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Tatiana [17]
3 years ago
8

How does increasing the tension of a spring affect a wave on the spring?

Physics
2 answers:
Liono4ka [1.6K]3 years ago
8 0
More energy is being used in the movement

Luden [163]3 years ago
7 0

For plato the answer is, a. wave velocity increases

posting this for future users!

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Question #2: Where does hope come from? Just you? The<br> people around you? Explain.
Tanya [424]

Answer:

it depends ether people can give you hope or you can have hope

Explanation:

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3 years ago
In longitudinal waves the places where the coils are bunched together are called *
Firdavs [7]
In longitudinal waves the places where the coils are bunched together are called *
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3 years ago
How can I answer question 19? please help me, I need the answer now
Crank
The spring starts out 22 cm long with nothing hanging on it.

Hanging 35 newtons of weight on it stretches the spring 1 meter.

Ellen is going to hang 250 grams of mass on the spring.
What's the weight of 250 grams of mass ?

   Weight = (mass) x (acceleration of gravity in the place where the mass is) .

On Earth, the acceleration of gravity is 9.8 m/s² .
250 grams is 0.25 of a kilogram.

Weight of 250 grams  =  (0.25 kilogram) x (9.8 m/s²)

                                 =    (0.25 x 9.8)  kg-m/s²

                                 =        2.45 newtons .

2.45 newtons of weight is  (2.45 / 35) of 35 newtons,
so it'll stretch the spring  (2.45 / 35) of a meter.

   2.45/35 = 0.07 of a meter  =  7 centimeters.

The spring was 22 cm long with nothing hanging on it,
and the 250-gm weight stretched it 7 cm.
So with the weight hanging on it, it's (22 + 7) = 29 cm long.

6 0
3 years ago
Help pls it’s urgent giving brainiest!!
Lorico [155]

Answer:

2nd no. positive.... 3rd no. negitive......

3 0
3 years ago
A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal
Natalija [7]

Answer : The specific heat of unknown sample is, 8748.78J/kg^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-[q_2+q_3]

m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]

where,

c_1 = specific heat of unknown sample = ?

c_2 = specific heat of water = 4186J/kg^oC

c_3 = specific heat of copper = 390J/kg^oC

m_1 = mass of unknown sample = 72.0 g  = 0.072 kg

m_2 = mass of water = 203 g  = 0.203 kg

m_2 = mass of copper = 187 g  = 0.187 kg

T_f = final temperature of calorimeter = 39.4^oC

T_1 = initial temperature of unknown sample = 80.0^oC

T_2 = initial temperature of water and copper = 11.0^oC

Now put all the given values in the above formula, we get

0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]

c_1=8748.78J/kg^oC

Therefore, the specific heat of unknown sample is, 8748.78J/kg^oC

7 0
3 years ago
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