All categories

3 years ago

A 7 cm thick and 20 cm long wedge is used to pierce a 3cm long log of diameter 20cm what is the velocity ratio of the wedge

Physics

1 answer:

stiks02 [169]3 years ago

6 0

Answer:

The velocity ratio of the wedge is 0.15

Explanation:

Use the following formula to calculate the velocity ratio

Velocity Ratio = Distance Moved by effort / Distance moved by Load

Where

Distance Moved by effort = 3cm

Distance moved by Load = 20 cm

Placing values in the formula

Velocity Ratio = 3cm / 20 cm

Velocity Ratio = 0.15

You might be interested in

I need help with this. Lab: Conservation of Linear Momentum Assignment: Lab Report Urgent!!

kifflom [539]

It won’t let me view it

8 0

3 years ago

Read 2 more answers

What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?

Kobotan [32]

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

7 0

3 years ago

HELP ME PLEASE! according ohm's law if the resistance remains constant what would happen if you decrease the amount of voltage b

SpyIntel [72]

B.
technically it would depend if the resistors were in series or parallel but B is the answer.

7 0

3 years ago

A force of 20 N produces an acceleration of 10 m/s² in mass m1 and an acceleration of 5 m/s² in

Scrat [10]

Explanation:

F = 20N m= m1 a=10m/s²

m=m2 a=5m/s²

F = ma

for the first one: 

f=m1 × a

20 = m1 ×10

20=10m1

m1=20/10

m1=2

for the second one :

f=m2×a

20=m2×5

m2= 20/5

m2= 4

since F=ma

F=(m1+m2) ×a

F =(4+2)×a

F =6×a

F=20(from the question above )

20=6×a

a=20/6

a=3.33

8 0

3 years ago

Read 2 more answers

An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),

Novosadov [1.4K]

Answer:

Explanation:

Given

altitude of the Plane $h=6\ miles$

When Airplane is $s=10\ miles$ away

Distance is changing at the rate of $\frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph$

From diagram we can write as

$h^2+x^2=s^2$

differentiate above equation w.r.t time

$2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}$

as altitude is not changing therefore $\frac{\mathrm{d} h}{\mathrm{d} t}=0$

$0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}$

at $s=10\ miles\ and\ h=6\ miles$

substitute the value we get $x=\sqrt{10^2-6^2}=8\ miles$

$8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290$

$\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph$

5 0

3 years ago