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Ad libitum [116K]
3 years ago
8

a rock of mass of 540 g in the air is found to have an apparent mass of 342 g when submerged in water (a) calculate the weight o

f the Rock given that the acceleration of the air g 10 m / s² to ignore the air resistance​
Physics
1 answer:
kupik [55]3 years ago
7 0

Answer: 5.4 N

Explanation:

Given

mass of rock in the air  is 540 g

Apparent mass of 342 gm when submerged in water

Weight of the rock is the product of mass and accleration due to gravity

\Rightarrow W=mg\\\Rightarrow 0.540\times 10=5.4\ N

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We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
Assume that the polymer material has a constant refractive index of 1.5. For light of 600nm wavelength at normal incidence, what
yaroslaw [1]

Answer:

Minimum thickness will be 100 nm

Explanation:

We have given refractive index is n = 1.5

Wavelength of the light incidence \lambda= 600 nm

We have to find the smallest thickness of the film so that there will be minimum light reflect

For minimum thickness of non reflecting film

t=\frac{\lambda }{4n} , here t is thickness, \lambda is wavelength and n is refractive index

Putting all values t=\frac{600}{4\times 1.5}=100nm

So minimum thickness will be 100 nm

8 0
3 years ago
An occupant of a car has a chance of surviving a crash if the deceleration during the crash is not more then 30 g. Calculate the
sweet [91]

Answer:

20,850 N

Explanation:

We can solve the problem by using second Newton's Law:

F=ma

where

F is the force

m is the mass

a is the acceleration

In this problem, we have:

m = 70 kg is the mass

a=-30 g=-30 (9.8 m/s^2)=-294 m/s^2 is the acceleration (which is negative, because it is a deceleration)

So, we can use the equation above to find the force:

F=(70 kg)(-294 m/s^2)=-20580 N

and the negative sign simply means that the force is in the opposite direction to the motion.

5 0
3 years ago
Three pendulums all have the same length and start from the same height. The first pendulum is very light and has a mass of 67 g
vovikov84 [41]

Answer:

All three pendulum will attain same velocity

Explanation:

All three pendulum will attain same velocity irrespective of their mass difference in isolated system (means where air drag are negligible) and at same length

As you know when velocity is calculated we can not take mass into account.  

3 0
3 years ago
A racing car travels on a circular track of radius 158 m, moving with a constant linear speed of 19.1 m/s. Find its angular spee
SOVA2 [1]

Answer:

\omega=0.12\frac{rad}{s}

Explanation:

In a uniform circular motion, since a complete revolution represents 2π radians, the angular velocity, which is defined as the angle rotated by a unit of time, is given by:

\omega=\frac{2\pi}{T}(1)

Here T is the period, that is, the time taken to complete onee revolution:

T=\frac{2\pi r}{v}(2)

Replacing (2) in (1):

\omega=\frac{2\pi}{\frac{2\pi r}{v}}=\frac{v}{r}\\\omega=\frac{19.1\frac{m}{s}}{158m}\\\omega=0.12\frac{rad}{s}

3 0
4 years ago
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