What is the strongest interparticle force is each substance ch3cl?

Calculate the distance in km that Charlie runs if he maintains the average speed from question 2 for 1 hour

egoroff_w [7]

Sorry, but what is question 2?

5 0

3 years ago

A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 5500

koban [17]

Idk I just need other answers

4 0

3 years ago

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Rudik [331]

Answer:

a) 0.41 m/s

b) 0.51 m/s

Explanation:

(a)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'

V' = 0.41 m/s

b)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = - 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (- 15.4) + (93.1 - 0.292) V'

V' = 0.51 m/s

6 0

3 years ago

The length of daylight on the moon is about __________.

Lyrx [107]

<span>The length of daylight on the moon is about 29.5 days.</span>

From the earliest days, the Moon has been there in the Solar System and there has never been a period when we couldn't gaze upward in the night sky and either observe the Moon hanging there, or realize that it would be back the precise one night from now (i.e. a New Moon).

A day on the Moon keeps going as long as 29.5 Earth days. We can say that it would take 29.5 days for the Sun to move the distance over the sky and come back to its unique position once more.

3 0

3 years ago

Read 2 more answers

A rocket is launched at an angle of 39◦ above the horizontal with an initial speed of 90 m/s. It moves for 7 s along its initial

vagabundo [1.1K]

Answer:

Y=1370.23m

Explanation:

The motion have two moments the first one the time the initial velocity is accelerating then when the engines proceeds to move as a projectile

$a=19 \frac{m}{s^{2} } \\voy=vo*sin(\alpha )\\voy=90*sin(39 )\\y_{o}=0m\\y_{f}=y_{o}+v_{oy}*t+\frac{1}{2}*a*t^{2}\\y_{f}=90*sin(39)*7s+\frac{1}{2}*19\frac{m}{s^{2} }*(7)^{2}\\y_{f}=861.97m$