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3 years ago

12

where is the center of the universe located? and any 16 or older guys on here who are down to talk?? i got a pad we can talk on

:) (i'm 17)

1 answer:

bazaltina [42]3 years ago

6 0

Answer:

the center of the universe is in your mom's stomach. why do I keep seeing people wanting to look for a date. the f. u. c. k.?

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How to convert acceleration to velocity.

Nana76 [90]

You can't. Velocity and acceleration measure two different things, so their units are incompatible. It's like asking, "How many meters does this book weigh?"

Maybe you mean "find" acceleration using given velocities, or a velocity function?

4 0

2 years ago

A box is at rest on a ramp at an incline of 22°. The normal force on the box is 538 N.

fomenos

Answer: 580 N

Refer to attached figure.

The angle of inclination is 22 degrees

weight (gravitational force) acts downwards.

Normal force is a contact force which acts perpendicular to the point of contact.

The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.

Gravitational force on an object = mg

The normal force $N= mg cos 22$

$\Rightarrow mg =\frac{N}{cos22}=\frac{538 N}{0.927}=580 N$

8 0

3 years ago

Read 2 more answers

5L=__ML<br> Express your answer in scientific notation<br> 5 x 109<br> 5x 106<br> 5x 103

Sphinxa [80]

Answer:

5000ml is the answer

Explanation:

multiply 5 time 1000

7 0

3 years ago

Read 2 more answers

When beryllium-7 ions (m = 11.65 × 10-27 kg) pass through a mass spectrometer, a uniform magnetic field of 0.205 T curves their

AysviL [449]

Answer:

ratio =0.3075 T

Explanation:

The magnetic field B creates a force on a moving charge such that

$F = qvB$

Now this causes a centripetal acceleration

$F = = mv^2/r$

so

$qvB = mv^2/r$ ...........(i)

$B = mv/(rq)$ ...............(ii)

If accelerating potential V is same and then kinetic energy equals the potential energy difference

$\frac{1}{2} mv^2 = Vq$

$v = \sqrt{(2Vq/m)}$ put these value in equation (ii)

$B = m\frac{\sqrt{(2Vq/m)} }{rq}$

simplifying we get

$B =m \frac{(\sqrt{ 2Vm/q})}{r}$

for same location r will be same in both case

$B_{7} = \frac{ \sqrt{(m_{7})(2V/q) }}{r}$ ..............(iii)

$B_{10} = \frac{ \sqrt{(m_{10})(2V/q) }}{r}$ ..........(iv)

dividing (iv) and (iii) equation we get

$\frac{B_{10}}{B_{7}} = \sqrt{\frac{m_{10}}{{m_7}} }$

${B_{10}} = B_{7} \sqrt{\frac{m_{10}}{{m_7}} }$

$B_{10} = 0.2574T\sqrt{\frac{ (1.663x10^-26}{(1.165x10^-26)}$

so on solving we get

=0.3075 T

5 0

3 years ago

How many significant figures are in 246.32

muminat

Answer:5

Explanation:

Decimal :2

Significant notation :2.4632× 10^2

8 0

3 years ago