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3 years ago

How many "one-sixth parts are needed to make onewhole?

Physics

1 answer:

TEA [102]3 years ago

8 0

6 one-sixth parts are needed to make a whole.

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A power supply is connected to a 59 Ohm resistor and a 53 Ohm resistor in series. The total current is found to be 0.15 A. What

kirill [66]

the answer is 47265.dug

Explanation:

im bot sure

4 0

2 years ago

A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its

True [87]

Answer:

The angular velocity is $w = 1.43\ rad/sec$

Explanation:

From the question we are told that

The mass of wooden gate is $m_g = 4.5 kg$

The length of side is L = 2 m

The mass of the raven is $m_r = 1.2 kg$

The initial speed of the raven is $u_r = 5.0m/s$

The final speed of the raven is $v_r = 1.5 m/s$

From the law of conservation of angular momentum we express this question mathematically as

Total initial angular momentum of both the Raven and the Gate = The Final angular momentum of both the Raven and the Gate

The initial angular momentum of the Raven is $m_r * u_r * \frac{L}{2}$

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is zero

Note: This above is the generally formula for angular momentum of square objects

The final angular velocity of the Raven is $m_r * v_r * \frac{L}{2}$

The final angular velocity of the Gate is $\frac{1}{3} m_g L^2 w$

Substituting this formula

$m_r * u_r * \frac{L}{2} = \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * v_r * \frac{L}{2} - m_r * u_r * \frac{L}{2}$

$\frac{1}{3} m_g L^2 w = m_r * \frac{L}{2} * [u_r - v_r]$

Where $w$ is the angular velocity

Substituting value

$\frac{1}{3} (4.5)(2)^2 w = 1.2 * \frac{2}{2} * [5 - 1.5]$

$6w = 4.2$

$w = \frac{6}{4.2}$

$w = 1.43\ rad/sec$

5 0

2 years ago

Does anyone know this

NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the right

Explanation:

To find the net force, we have to analyze all the forces acting on the box.

We have:

Force to the right: $F_a = 20 N$ , the applied force
Force to the left: $F_f = 4 N$ , the force of friction
Force to the bottom: $F_g = 400 N$ , the weight of the box (the weight is always downward vertically)
Force to the top: $F_N = 400 N$ . This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

$F_x = F_a - F_f = 20 - 4 = 16 N$ (to the right)

While the vertical force is

$F_y = F_N - F_g = 400 - 400 = 0$

So the net force is 16 N to the right.

In this case, we have the following forces:

$F_g = 4 N$ downward, the weight of the ball
$F_a = 100 N$ to the left, the force that kicks the ball
$F_f = 2 N$ to the right, the force of friction
$F_N = 4 N$ upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

$F_x =F_a - F_f = 100 -2 = 98 N$ (to the left)

While the vertical force is

$F_y = F_g - F_N = 4 - 4 = 0$ (downward)

And so, the net force is 98 N to the left.

The force acting on the squirrel in this problem are:

$F_g = 8 N$ downward, the weight of the squirrel
$F_f = 7.5 N$ upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

$F_y = F_g - F_f = 8 - 7.5 = 0.5 N$

And since the weight is larger than the air resistance, the direction of the net force is downward.

The forces acting on Monkey are:

$F_1=95 N$ is the force applied to the right by Bunny
$F_2 = 75 N$ is the force applied by Deer from the left (so, also on the right)
$F_g = 50 N$ is the weight of Monkey, downward
$F_N = 50 N$ is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

$F_x = F_1 + F_2 = 95+75=170 N$ (to the right)

While the net force in the vertical direction is

$F_y = F_N - F_g = 50 - 50 = 0$

And therefore the net force is 170 N to the right

The forces acting on Deer are:

$F_a = 100 N + 100 N = 200 N$ to the right, the combined force applied by Bunny and Monkey
$F_f = 25 N$ to the left, the force of friction
$F_g = 150 N$ downward, the weight of the deer
$F_N = 150 N$ upward, the normal reaction from the surface that balances the weight

So the net horizontal force is

$F_x = F_a - F_f = 200 - 25 = 175 N$ to the right

While the net vertical force is

$F_y = F_N - F_g = 150 - 150 = 0$

So the net force is 175 N to the right.

Learn more about vector addition:

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3 0

2 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

olasank [31]

2) 20.2 m/s

In the first 4.4 seconds of its motion, the blue car accelerates at a rate of

$a=4.6 m/s^2$

So its final velocity after these 4.4 seconds is

$v=u+at$

where

u = 0 is the initial velocity (the car starts from rest)

a is the acceleration

t is the time

Substituting t = 4.4 s, we find

$v=0+(4.6)(4.4)=20.2 m/s$

After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until

$t=4.4 + 8.5 = 12.9 s$

Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.

3) 216.2 m

The distance travelled by the car during the first 4.4 s of the motion is given by

$d_1 = ut_1 + \frac{1}{2}at_1^2$

where

u = 0 is the initial velocity

$t_1 = 4.4 s$ is the time

$a=4.6 m/s^2$ is the acceleration

Substituting,

$d_1 = 0 +\frac{1}{2}(4.6)(4.4)^2=44.5 m$

The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is

$d_2 = vt_2$

where

$v_2 = 20.2 m/s$ is the new velocity

$t_2 = 8.5 s$ is the time

Substituting,

$d_2 = (20.2)(8.5)=171.7 m$

So the total distance travelled before the brakes are applied is

$d=44.5 m+171.7 m=216.2 m$

4) $-6.62 m/s^2$

We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is

$d_3 = 247 m -216.2 m=30.8 m$