Block 1, of mass m1

At a department store, you adjust the mirrors in the dressing room so that they are parallel and 6.2 ft apart. You stand 1.8 ft

barxatty [35]

Answer:

a) 3.6 ft

b) 12.4 ft

Explanation:

Distance between mirrors = 6.2 ft

difference from from the mirror you face = 1.8 ft

a) you stand 1.8 ft in front of the mirror you face.

According to plane mirror rules, the image formed is the same distance inside the mirror surface as the distance of the object (you) from the mirror surface. From this,

your distance from your first "front" image = 1.8 ft + 1.8 ft = 3.6 ft

b) The mirror behind you is 6.2 - 1.8 = 4.4 ft behind you.

the back mirror will be reflected 3.6 + 4.4 = 8 ft into the front mirror,

the first image of your back will be 4.4 ft into the back mirror,

therefore your distance from your first "back" image = 8 + 4.4 = 12.4 ft

8 0

3 years ago

Which answer choice describes the noble gases

Alexxx [7]

Answer:

b

Explanation:

7 0

3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 11.0 s. At th

sleet_krkn [62]

Answer

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

11 α = 6 - 0

= 0.545 rev/s²

The angular displacement

θ₁= ωi t + (1/2) α t²

θ₁= 0 + (1/2) (0.545)(11)^2

θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

14 α = 0 - 6

= - 0.428 rev/s²

The angular displacement

θ₂= ωi t + (1/2) α t²

θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

θ₂= 42 rev

total revolution in 25 s is equal to

θ = θ₁ + θ₂

θ = 33 + 42

θ = 75 rev

3 0

3 years ago

A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring project

Gwar [14]

Answer:

$x' = 10 x$

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

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

so we will have

$v = \sqrt{\frac{k}{m}}(x)$

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

$5v = \sqrt{\frac{k}{4m}}(x')$

now from above two equations we have

$\frac{5v}{v} = \frac{x'}{2x}$

so we have

$x' = 10 x$

3 0

3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago