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Anon25 [30]
3 years ago
10

Under which conditions are particles in a medium said to be in phase with one another?

Physics
1 answer:
rewona [7]3 years ago
7 0

Answer:

Corvette Corvette

Explanation:

teehee

The answer is A btw :)

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Please help me I have to send for my teacher​
expeople1 [14]

Answer:

for the fill in the blanks

1- static

2-kinetic

3-coeffiecient

4-opposite to

5-sin theta

6-cos theta

im not sure however what to do with the top part or if its even part of what you need help with

8 0
3 years ago
An ideal heat engine absorbs heat at 100°C and exhausts heat at 0°C. What is its percent efficiency?
vampirchik [111]

Answer:

26.8 %

Explanation:

TH = 100°C = 100 + 273 = 373 K

Tc = 0°C = 273 K

The efficiency of the heat engine is given by

\eta = 1-\frac{T_{c}}{T_{H}}

\eta = 1-\frac{273}{373}

η = 0.268

η = 26.8 %

Thus, the efficiency of the heat engine is 26.8 %.

6 0
3 years ago
During a football game, a receiver has just caught a pass and is standing still. Before he can move, a tackler, running at a vel
amm1812

Answer:

<h2>122kg</h2>

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

<u>Before collision;</u>

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

<u>After collision</u>

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

5 0
3 years ago
Calcular la velocidad tangencial y la velocidad angular de un movil que describe una circunferencia de 18 centimetros de radio e
Hoochie [10]

Answer:

1.41 m/s, 7.85 rad/s

Explanation:

We can start by calculating the tangential velocity, which is given by:

v=\frac{2\pi r}{T}

where

r = 18 cm = 0.18 m is the radius of the circumference

T = 0.8 s is the period

Substituting,

v=\frac{2\pi(0.18)}{0.8}=1.41 m/s

Now we can also calculate the angular velocity,  which is given by:

\omega=\frac{2\pi}{T}

where again,

T = 0.8 s is the period

Substituting,

\omega=\frac{2\pi}{0.8}=7.85 rad/s

3 0
3 years ago
Which trial’s cart has the greatest momentum at the bottom of the ramp?
Licemer1 [7]
The momentum of each cart is given by:
p=mv
where
m is the mass of the cart
v is its velocity (at the bottom of the ramp)

To answer the problem, let's calculate the momentum of each of the 4 carts:
1) p=(200 kg)(6.5 m/s)=1300 kg m/s
2) p=(220 kg)(5.0 m/s)=1100 kg m/s
3) p=(240 kg)(6.4 m/s)= 1536 kg m/s
4) p=(260 kg)(4.8 m/s)=1248 kg m/s

Therefore, the cart with greatest momentum is cart 3, so the right answer is
<span>- trial 3, because this trial has a large mass and a large velocity</span>
8 0
3 years ago
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