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3 years ago

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A 5.3 kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force

of magnitude 21 N. Find the block's speed after it has moved through a horizontal distance of 6.4 m.

2 answers:

Anni [7]3 years ago

6 0

Whatever the person said

dimulka [17.4K]3 years ago

3 0

Answer:

F= MASS *ACCELERATION

12=6*a

a=2 m/sec^2

v^2-u^2=2*a*s

v^2=2*2*3

v=sqrt(12)

v=3.5 m/sec

work done by force= change in kinetic energy

F*S=0.5*M*U^2-0.5*M*V^2

12*3=0 0.5*6*V^2

V=3.46 M/SEC

Explanation:

#if you need any queshtions answered within secs mins hit me up and I got you!

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A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

The radius of the loop is $r = 6 \ cm = 0.06 \ m$

The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

$B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

Where $\mu$ is the permeability of free space with value $\mu _o = 4\pi * 10^{-7} N/A^2$

$0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=> $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

$I = 6.68 \ A$

3 0

3 years ago

Type the correct answer in the box. Spell the word correctly.

serious [3.7K]

Answer:

The Lambda-CDM model contains a cosmological constant, denoted by a lambda (λ), which is associated with dark energy and <u>cold dark matter</u>.

^Also works for Plato users.

3 0

3 years ago

Read 2 more answers

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

A 1. 18 kg gold cube hangs at the end of a 4. 00 m long string. Rhogold = 19. 3 × 103 kg/m3; rhomercury = 13. 6 × 103 kg/m3. Whe

VashaNatasha [74]

When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.

<h3>What is tension?</h3>

Tension is the force acting on the linear object like string, chain or rope due to pulling.

Volume of gold V = mass / density

V = 1.18 /19.3x 10³ =61.1 x 10⁻⁶ m³

Tension in the string after immersing will be

T = [ρ(Gold) -ρ(Hg)] g. V

T =[ 19.3x 10³ - 13.6 x 10³] x 9.81 x 61.1 x 10⁻⁶

T =3.416 N
Thus, the tension in the string is 3.42 N.

Learn more about tension.

brainly.com/question/4087119

#SPJ4

6 0

2 years ago

Observer 1 rides in a car and drops a ball from rest straight downward, relative to the interior of the car. The car moves horiz

steposvetlana [31]

a) 10.5 m/s

While for observer 1, in motion with the car, the ball falls down straight vertically, according to observer 2, which is at rest, the ball is also moving with a horizontal speed of:

$v_x = 3.80 m/s$

As the ball falls down, it also gains speed along the vertical direction (due to the effect of gravity). The vertical speed is given by

$v_y = u_y + gt$

where

$u_y =0$ is the initial vertical speed

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Therefore, after t = 1.00 s, the vertical speed is

$v_y = 0 + (9.8)(1.00)=9.8 m/s$

And so the speed of the ball, as observed by observer 2 at rest, is given by the resultant of the horizontal and vertical speed:

$v=\sqrt{v_x^2 +v_y^2}=\sqrt{(3.8)^2+(9.8)^2}=10.5 m/s$

b) $\theta = -68.8^{\circ}$

As we discussed in previous part, according to observer 2 the ball is travelling both horizontally and vertically.

The direction of travel of the ball, according to observer 2, is given by

$\theta = tan^{-1} (\frac{v_y}{v_x})=tan^{-1} (\frac{-9.8}{3.8})=-68.8^{\circ}$

We have to understand in which direction is this angle measured. In fact, the car is moving forward, so $v_x$ has forward direction (we can say it is positive if we take forward as positive direction).

Also, the ball is moving downward, so $v_y$ is negative (assuming upward is the positive direction). This means that the direction of the ball is forward-downward, so the angle above is measured as angle below the positive horizontal direction:

$\theta = -68.8^{\circ}$

4 0

3 years ago