car traveled in this time=4.75 Km

Explanation:

We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:

$\Delta V=V_{car}-V_{train}=95km/h-75km/h=20km/h$

We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:

$V=\dfrac{x}{t}$

Where x is the distance, t is the time and v is the speed. using the data that we have:

$V=\dfrac{x}{t}=\dfrac{1km}{20km/h}=0.05h$

This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:

$V=\dfrac{x}{t}\\x=v\cdot t\\x=95km/h\cdot 0.05h\\x=4.75km$

7 0

3 years ago

Find the ratio of the radii of a baseball to the Earth, knowing that the radius of a baseball is .09 m, and that the Earth's rad

Xelga [282]

0.09 / 6.37 x 10⁶ = 1.4129 x 10⁻⁸

The radius of the baseball is 1.4129 x 10⁻⁸ the radius of the Earth.

6 0

3 years ago

(a) (i) Find the gradient of f. (ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rat

vitfil [10]

Question:

Problem 14. Let f(x, y) = (x^2)y*(e^(x−1)) + 2xy^2 and F(x, y, z) = x^2 + 3yz + 4xy.

(a) (i) Find the gradient of f.

(ii) Determine the direction in which f decreases most rapidly at the point (1, −1). At what rate is f decreasing?

(b) (i) Find the gradient of F.

(ii) Find the directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k.

Answer:

The answers to the question are

(a) (i) the gradient of f = ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) The direction in which f decreases most rapidly at the point (1, −1), ∇f(x, y) = -1·i -3·j is the y direction.

The rate is f decreasing is -3 .

(b) (i) The gradient of F is (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2 i + 3 j − √ 3 k is ñ∙∇F = 4·x +⅟4 (8-3√3)y+ 9/4·z at (1, 1, −5)

4 +⅟4 (8-3√3)+ 9/4·(-5) = -6.549 .

Explanation:

f(x, y) = x²·y·eˣ⁻¹+2·x·y²

The gradient of f = grad f(x, y) = ∇f(x, y) = ∂f/∂x i+ ∂f/∂y j = = (∂x²·y·eˣ⁻¹+2·x·y²)/∂x i+ (∂x²·y·eˣ⁻¹+2·x·y²)/∂y j

= ((y·x² + 2·y·x)·eˣ⁻¹ + 2·y² )i + (x²·eˣ⁻¹+4·y·x) j

(ii) at the point (1, -1) we have

∇f(x, y) = -1·i -3·j that is the direction in which f decreases most rapidly at the point (1, −1) is the y direction.

The rate is f decreasing is -3

(b) F(x, y, z) = x² + 3·y·z + 4·x·y.

The gradient of F is given by grad F(x, y, z) = ∇F(x, y, z) = = ∂f/∂x i+ ∂f/∂y j+∂f/∂z k = (2·x+4·y)i + (3·z+4·x)j + 3·y·k

(ii) The directional derivative of F at the point (1, 1, −5) in the direction of the vector a = 2·i + 3·j −√3·k

The magnitude of the vector 2·i +3·j -√3·k is √(2²+3²+(-√3)² ) = 4, the unit vector is therefore

ñ = ⅟4(2·i +3·j -√3·k)

The directional derivative is given by ñ∙∇F = ⅟4(2·i +3·j -√3·k)∙( (2·x+4·y)i + (3·z+4·x)j + 3·y·k)

= ⅟4 (2((2·x+4·y))+3(3·z+4·x)- √3∙3·y) = 4·x +⅟4 (8-3√3)y+ 9/4·z at point (1, 1, −5) = -6.549

8 0

3 years ago