Which pair of objects would have the LEAST gravitational attraction to each other?

a spring with a constant of 80N/m is stretched by a force of 240N. how much the displacement of the spring from equilibrium?

Inessa05 [86]

Answer:

1200N/m

Explanation:

given parameters:

force on the motorcycle spring is 240N

Extension 2cm or 0.02m

unknown _

spring constant:

:?

solution:

to a spring a force applied is given as :

f=ke

f is applied as force

k is spring constant

e is the Extension

240= kx0.02

k=1200N/m

8 0

3 years ago

A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 10

anzhelika [568]

Answer:

Explanation:

Given the following :

Speed (V) = speed of 2.30×10^7 m/s

Acceleration (a) = 1.70×10^13 m/s^2

Using the right hand rule provided by Lorentz law:

B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

SinΘ = sin90° = 1

Note F = ma

Therefore,

B = ma / qvSinΘ

Mass of proton = 1.67 × 10^-27

Charge = 1.6 × 10^-19 C

B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

B = 2.839 × 10^-14 / 3.68 × 10^-12

B = 0.7715 × 10^-2

B = 7.72 × 10^-3 T

2) Magnetic field will be in the negative y direction according to the right hand thumb rule.

Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.

5 0

3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago

Which of the following statements is an explanation? A. Aggressive chimpanzees should be monitored during meal times or fed sepa

ratelena [41]

Answer:

C?

Explanation:

My best guess would be C as it's the only answer that gives a reason behind the statement.

5 0

3 years ago

Read 2 more answers

A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

So

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

6 0

3 years ago