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vodomira [7]
2 years ago
10

What is the wavelength of a wave traveling with a wave speed of 1348 m/s

Physics
1 answer:
saw5 [17]2 years ago
5 0

Answer:    f = 4 Hz i think

Explanation:

sorry i dont rember (cant type lol)

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Strontium has how many neutrons
Lilit [14]

Answer:

50

Explanation:

7 0
3 years ago
Read 2 more answers
A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

4 0
3 years ago
When you step inside a warm ski lodge on a cold day, you find your glasses fog. why does this occur?
marin [14]
Answer: Condensation of moisture in the air onto the glasses.

Explanation:
When you come from a cold environment into a warm ski lodge, the glasses are at a much lower temperature than that of the lodge.
Because the air in the room contains moisture (water vapor), it condenses on a surface whose temperature is lower than the dew point of the room. 
5 0
3 years ago
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
N76 [4]

Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

The linear acceleration of a yo-yo, a = 5.7 m/s²

We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.

(a) Tension :

The net force acting on the string is :

ma=mg-T

T=m(g-a)

Putting all the values,

T = 0.08(9.8-5.7)

= 0.328 N

(b) Angular acceleration,

The relation between the angular and linear acceleration is given by :

\alpha =\dfrac{a}{r}\\\\\alpha =\dfrac{5.7}{0.027}\\\\=211.12\ m/s^2

(c) Moment of inertia :

The net torque acting on it is, \tau=I\alpha, I is the moment of inertia

Also, \tau=Fr

So,

I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

Hence, this is the required solution.

3 0
3 years ago
A light spring of constant 176 N/m rests vertically on the bottom of a large beaker of water.
irakobra [83]

Answer:

14.0 cm

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.

Sum of forces in the y direction:

∑F = ma

ρVg − mg − k∆L = 0

(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0

∆L = 0.140 m

∆L = 14.0 cm

8 0
3 years ago
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