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tresset_1 [31]
4 years ago
5

While chopping down his father's cherry tree, george discovered that if he swung the axe with a speed of 25 m/s, it would embed

itself 2.3 cm into the tree before coming to a stop. if the axe head had a mass of 2.5 kg, how much force was the axe exerting on the tree?
Physics
1 answer:
wel4 years ago
6 0
<span>First we must find the amount of acceleration needed using the equation Vfinal^2 = Vinitial^2 + 2*Acceleration*distance => 0 = 25^2 + 2*Acceleration*0.025m => Acceleration = 13,587m/s^2. Plug this into Force = Mass*Acceleration equation => Force = 2.5*13,587 = 33,967.5N.</span>
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A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. How far away is the reflecting object? (c
Assoli18 [71]

A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time =  20 μs = 20 * 10^{-6} s

let distance to the reflecting surface be = x

total distance travelled by pulse will be  = 2x

speed = 3.0 × 10^{8} m/s

distance = speed * time

2x = 3.0 × 10^{8} * 20 * 10^{-6}

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The reflecting object would be 3000 m away

To learn more about echo here

brainly.com/question/14861578?referrer=searchResults

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What happens to the temperature of a substance during a phase change?
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8 0
3 years ago
A block of mass M1 = 288.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and
AlekseyPX

Answer:

M2 = 278.06 kg

Explanation:

We calculate the weight of M1

W=m*g

Where

m: mass (kg)

g: acceleration due to gravity (m/s²)

W₁=288* 9.8= 2822.4 N

Look at the attached graphic

We calculate the x-y components of the weight :

W₁x= 2822.4*sin41° N =1851.66 N

W₁y= 2822.4 *cos41° N = 2130.09 N

We apply Newton's first law for the balance in M1:

Σ Fy=0

Fn-W₁y=0  ,   Fn: normal force

Fn=W₁y=2130.09N

Friction Force = Ff=μs *Fn = 0.41*2130.09 =873.34 N

Σ Fx=0

T- W₁x- Ff=0

T= 1851.66 + 873.34

T= 1851.66 + 873.34

T=2725 N

We apply Newton's first law for the balance in M2:

Σ Fy=0

T- W₂ =0

W₂ = T = 2725 N

W₂ = M2*g

M2 = W₂/g

M2 = 2725/9.8

M2 = 278.06 kg

6 0
3 years ago
Why is our (a person's) gravitational pull NOT as strong as the Earth's gravitational pull
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Explanation:

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