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Andrew [12]
3 years ago
12

1.) What things signal the coming of spring?

Physics
1 answer:
olasank [31]3 years ago
7 0

Answer:

In Explanation

Explanation:

  1. The signs of the coming of spring include firstly the sight of Robins. It is believed that the sight and singing of the Robins is considered arrival of spring by many in the US. Secondly, the winter is not yet over but the blooming of Daffodils first of all flowers is indication of the spring. As the air gets warmer the flowers bud and new leaves start growing on trees since the days also begin to get longer providing much warmth which also an indicator of the spring.
  2. Without the axial tilt of our Earth, firstly, the Sun would have remained above the equator (Equinox) forever but fortunately due to the tilt we get equinoxes twice a year and not forever. Moreover, the temperature and precipitation levels would not vary much and the equator would remain always warm and the poles always cold. Secondly, due to the tilt as the Earth orbits around the Sun and with changes in solstices we get variety of the seasons which would have been impossible otherwise.
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Which is an example of a solution?
vladimir1956 [14]

Answer:

C

Explanation:

plato

6 0
3 years ago
A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
Neko [114]
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
P_{2} = \frac{ P_{1}  T_{2} }{ T_{1} } = \frac{40*305}{289}  =42.21 psi
3 0
3 years ago
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
Question is in the picture
Svetradugi [14.3K]

the more pressure put on the string, the more frequency and higher pitch.

4 0
4 years ago
A mass weight of 120N is hung from two strings. what is the tension?
kramer
The weight should be shared between the two string equally. Therefore, tension in each string, T is;

T = 120 N/2 = 60 N
7 0
3 years ago
Read 2 more answers
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