Una tortuga se desplaza en línea recta 100cm al norte y luego 80cm al este, determine su desplazamiento

If a cat falls off a ledge, the Earth pulls the cat to the ground with the force of gravity. According to Newton's Third Law the

elena55 [62]

Answer:

The answer Is B. 

Explanation:

The cats acceleration is decreased

by air resistance. 

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8 0

2 years ago

Read 2 more answers

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

Alisiya [41]

Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

7 0

2 years ago

Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a

Tema [17]

Note: The diagram referred to in the question is attached here as a file.

Answer:

The magnitude of the magnetic field is $B = \frac{0.071 \mu I}{d}$

Explanation:

The magnetic field can be determined by the relationship:

$B = \frac{\mu I}{2\pi R}$ ...............(1)

Were I is the current flowing through the wires

The distance R from point 1 to m is calculated using the pythagora's theorem

$R = \sqrt{d^{2} + (2d)^{2} }$

$R = \sqrt{5d^{2} } \\R = d\sqrt{5}$

Substituting R into equation (1)

$B = \frac{\mu I}{2\pi d\sqrt{5} }$

$B = \frac{0.071 \mu I}{d}$

3 0

2 years ago

A straight wire carries a current of 3 A which is in the plane of this page, pointed toward the top of the page. A particle of c

Ilia_Sergeevich [38]

Answer:

The magnitude of the magnetic force exerted on the moving charge by the current in the wire is 2.18 x $10^{-8}$ N

The direction of the magnetic force exerted on the moving charge by the current in the wire is radially inward

Explanation:

given information:

current, I = 3 A

$q_{0}$ = +6.5 x $10^{-6}$ C

r = 0.05 m

v = 280 m/s

and direction of the magnetic force exerted on the moving charge by the current in the wire, we can use the following formula:

F = qvB sin θ

where

F = magnetic force (N)

q = electric charge (C)

v = velocity (m/s)

θ = the angle between the velocity and magnetic field

to find B we use

B = μ $_{0}$ I/2πr

μ $_{0}$ = 4π x $10^{-7}$ or 1.26 x $10^{-6}$ N/ $A^{2}$ , thus

B = 4π x $10^{-7}$ x 3 / 2π(0.05)

= 1.2 x $10^{-5}$ T

Now, we can calculate the magnitude force

F = qvB sin θ

θ = 90°, because the speed and magnetic are perpendicular

F = 6.5 x $10^{-6}$ x 280 x 1.2 x $10^{-5}$ sin 90°

= 2.18 x $10^{-8}$ N

Using the hand law, the magnetic direction is radially inward

8 0

2 years ago

You weigh yourself at the top of a high mountain and the scale reads 728 N. If your

SpyIntel [72]

The gravitational strength is equal to $9.71\frac{N}{kg}$

Why?

Since we are given the gravity force acting (728N) and the mass (75Kg), we can calculate the gravitational strength using the following formula:

$gravityforce=mass*gravitationalStrength\\\\gravitationalStrength=\frac{gravityforce}{mass} \\\\gravitationalStrength=\frac{728N}{75kg}=9.71\frac{N}{kg}$

Have a nice day!

4 0

2 years ago