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3 years ago

Plzz answer this question correctly

Physics

1 answer:

VikaD [51]3 years ago

3 0

Answer:

the acceleration of A is three times that of B

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two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f

Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

$E = 1.44 \times 10^7 N/C$

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

$E = 2\frac{kq}{r^2}$

now we have

$q = 2\times 10^{-6} C$

$r = 5 cm = 0.05 m$

now we have

$E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}$

$E = 1.44 \times 10^7 N/C$

6 0

3 years ago

Can someone please give me the (Answers) to this? ... please ... I need help….

IceJOKER [234]

#1.

Car 1 weighs 300 kilograms and is moving right at 3 meters per second (m/s)

v1 (before) = 3 m/s

v2 (before) = -1 m/s

v1 (after) = 0.5 m/s

#2.

Law of conservation of momentum

momentum before collorion = momentim after collosion

MV + mv = MV' + mv'

1500x25+ 1000x5

37500 + 15000

6 0

2 years ago

When you set a heavy bag down on the ground, you are doing _______ work on it.

puteri [66]

When you set a heavy bag down on the ground, you are doing negative work on it.

4 0

3 years ago

A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over

kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air $\nu\ of air =19.31*10^{-6} m2/s$

$\rho = 1.048 kg/m3$

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

$Re =\frac{VD}{\nu}$

$= \frac{ 15*0.025}{19.31*10^{-6}}$

= 1.941 *10^4

drag coffecient is given as

$C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}$

solving for f_D

$f_D = C_D A_f*\frac{\rho v^2}{2}$

$=C_D D*\frac{\rho v^2}{2}$

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

$f_D = 1.1*0.025 *\frac{1.048*15^2}{2}$

f_D = =3.24 N/m

4 0

3 years ago

A rocket, blasting into space, accelerates at 2 m/s^2 for 20 seconds. What is its speed at the end of 20 seconds?

OLEGan [10]

Answer:

Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (beginning speed)

= (3,600 mi/hr) - ( 0 )

= 3,600 mi/hr

= same as 1 mile/second.

Acceleration = (1 mi/sec) / (10 sec)

= 0.1 mi/sec² .

But 1 mile = 5,280 ft,

0.1 mi/s² = 528 ft/s²

Explanation:

have a great day

4 0

3 years ago