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matrenka [14]
4 years ago
11

Two skaters collide and grab onto each other on frictionless ice. One of them (mass = 70.0 kg) is moving to the right at 4.00 m/

s, while the other (mass = 65.0 kg) is moving to the left at 2.50 m/s. What are the magnitude and direction of the velocity of these skaters just after they collide?

Physics
1 answer:
ahrayia [7]4 years ago
8 0
All working shown on the photo as well as answer

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Uma bola é lançada para cima com a velocidade inicial de 20m/s. A sua aceleração é constante e igual a 10m/s² para baixo. Quanto
Vadim26 [7]

Answer:

sorry but i cannot understand your language, can you speak in english

7 0
3 years ago
Families often require the services of _______ to best meet their family needs or the specific needs of their child. A. public l
KengaRu [80]

Answer: B. Community agencies

Explanation:

Family requires the services of community agencies for meeting their family and child needs. The community agencies provide with food items, dairy items, medicines, stationary stuff, clothes required as daily needs.

7 0
3 years ago
gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0
3 years ago
13. Direct instruction takes place when
jekas [21]
A skill set is explicitly taught or an activity is completed
3 0
4 years ago
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. at time t = 0.00
Masteriza [31]
The cart is moving by simple harmonic motion, and its position at time t is described by
x(t) = A \cos (\omega t)
where
A is the amplitude of the oscillation
\omega is the angular frequency

The amplitude of the oscillation corresponds to the maximum displacement of the spring, which corresponds to the initial position where the spring was released: 
A=0.250 m

The period of the motion is T=5.67 s, and the angular frequency is related to the period by
\omega =  \frac{2 \pi}{T}= \frac{2 \pi}{5.67 s} =1.11 rad/s

Therefore now we can calculate the position of the system at the time t=29.6 s:
x(29.6 s)=(0.250 m)\cos ((1.11 rad/s)(29.6 s))=+0.033 m
3 0
3 years ago
Read 2 more answers
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