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3 years ago

In beachville, the last high tide occurred at 1:00 p.m. the next high tide will happen around _____. 1:25

Physics

2 answers:

Vladimir [108]3 years ago

8 0

At Around 1:25 a.m. it will happen

xxTIMURxx [149]3 years ago

4 0

The answer is 1:25am

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A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?

Aleks04 [339]

25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N

7 0

3 years ago

Read 2 more answers

A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i

marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0

2 years ago

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

$\bar F \cdot \Delta t = m \cdot (v_{2}-v_{1})$ (1)

Where:

$\bar F$ - Average impact force, in newtons.

$\Delta t$ - Duration of the impact, in seconds.

$m$ - Mass of the sledge hammer, in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final velocity, in meters per second.

If we know that $\Delta t = 0.0020\,s$ , $m = 4\,kg$ , $v_{1} = -6\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then we estimate the average impact force is: