Whats the rounded Atomic mass of iron?

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

Travka [436]

Yes you are correct...I dont think you need to do this? You are getting them all right but now your just wasting points..

But anyways you are correct. ♡♡

5 0

4 years ago

A stiff wire 44.0 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l

Helga [31]

The magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

Explanation:

It is known that the force acting on a current carrying conductor placed in a magnetic field is

$F=BIL sin theta$

Here B is the magnetic field, I is the current flowing through the wire and L is the length of the wire which is given as 44 cm.

Since the wire is bended in the middle at right angle so the length of the two sides of the wire will be 22 cm each. Also one part is lying over z axis and another part lies in the plane of xy in the equation of line y = 2x. So the slope of this wire will be

$\frac{y}{x} =2$

This will be equal to tan θ.

So θ = tan⁻¹ (2) =63.4°

Then, the length of the wire will be written as components of i, j and k.

$L = (-22)k+(22) cos ( 63.4) i+(22) sin (63.4)j$

$L = 0.098 i+0.197 j-0.22k$

Then,

F = I (L × B)

$F = 20.5 ((0.0985 i + 0.197 j -0.22k) * (0.316 i))$

$F = 20.5 (\left[\begin{array}{ccc}i&j&k\\0.098&0.197&-0.22\\0.316&0&0\end{array}\right] )$

$F = 20.5(i(0)-j(0-(-0.22*0.316))+k(0-(0.316*0.197))) = 20.5(-0.069 j-0.062 k)$

$F = -1.415 j-1.271 k$

The magnitude of force on the wire will be

$F = \sqrt{(-1.415)^{2}+(-1.27)^{2} } = \sqrt{3.615}=1.90 N$

And the direction can be found by the tan inverse of the ratio of k component to j component of the force.

$theta = tan-1(\frac{-1.271}{-1.415})= 41.9 degrees$

So the magnitude of force acting on wire will be 1.90 N and the direction of the force acting on the wire will be 41.9 degrees below the negative y axis.

5 0

4 years ago

Riding a bicycle is this kind of long-term memory ______________

Schach [20]

answer

procedural memory ❤

8 0

3 years ago

A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0

3 years ago

Read 2 more answers

A load of 250 kg is hung by a crane's cable. The load is pulled by a horizontal force such that the cable makes a 30 angle to th

Anna [14]

Answer:

1250 $\sqrt{3}$

Explanation:

Let's look at this in a simple manner, because it is.

The crane weights 250Kg. Okay.

Since it is hung, there is the acceleration of gravity being applied on it (10m/s²)

Since F = m * a

F = 250 * 10

F = 2500

Now we know that the downward Force is 2500N.

To find the force that is being applied on that 30° angle, we can multiply our 2500N by cos30°, which happens to be $\frac{\sqrt{3}}{2}$ .

Therefore, the force pulling the box in the cable's direction is:

$\frac{2500\sqrt{3} }{2} = 1250\sqrt{3}$

If you have any questions feel free to comment.

Have a great one and mark brainliest if it helped, please

8 0

2 years ago