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Anastaziya [24]
3 years ago
13

Mr. Beall paddles his canoe at 8.0 m/s North in a river that flows at 6.0 m/s to the South. What is the magnitude and direction

of the resulting velocity?
a)

14 m/s South


b)

14 m/s North


c)

10 m/s North


d)

2 m/s North
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

2 /s north

Explanation:

Given that,

Velocity due North is 8 m/s and due south is 6 m/s

We need to find the magnitude and the direction of the resulting velocity.

Let North is positive and South is negative. When two velocities are in opposite direction, they adds up. So,

v=8+(-6)\\\\v=2\ m/s

It is positive. So, it is in North direction.

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Explanation:

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4 0
3 years ago
A 21 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar
Georgia [21]

Answer:

a)15 N

b)12.6 N

Explanation:

Given that

Weight of block (wt)= 21 N

μs = 0.80 and μk = 0.60

We know that

Maximum value of static friction given as

Frs = μs m g = μs .wt

by putting the values

Frs= 0.8 x 21 = 16.8 N

Value of kinetic friction

Frk= μk m g = μk .wt

By putting the values

Frk= 0.6 x 21 = 12.6 N

a)

When T = 15 N

Static friction Frs= 16.8 N

Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.

Friction force = 15 N

b)

When T= 35 N

Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N

Friction force = 12.6 N

8 0
3 years ago
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
Technician A says that the device that controls shift points on an automatic transmission is the valve body. Technician B says t
alexira [117]

Answer: Technician B only

Explanation:

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3 0
3 years ago
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