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balu736 [363]
3 years ago
8

Someone is whirling a hammer that has a mass of 8.5 kg in the air that is tied to a Chain 1.5 m long in a circle that makes 1 re

volution in 2 seconds a. What is the centripetal acceleration f the hammer B what is the tension of the Chain
Physics
1 answer:
Basile [38]3 years ago
6 0
This is o
Explanation
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In an NMR experiment, the RF source oscillates at 34 MHz and magnetic resonance of the hydrogen atoms in the sample being in- ve
choli [55]

Answer:

B_{int}=-0.015T

Explanation:

From the question we are told that:

RF source oscillation speed \sigma= 34 MHz

The external field Bext =0.78 T.

Pro- ton magnetic moment component \mu=1.41 X 10-26 J/T

Generally the equation for magnitude of B_{int} is mathematically given by

B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}

B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}

B_{int}=0.78-0.7957

B_{int}=-0.015T

6 0
3 years ago
True or False? When a pot of water is heated on a stove, it will begin to bubble. This is a sign that there was a chemical react
Nata [24]

Answer:

False

Explanation:

Boiling water is not a chemical change it is a Physical change

<h3>Hope this helps PLEASE MARK AS BRAINLIEST !!</h3>
4 0
3 years ago
Read 2 more answers
Calculate the pressure on the bottom of a swimming pool 3. 5 m deep. How does the pressure compare with atmospheric pressure, 10
Nostrana [21]

Answer:

  1.343 atm

Explanation:

The mass of water above 1 square meter of swimming pool bottom is ...

  M = (3.5 m)·(1000 kg/m^3) = 3500 kg/m^2

Then the force exerted by the water on the pool bottom is ...

  F = Mg = (3500 kg/m^2)(9.8 m/s^2) = 34300 N/m^2 = 34300 Pa

Compared with atmospheric pressure, this is ...

  34,300/10^5 = 0.343 . . . . atmospheres

Added to the atmospheric pressure on the water's surface, the total pressure on the pool bottom is 1.343 atmospheres.

5 0
4 years ago
A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were f
STatiana [176]

Answer:

 y = 1.19 m  and λ = 8.6036 10⁻⁷ m

Explanation:

This is a slit interference problem, the expression for destructive interference is

          d sin θ = m λ

indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m

          λ = d sin  θ / m

let's calculate

          λ = 4.50 10⁻⁶ sin 35  /3

          λ = 8.6036 10⁻⁷ m

for the separation distance from the central stripe, we use trigonometry

         tan θ= y / L

         y = L tan θ

the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits

         L = 2 -0.30

         L = 1.70 m

           

let's calculate

        y = 1.70 tan 35

        y = 1.19 m

3 0
3 years ago
A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rat
Inessa05 [86]

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s

4 0
3 years ago
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