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3 years ago

8

A crew team rows a boat at a rate of 20 km/h in still water. In practice on a river, the team rows for 30 minutes up the river (

against the current), and then for 30 minutes down the river (with the current). The speed of the river current is 1.5 km/h. How much farther did they travel in the second 30 minutes?

1 answer:

sweet-ann [11.9K]3 years ago

6 0

In the second 30 mins, the speed should be 20 + 1.5 = 21.5 km/h

So S = 21.5 * 30/60 = 10.75 km

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I need the answer right now pls help me

yaroslaw [1]

The AREA of the shaded region is the moving object's displacement.

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3 years ago

What is the best way to support your hypothesis

Diano4ka-milaya [45]

Answer:

Making a Hypothesis

Explanation:

-Research the subject of your question. Review the literature and find out as much as you can about previous information and discoveries surrounding your question.

-Develop an educated guess that answers your initial question. This is your hypothesis. Make a prediction based on your hypothesis and state it as a cause-effect relationship.

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3 years ago

What average net force is required to stop a 1950 kg car in 10.5 s if it’s initially traveling at 28m/s

nika2105 [10]

Answer:

<em>An average net force of 5200 N is needed to stop the car</em>

Explanation:

<u>Cinematics and Dynamics</u>

Cinematics describes the variables involved in the movement without dealing with its causes. There are four main concepts in cinematics: Velocity (or its scalar equivalent, the speed), acceleration, time, and displacement (or the scalar equivalent, distance).

The acceleration can be calculated by:

$\displaystyle a=\frac{v_f-v_o}{t}$

The initial speed is vo=28 m/s, it stops (vf=0) in t=10.5 seconds, thus the acceleration is:

$\displaystyle a=\frac{0-28}{10.5}$

$a = -2.67~m/s^2$

The acceleration is negative because the car loses speed.

Knowing the mass of the car m=1950 Kg, we can calculate the net force required to stop the car by using the formula:

F = m.a =1950*2.67

F = 5200 N

An average net force of 5200 N is needed to stop the car

7 0

2 years ago

Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally

Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0

east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero

distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m

3 0

3 years ago

7) A long straight wire of radius 1.00x10-3 m (1 mm) carries uniform current density J such that the magnetic field at the edge

Vedmedyk [2.9K]

Answer:A)0.5 and 2 mm

Explanation:

Given

radius R of wire is 1 mm

magnetic Field at edge(surface) $=10^{-5} T$

Magnetic Field at a distance r' from Center is $B'=5\times 10^{-6} T$

and we know

$B=\frac{\mu _0I}{2\pi r}$

where $I= current$

$\mu _o=$ Permeability of free space

$r=$ distance from center

For $r=R$

$10^{-5}=\frac{\mu _0I}{2\pi R}$ ---1

For $r=r'$

$5\times 10^{-6}=\frac{\mu _0I}{2\pi r'}$ ---2

Divide 1 and 2

$\frac{10^{-5}}{0.5\times 10^{-5}}=\frac{r'}{R}$

$r'=2 R=2 mm$

If r is inside the wire then

$B=\frac{\mu _0rI}{2\pi R^2}$

for $r=R$

$10^{-5}=\frac{\mu _0R\cdot I}{2\pi R^2}$ ---3

for $r=r"$

$5\times 10^{-6}=\frac{\mu _0r"I}{2\pi R^2}$ ----4

Divide 3 and 4

$2=\frac{R}{r"}$

$r"=0.5 mm$

3 0

2 years ago