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Mumz [18]
3 years ago
12

Help Please

Physics
1 answer:
densk [106]3 years ago
3 0

Answer:

Energy=3.1times 10^-17 J

Rest mass: 6.2 kg

Speed: 47.5 m/s

Wavelength: 2.659 times 10^-6

Momentum: 67.3 kg(m/s)

Explanation:

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Nata [24]
Waves carry energy from one place to another. Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals. ... Some types of waves need to be transmitted through matter, either a solid, liquid or a gas. For example, water waves have to travel in water.
3 0
3 years ago
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PLEASE HELP URGENT!!!!!!!!!!!!!
Nimfa-mama [501]

c i think sorry if wrong

4 0
3 years ago
(ASAP) would it be 125 m/s2 to calculate for her speeding up?
serg [7]

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
2 years ago
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
3 years ago
A<br> В<br> C<br> D<br> At which point will an image be formed?<br> ОА<br> Ов<br> Ос<br> OD
Anettt [7]

Answer:

oa

Explanation:

it may be oa is the right answer for this question

but I don't know properly

8 0
2 years ago
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