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2 years ago

How much is the velocity of a body when it travels 600m in 5 minutes?

Physics

2 answers:

maks197457 [2]2 years ago

4 0

answer

so unit of velocity is m/s

displacement=600m

5minutes should be converted to seconds

5×60=300 seconds

so,

velocity= displacement÷time

= 600m ÷300s

=2m/s or 2ms^-1

mr_godi [17]2 years ago

3 0

Answer:

2 m/s

Explanation:

we have given

distance traveled by body =600m

time taken = 5*60 =300 sec( si unit of time is second)

velocity =?

we know that

velocity =distance traveled/ time taken

=600/300

= 2 m/s

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One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

F - fr = 0

F = fr

fr = 52.1 N

Now we can work with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

N - W₁ -W₂ = 0

N = W₁ + W₂

as the two blocks are identical

N = 2W

X axis

F - fr₁ - fr₂ = 0

F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

fr₁ = 52.1 N

fr₁ = μ N

s the normal in the lower block of twice the friction force is

fr₂ = μ 2N

fr₂ = 2 μ N

fr₂ = 2 fr₁

we substitute

F = fr₁ + 2 fr₁

F = 3 fr₁

F = 3 52.1

F = 156.3 N

7 0

2 years ago

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