Someone please help me understand this question!

An athlete walks 8 meters East , 2 meters South, 8 meters West, and finally 2 meters North. What is the total distance traveled?

serg [7]

Explanation:

20 meters is travelled by him

please mark me as brainliest

6 0

2 years ago

Read 2 more answers

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s ,

Airida [17]

Answer:

Complete question:

c.If the current in the second coil increases at a rate of 0.365 A/s , what is the magnitude of the induced emf in the first coil?

a. $M= 6.53\times10^{-3} H$

b.flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

Explanation:

a. rate of current changing = $\frac{di}{dt}=[tex]M=\frac{1.60\times10^{-3} V}{0.240\frac{A}{s} }}$ [/tex]

Induced emf in the coil =e= $1.60\times10^{-3} V$

For mutual inductance in which change in flux in one coil induces emf in the second coil given by the farmula based on farady law

$e=-M\frac{di}{dt}$

$M=\frac{e}{\frac{di}{dt} }$

$M=\frac{1.60\times10^{-3} }{-0.245}$

$M= 6.53\times10^{-3} H$

b.

Flux through each turn=?

Current in the first coil =1.25 A

Number of turns = 20

using MI = NФ

flux through each turn = Ф = $\frac{6.53\times10^{-3}\times1.25}{20}$

flux through each turn = Ф = $4.08\times10^{-4} Wb$

c.

second coil increase at a rate = 0.365 A/s

magnitude of the induced emf in the first coil =?

using $e=-M\frac{di_{2} }{dt}$

$e= 6.53\times10^{-3} \times 0.365$

magnitude of the induced emf in the first coil = e= $2.38\times10^{-3} V$

4 0

3 years ago

when astronauts work in space, they are sometimes attached to the spacecraft by a line called an umbilical. Why do you think the

wariber [46]

Probably for the umbilical cord that connects babies (from their early stages in the womb to their removal) to their mothers. The cord is cut, forming the belly button. This is analogous to astronauts in space.

3 0

4 years ago

Read 2 more answers

an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni

stiv31 [10]

The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N

3 0

3 years ago

How much force must be applied on a blade of length 4cm and thickness of 0.1mm to exert a pressure of 4000000pa?

Viktor [21]

Answer:

F= 403429 kpa

Explanation:

Pressure is the product of force and area

Mathematically,

P=F*A -------where F is force and A is area.

A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

7 0

3 years ago