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2 years ago

6

Fig. 19-4 shows two wires, in cross section, carrying 7.0 Amperes out of the page. Determine the total magnetic field, due to bo

th wires, at point X midway between the wires.

1 answer:

mel-nik [20]2 years ago

6 0

I habe no idea i just need points sorry

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For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar

Gala2k [10]

Distance covered by the squirrel to look for an acorn :

d = ( 3 m/s ) × 10 s = 30 m.

Time taken to eat an Acron is 5 seconds.

Time taken to cover distance of 30 m with 2 m/s speed is :

$T=\dfrac{30}{2}\ s= 15 \ s$

Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.

Hence, this is the required solution.

7 0

2 years ago

Explain how cognitive psychologists combine traditional conditioning models with cognitive processes.

user100 [1]

Behaviorists generally claimed that conditioning occurred without thinking or reasoning ans was simply a result of consequences or reinforcement. Cognitive psychologists demonstrated that thinking and reasoning (cognition) influences the conditioning processes and that many behaviors that are conditioned depend on the type of cognitive reasoning that occurs during conditioning. Therefore, as one is being conditioned to respond to environmental stimuli or is responding to a consequence, they are also pondering and thinking about the process occuring. Cognition is often the reason individuals are not all conditioned in the same manner.

4 0

1 year ago

Read 2 more answers

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

7 0

2 years ago

|| Climbing ropes stretch when they catch a falling climber, thus increasing the time it takes the climber to come to rest and r

Otrada [13]

To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.

Newton's second law is defined as

$F = ma$

Where,

m = mass

a = acceleration

From this equation we can figure the acceleration out, then

$a = \frac{F}{m}$

$a = \frac{11*10^3}{80}$

$a = 137.5m/s$

From the cinematic equations of motion we know that

$v_f^2-v_i^2 = 2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial velocity

a = acceleration

x = displacement

There is not Final velocity and the acceleration is equal to the gravity, then

$v_f^2-v_i^2 = 2ax$

$0-v_i^2 = 2(-g)x$

$v_i =\sqrt{2gx}$

$v_i = \sqrt{2*9.8*4.8}$

$v_i = 9.69m/s$

From the equation of motion where acceleration is equal to the velocity in function of time we have

$a = \frac{v_i}{t}$

$t = \frac{v_i}{a}$

$t =\frac{9.69}{137.5}$

$t = 0.0705s$

Therefore the time required is 0.0705s

4 0

2 years ago

Read 2 more answers

Light from a distant star is collected by a concave mirror that has a radius of curvature of 150 cm. How far from the mirror is

andrey2020 [161]

The focal point of a mirror is half of the radius of curvature. We can use the formula, 1/f = 1/v + 1/u where f is the focal length , v is the image distance and u is object distance The distance of the star is assumed to be incredibly far away and 1 divided by a really big number is approximately zero, thus; 1/f = 1/v = 1/75 = 1/v therefore; the image is formed 75 cm infront of the mirror

4 0

2 years ago