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2 years ago

1. Which object is farthest from the origin at t=2sec

Physics

1 answer:

Stolb23 [73]2 years ago

6 0

Answer:

that one i know only pe not that sorry again

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A pump is used to transport water to a higher reservoir. if the water temperature is 15ºc, determine the lowest pressure that ca

Reika [66]

Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called cavitation.

So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at $15^{\circ}$ is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.

7 0

3 years ago

Newton's laws of motion are true on earth and space.

frez [133]

Answer:

False

Explanation:

It is a common misunderstanding that objects in space have no weight. If that were true, they would just float away from the Earth, the Sun and the other planets. Objects in low Earth orbit experience about 90% of the weight that they feel on the surface of the Earth.

8 0

2 years ago

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

goldfiish [28.3K]

Answer

The combined displacement is 2km north

Explanation

Since displacement is a vector quantity, we take into account the direction.

Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.

We now add to obtain,

$4+-2+5+-5$

This will simplify to

$=4-2+5-5=2$

Therefore the combined displacement is 2km north

5 0

3 years ago

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin

JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

$= 343[\dfrac{508}{490}-1]$

$v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

$=\dfrac{12.6^2}{2\times 9.8}$

=8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

$= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

=0.293 m

total distance

= 8.1 + 0.293 = 8.392 m

3 0

3 years ago

The burner on a stove is 325° f. given that the burner emits electromagnetic radiation as a blackbody, what is the maximum wavel

lesantik [10]

To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.

C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C

C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K

λmax = 2898/435.78 = 6.65 μm

5 0

3 years ago