If τ=r×F then F.τ is equal

Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the

Ugo [173]

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

5 0

3 years ago

At its widest point, the diameter of a bottlenose dolphin is 0.50 m. Bottlenose dolphins are particularly sleek, having a drag c

fiasKO [112]

Answer:

497.00977 N

3742514.97005

Explanation:

$\rho$ = Density of water = 1000 kg/m³

C = Drag coefficient = 0.09

v = Velocity of dolphin = 7.5 m/s

r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m

A = Area

Drag force

$F_d=\frac{1}{2}\rho CAv^2\\\Rightarrow F_d=\frac{1}{2}\times 1000 \times 0.09(\pi 0.25^2)7.5^2\\\Rightarrow F_d=497.00977\ N$

The drag force on the dolphin's nose is 497.00977 N

at 20°C

$\mu$ = Dynamic viscosity = $1.002\times 10^{-3}\ Pas$

Reynold's Number

$Re=\frac{\rho vd}{\mu}\\\Rightarrow Re=\frac{1000\times 7.5\times 0.5}{1.002\times 10^{-3}}\\\Rightarrow Re=3742514.97005$

The Reynolds number is 3742514.97005

8 0

3 years ago

100 POINTS!! PLEASE DONT ANSWER UNLESS YOU KNOW BOTH ANSWERS PLEASE.

JulijaS [17]

Answer:

1. is B and 2. is A

Explanation:

6 0

2 years ago

Read 2 more answers

When a board with a box on it is slowly tilted to a larger and larger angle, common experience shows that the box will at some p

Savatey [412]

Answer:

c. The coefficient of kinetic friction is less than the coefficient of static friction

Explanation:

When the box finally does break loose. Then the component of the box weight which is parallel to the board weight parallel component, is equal to the $\mu_gn$ .

$w_{box}=\mu_gn$

For the box to acce;erate thee must be non-zero net force acting on the box parallel to the board. Or we can say,

$w_1>f_g\\w_1>\mu_gn$

Therefore the force of kinetic friction must be less than the force of static friction. Thus,

$\mu_g$

3 0

2 years ago

A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m

kenny6666 [7]

Answer:

61.76 N.

Explanation:

Given the mass of the car, m = 1.60 kg.

The speed of the car, v = 12.0 m/s.

The radius of the circle, r = 5 m.

As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.

Let N is the normal force.

So, $N - mg = F_c$

$N-mg=\frac{mv^2}{r}$

Now substitute the given values, we get

$N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}$

$N=15.68+46.08$

N = 61.76 N.

Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.

5 0

2 years ago