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ruslelena [56]
3 years ago
9

A current-carrying loop of wire lies flat on a horizontal tabletop. When viewed from above, the current moves around the loop in

a counterclockwise sense. For points on the tabletop outside the loop, the magnetic field lines caused by this current
a. circle the loop in a counterclockwise direction.
b. point straight up. point straight down.
c. circle the loop in a clockwise direction.
Physics
1 answer:
Oksana_A [137]3 years ago
6 0

Answer: i dont do physics yet lol

Explanation:

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In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves
11111nata11111 [884]
Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m 
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N 
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1) What is the magnitude of the normal force on the care when it is at the bottom of the circle. 

<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>

<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>

<span>Centripetal force = 7119.55 N </span>


<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>

<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>

<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>

√<span>(gr) </span>

√<span>(9.8 x 11.5) = 10.62 m/s</span>
7 0
3 years ago
A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

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Andreyy89

Answer:

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Explanation:

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