Why does it takes the outer planets so long to orbit the Sun? Try to come up with two reasons.

A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is co

kupik [55]

Answer:

A) μ = A.m²

B) z = 0.46m

Explanation:

A) Magnetic dipole moment of a coil is given by; μ = NIA

Where;

N is number of turns of coil

I is current in wire

A is area

We are given

N = 300 turns; I = 4A ; d =5cm = 0.05m

Area = πd²/4 = π(0.05)²/4 = 0.001963

So,

μ = 300 x 4 x 0.001963 = 2.36 A.m².

B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;

B = (μ_o•μ)/(2π•z³)

Let's make z the subject ;

z = [(μ_o•μ)/(2π•B)] ^(⅓)

Where u_o is vacuum permiability with a value of 4π x 10^(-7) H

Also, B = 5 mT = 5 x 10^(-6) T

Thus,

z = [ (4π x 10^(-7)•2.36)/(2π•5 x 10^(-6))]^(⅓)

Solving this gives; z = 0.46m =

3 0

3 years ago

A car is traveling at 100 km/h when the driver sees an accident 80 m ahead and slams on the brakes. what minimum constant decele

Effectus [21]

Assuming the driver starts slamming the brakes immediately, the car moves by uniformly decelerated motion, so we can use the following relationship
$2aS = v_f^2 - v_i^2$ (1)
where
a is the deleceration
S is the distance covered after a time t
$v_f$ is the velocity at time t
$v_i=100 km/h = 27.8 m/s$ is the initial speed of the car

The accident is 80 m ahead of the car, so the minimum deceleration required to avoid the accident is the value of a such that S=80 m and $v_f=0$ (the car should stop exactly at S=80 m to avoid the accident). Using these data, we can solve the equation (1) to find a:
$a=- \frac{v_i^2}{2 S}= -\frac{(27.8 m/s)^2}{2 \cdot 80 m} =-4.83 m/s^2$
And the negative sign means it is a deceleration.

4 0

3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$

Factorising out $10^{24}$

$1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0$

Solving for d in the quadratic equation above:

d = 346 * 10⁶ m

4 0

3 years ago

Please help my test is going on. Please answer both (i) and (ii).

Eduardwww [97]

Answer:

i. 7.5 m

ii. 15000 N

Explanation:

Area under vt graph shows the displacement

so area of triangle ABE= 1/2×15×2

=7.5 m

ii) F=ma

here, m=1000kg

and a=v-u/t

=15-0/1

a=15

F=1000×15=15000N

8 0

3 years ago

A lumberjack (mass = 103 kg) is standing at rest on one end of a floating log (mass = 255 kg) that is also at rest. The lumberja

mina [271]

Answer:

(a) -1.18 m/s

(b) 0.84 m/s

Explanation:

(a)

The total linear momentum before the lumberjack begins to move is zero because all parts of the system are at res

From the law of conservation of momentum

m1v1+m2v2=0 hence m1v1=-m2v2 where m1 is mass of lumberjack, v1 is velocity of lumberjeck, m2 is mass of floating log, v2 is velocity of the floating log.

Substituting M1 for 103 Kg, V1 for 2.93 m/s, M2 for 255 Kg into the above equation we obtain

103Kg*2.93 m/s=-255Kg*V2

V2=-(103 kg*2.93 m/s)/255=-1.183490196 m/s

Hence V2=-1.18 m/s

(b)

For the second log

V(M1+M2)=m1v1 where V is the common velocity

V(103 Kg+255 Kg)=103 Kg*2.93 m/s

V=(103 Kg*2.93 m/s)/(103 Kg+255 Kg)=0.842988827 m/s

V=0.84 m/s

6 0

2 years ago