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Harman [31]
3 years ago
10

Which of following is an example of deduction?

Physics
1 answer:
Jobisdone [24]3 years ago
6 0
A, because Antoine came to a conclusion that it must have rained based on his visuals. 
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Consider the vector field. f(x, y, z) = xy2z2i x2yz2j x2y2zk (a) find the curl of the vector field?
Marat540 [252]

Observe that the given vector field is a gradient field:

Let f(x,y,z)=\nabla g(x,y,z), so that

\dfrac{\partial g}{\partial x} = x y^2 z^2

\dfrac{\partial g}{\partial y} = x^2 y z^2

\dfrac{\partial g}{\partial z} = x^2 y^2 z

Integrating the first equation with respect to x, we get

g(x,y,z) = \dfrac12 x^2 y^2 z^2 + h(y,z)

Differentiating this with respect to y gives

\dfrac{\partial g}{\partial y} = x^2 y z^2 + \dfrac{\partial h}{\partial y} = x^2 y z^2 \\\\ \implies \dfrac{\partial h}{\partial y} = 0 \implies h(y,z) = i(z)

Now differentiating g with respect to z gives

\dfrac{\partial g}{\partial z} = x^2 y^2 z + \dfrac{di}{dz} = x^2 y^2 z \\\\ \implies \dfrac{di}{dz} = 0 \implies i(z) = C

Putting everything together, we find a scalar potential function whose gradient is f,

f(x,y,z) = \nabla \left(\dfrac12 x^2 y^2 z^2 + C\right)

It follows that the curl of f is 0 (i.e. the zero vector).

5 0
2 years ago
If the maximum acceleration that is tolerable for passengers in a subway train is 1.39 m/s2 and subway stations are located 780
a_sh-v [17]

Answer:

The maximum speed a subway train can attain between stations is 32.93 m/s.

Explanation:

Given;

maximum tolerable acceleration = 1.39 m/s²

distance between subway train, d = 780 m

The distance available to accelerate between stations = ¹/₂ x 780 m = 390 m

Apply the following kinematic equation to determine the maximum speed;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2(1.39)(390)

v² = 1084.2

v = √1084.2

v = 32.93 m/s

Therefore, the maximum speed a subway train can attain between stations is 32.93 m/s.

6 0
4 years ago
A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

The height of the bullet at this time is calculated as follows;

y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft

Learn more about height of projectiles here: brainly.com/question/10008919

6 0
3 years ago
A 75.0kg teacher (including the parachute) is skydiving! As the parachute opens, the system experiences a 1000N drag.
Ronch [10]

Answer:

Net force = 250 N

Explanation:

Given that,

The mass of a teacher, m = 75 kg

As the parachute opens, the system experiences a 1000N drag.

We need to find the magnitude of net force acting on the teacher. 2 forces are acting on her i.e. drag force and weight.

Net force,

F = mg-D

= 75(10)-1000

= -250 N

Hence, the net force is 250 N and it is acting in the upward direction.

3 0
3 years ago
A dog is 10 m from a cat, whose speeds are 6 and 5 m / s, respectively. What time does the dog require to catch the cat?
scoray [572]

Answer:

10 seconds

Explanation:

because the cat is moving one m/s slower than the dog, the dog has a relative speed of 1 m/s. 10 meters would take 10 seconds for the dog to cover

4 0
4 years ago
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