3. Which of the following is the correctly abbreviated SI unit describing the mass of an object?

How do u work out b)<br> ii)

Yakvenalex [24]

Explanation:

potential difference = current × risstance

see part a point3 current is same

v= IR

v = 8×0.2

v= 1.6

see part a point 5

this potential difference is less than cell

7 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

6 0

3 years ago

In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2 (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

4 years ago

Pushing on the pump of a bottle of hand washing solution compresses a small spring which obeys Hooke's Law. If the potential ene

OLEGan [10]

U = 1/2 * k * x²

U = 0.003J
x = 0.0055m
k spring constant

6 0

3 years ago

Why would a dewar flask not need a lid?

snow_tiger [21]

Answer:

because it does ent

Explanation:

5 0

3 years ago