What is the net worth of a car moving 30 miles per hour

A dust particle floats in front of a silent loudspeaker as shown in the figure. The loudspeaker is turned on and plays a constan

dybincka [34]

Answer:tbh idk

Explanation:

tbh idk

8 0

3 years ago

Read 2 more answers

What is CHA-CHA-CHA.

polet [3.4K]

Answer:

The cha-cha-cha, is a dance of Cuban origin. It is danced to the music of the same name introduced by Cuban composer and violinist Enrique Jorrin in the early 1950s. This rhythm was developed from the danzón-mambo

5 0

2 years ago

A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin

ollegr [7]

Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
$pV=k$
which can also be rewritten as
$p_1 V_1 = p_2 V_2$ (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, $p_1 = 1.5 atm$ , $V_1 =6.5 L$ and $V_2 =13.0 L$ , so the final pressure of the gas can be found by re-arranging eq.(1):
$p_2 = p_1 \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm$

Therefore the correct answer is
<span>1. 0.75 atm</span>

8 0

3 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

love history [14]

Answer:

13.51 nm

Explanation:

To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians

y/L=tan θ ≈ θ

and ∆θ ≈∆y/L

Where ∆y= wavelength distance= 2.92 mm =0.00292m

L=screen distance= 2.40 m

=0.00292m/2.40m

=0.001217 rad

The grating spacing is d = (90000 lines/m)^−1

=1.11 × 10−5 m.

the small-angle

approx. Using difraction formula with m = 1 gives:

mλ = d sin θ ≈ dθ →

∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad

=0.000000001351m

= 13.51 nm

6 0

3 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

4 0

3 years ago