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3 years ago

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A wire with a current of 5.2 A is at an angle of 45° relative to a magnetic field of 0.87 T. What is the force exerted on a 1.2

m length of the wire

1 answer:

julsineya [31]3 years ago

7 0

Answer:

F = I L X B = I L B sin theta

F = 5.2 * 1.2 * .87 sin 45 = 3.84 N in the direction of (L X B)

Note: L and B must be at right angles for F = I L B

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A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 260 kg · m2 and is rotating at 12.0 rev/min about a

MrRa [10]

Answer:

The new angular speed of the merry-go-round is 8.31 rev/min.

Explanation:

Because the merry-go-round is rotating about a frictionless axis there’re not external torques if we consider the system merry-go-round and child. Due that we can apply conservation fo angular momentum that states initial angular momentum (Li) should be equal final angular momentum (Lf):

$L_f=L_i$ (1)

The initial angular momentum is just the angular momentum of the merry-go-round (Lmi) that because it's a rigid body is defined as:

$L_i=L_{mi}=I\omega_i$ (2)

with I the moment of inertia and ωi the initial angular speed of the merry-go-round

The final angular momentum is the sum of the final angular momentum of the merry-go-round plus the final angular momentum of the child (Lcf):

$L_f=L_{mf}+L{cf}=I\omega_f+L{cf}$ (3)

The angular momentum of the child should be modeled as the angular momentum of a punctual particle moving around an axis of rotation, this is:

$L{cf}=mRv_f$ (4)

with m the mass of the child, R the distance from the axis of rotation and vf is final tangential speed, tangential speed is:

$v_f=\omega_f R$ (5)

(note that the angular speed is the same as the merry-go-round)

using (5) on (4), and (4) on (3):

$L_f=I\omega_f+m\omega_f R^2$ (6)

By (5) and (2) on (1):

$I\omega_f+m\omega_f R^2=I\omega_i$

Solving for ωf (12.0 rev/min = 1.26 rad/s):

$\omega_f= \frac{I\omega_i}{]I+mR^2}=\frac{(260)(1.26)}{260+(24.0)(2.20)^2}$

$\omega_f=0.87\frac{rad}{s}=8.31 \frac{rev}{min}$

8 0

3 years ago

A 1500 kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h

Mandarinka [93]

Answer:

Incomplete question check attachment for diagram

Explanation:

Given that,

Mass of car

M = 1500kg

Enter curve at Point A with speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car was slow down at a constant rate till it gets to point C at speed of

Vc = 50km/r = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as given in the attachment is

S=200m

We want to find the total horizontal forces at point A, B and C exerted by the road on the tire

The constant tangential acceleration can be calculated using equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative because the car was decelerating

Since the car is slowing down at a constant rate, the tangential acceleration is equal at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can calculate the normal component of acceleration(centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration is calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is point of inflection

Then,

an = Vb²/p = Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Since tangential acceleration is constant, then, this is the tangential force at each point A, B and C

Now, normal force

Point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Then, the horizontal force acting at each point is

Using Vector of right angle triangle

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

6 0

3 years ago

An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f

IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

$v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}$

Similarly, The equation of mption:

$x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})$

replacing our assumed values:

where $v_=0 \ and \ g= 9.81$

$x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})$

$x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m$

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

$\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}$

$1000= 0.981t-0.981(1-e^{-(10)t}) \ m$

By diregarding $e^{-(10)t} \ m$

$1000= 0.981t-0.981$

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

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3 years ago

olga nikolaevna [1]

The answer is B. Weak electrolytes are weak because they only partially dissociate into ions in water.

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3 years ago

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You have DNA samples from five different suspects. You know that all humans have very similar DNA. What feature of DNA allows yo

Aneli [31]

Answer:

The answer is positively B.

There are so many base pairs that there are individual differences.

Confirmed by my Forensics test today.

Explanation:

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3 years ago