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Vladimir79 [104]
2 years ago
5

A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the compo

nent of the box’s weight perpendicular to the incline?
(1)19 N
(2)21 N
(3)42 N
(4)46 N
Physics
1 answer:
saul85 [17]2 years ago
8 0

The magnitude of the component of the box’s weight perpendicular to the incline can be olve using the formula:

F = wcos(a)

Where F is the box’s weight perpendicular to the incline

W is the weight of the box

A is the angle of the incline

F = (46)cos(25)

F = 42 N

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A very long, straight wire has charge per unit length 3.50×10^−10 C/m . At what distance from the wire is the electricfield magn
Dafna11 [192]

Answer:

r= 2.17 m

Explanation:

Conceptual Analysis:

The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:

E= 2k*(λ/r) Formula (1)

Where:

E: electric field .( N/C)

k: Coulomb electric constant. (N*m²/C²)

λ: linear charge density. (C/m)

r : distance from the charge line to the surface where E calculates (m)

Known data

E= 2.9  N/C

λ = 3.5*10⁻¹⁰ C/m

k= 8.99 *10⁹ N*m²/C²

Problem development

We replace data in the formula (1):

E= 2*k*(λ/r)

2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)

r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)

r= 2.17 m

5 0
2 years ago
Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent
Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

6 0
2 years ago
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You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa
anzhelika [568]

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

5 0
2 years ago
Two identical boxes are being pulled across a horizontal floor at a constant velocity by a horizontal pulling force of 176 N tha
MAXImum [283]

Answer:

the tension in the rope between the boxes is equal to 88 N

Explanation:

given,

the force applied on one body F = 176 N

When two bodies are moving on horizontal plane at constant velocity then their kinetic friction (f k) is equal to applied force F

According to newton third law the resultant force acting on one body is equal to the resultant force acting on the another body.

T is the tension in the rope

T- f_k = - (T- f_k)

T - F = - (T - F)

T - 176 = - (T - 0)

2 T = 176

T = 176/2 = 88 N

so, the tension in the rope between the boxes is equal to 88 N

3 0
2 years ago
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A chair of mass 30.0 kg is at rest on a horizontal floor. The floor is not frictionless. You push on the chair with a force of 8
miv72 [106K]
First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.

To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N

Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.

Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.

Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N

You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc
7 0
1 year ago
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