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rosijanka [135]
2 years ago
5

Which situation would create a field like the one shown here? A. an interaction between two magnetic south poles B. an interacti

on between two magnetic north poles C. an interaction between a magnetic south pole and a north pole D. an interaction between a magnetic south pole and an aluminum bar E. an interaction between a magnetic north pole and a zinc bar
Physics
2 answers:
alexgriva [62]2 years ago
4 0

Answer:

you have no picture

Explanation:

if it shows the arrows going towards eachother it is north and south. if it is not it will be either north and north or south and south

andriy [413]2 years ago
3 0

Answer:

C. an interaction between a magnetic south pole and a north pole

Explanation:

  • According to the situation the magnetic properties of the fields of the south and the north by the nature of the electrically charged electromagnets. That exerts a force and likewise, the earth also has two magnetic poles that have positive and negative properties and both oppose the each other properties.
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Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is
vaieri [72.5K]
1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle)  to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
 
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos 

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. :  sin(40) x '?' = 4905
-divide both sides by sin(40):  '?' = 4905/ sin(40)
--> Solve.

8 0
2 years ago
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
aleksklad [387]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

6 0
2 years ago
The flat-bed trailer carries two 1500-kg beams with the upper beam secured by a cable. The coefficients of static friction betwe
Novosadov [1.4K]

Answer:

a) a= 8.33 m/s²,    T = 12.495 N , b)    a = 2.45 m / s²

Explanation:

a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.

We apply Newton's second law to the lower charge

            fr₁ + fr₂ = ma

The equation for the force of friction is

          fr = μ N

Y Axis

         N - W₁ –W₂ = 0

         N = W₁ + W₂

         N = (m₁ + m₂) g

Since the beams are the same, it has the same mass

        N = 2 m g

We replace

           μ₁ 2mg + μ₂ mg = m a

          a = (2μ₁ + μ₂) g

          a = (2 0.30 + 0.25) 9.8

          a= 8.33 m/s²

Let's look for cable tension with beam 2

          T = m₂ a

          T = 1500 8.33

          T = 12.495 N

b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal

           fr = m₂ a₂

           N- w₂ = 0

          N = W₂ = mg

          μ₂ mg = m a₂

          a = μ₂ g

          a = 0.25 9.8

          a = 2.45 m / s²

4 0
2 years ago
true or false:acceleration toward the center of a curved or circular path is called gravitational acceleration.
nalin [4]
Nope. It's called 'centripetal' acceleration. The force that created it MAY be gravitational, but it doesn't have to be. For things on the surface of the Earth moving in circles, it's never gravity.
5 0
2 years ago
The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
2 years ago
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