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Which student is doing work?

Vilka [71]

Answer:

d

hope this helps

Explanation:

8 0

3 years ago

Read 2 more answers

A .5 kg toy train car moving forward at 3 m/s collides with and sticks to a .8 kg toy car that is traveling at 2 m/s what is the

Viktor [21]

Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
$m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1}+ m_{2} * v'_{2}$

In case of perfectly inelastic collision v'1 and v'2 are same.

We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x

0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s

4 0

2 years ago

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d

kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0

3 years ago

Continue taking reading until the water starts to be boil

Delicious77 [7]

Answer:

pogggs

Explanation:

4 0

2 years ago