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2 years ago

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:

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