Sean applies a force of 100N to move a box 5 meters. How much work did he do?

A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

RoseWind [281]

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

3 0

2 years ago

1. Determine the image distance in each of the following.

miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0

2 years ago

C, N, Ne, Ar which contains a metal, nonmetal, noble gas , metalloid

postnew [5]

Answer:

c,carbono nonmetal

n, nitrogen nonmetal

ne, neón noble gas

ar,argon noble gas

4 0

2 years ago

How much thermal energy is required to raise the temperature of 5 kg of water 50℃? The specific heat of water is 4168 J/kg*C.

sasho [114]

Answer/solution:

Given :

Mass =5kg

T 1 =20 C,T 2 =100 ∘C

ΔT=100−20=80 ∘C

Q=m×C×ΔT

where C= specific heat capacity of water

=4200J/(kgK)

Q=5×4200×80

=1680000 Joule.

=1680KJ

3 0

2 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

2 years ago