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Simora [160]
1 year ago
15

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

you and all talking equally loud at the same time. The threshold of hearing is 1.0 × 10-12 W/m2. What sound level are you being exposed to? You can neglect any absorption, reflection, or interference of the sound. The threshold of hearing is 1.0 × 10-12 W/m2.
Physics
1 answer:
Hunter-Best [27]1 year ago
7 0

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

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To make the jump, Neo and Morpheus have pushed against their respective launch points with their legs applying a _____ to the la
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Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
7 0
2 years ago
A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.
Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

m = 100 * 11.34 = 1134 g

m = 1.134 kg

so its weight in air will be given as

W = mg = 1.134* 9.8 = 11.11 N

now the buoyant force on the lead is given by

F_B = W - F_{net}

F_B = 11.11 - 11 = 0.11 N

now as we know that

F_B = \rho V g

0.11 = 1000* V * 9.8

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water


#2

buoyant force on the lead block is balancing the weight of it

F_B = W

\rho V g = W

13* 10^3 * V * 9.8 = 11.11

V = 87.2 cm^3

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight =  11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury


#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0
1 year ago
A wheel rotates without friction about a stationary horizontal axis at the center of the wheel. A constant tangential force equa
love history [14]

Answer:

I = 16 kg*m²

Explanation:

Newton's second law for rotation

τ = I * α   Formula  (1)

where:

τ : It is the moment applied to the body.  (Nxm)

I :  it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)

α : It is angular acceleration. (rad/s²)

Kinematics of the wheel

Equation of circular motion uniformly accelerated :  

ωf = ω₀+ α*t  Formula (2)

Where:  

α : Angular acceleration (rad/s²)  

ω₀ : Initial angular speed ( rad/s)  

ωf : Final angular speed ( rad

t : time interval (rad)

Data  

ω₀ = 0

ωf = 1.2 rad/s

t = 2 s

Angular acceleration of the wheel  

We replace data in the formula (2):  

ωf = ω₀+ α*t

1.2= 0+ α*(2)

α*(2) = 1.2

α = 1.2 / 2

α = 0.6 rad/s²

Magnitude of the net torque (τ )

τ = F *R

Where:

F  = tangential force (N)

R  = radio (m)

τ = 80 N *0.12 m

τ = 9.6 N *m

Rotational inertia of the wheel

We replace data in the formula (1):

τ = I * α

9.6 = I *(0.6 )

I = 9.6 / (0.6 )

I = 16 kg*m²

8 0
1 year ago
An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect
melisa1 [442]

Answer:

The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

Distance d' =15 cm

Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

W=\Delta K.E

W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2

Here, final velocity is zero

W=0-\dfrac{1}{2}mv_{i}^2...(I)

We know that,

W=-Fd

W=-E\times e\times d

W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d...(II)

From equation (I) and (II)

-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d

Charge is negative for electron

\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}

Put the value into the formula

\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}

\lambda=-8.056\times10^{-9}\ C/m^2

Hence, The plate's surface charge density is -8.056\times10^{-9}\ C/m^2

3 0
2 years ago
A singly-charged ion enters a velocity selector that has a 0.18-T magnetic field perpendicular to an electric field of 1.6 kV/m,
lesya692 [45]

Answer: a) 8.9km/s b) 4.05 x 10^-25 kg

Explanation:

a)

To find the velocity, 

V = E/B plug in the numbers we get

V = 1.6kV/m / 0.18 T = 8.9 km/s 



b)

To find the ion mass, 

m = erB^2/E = 0.125 m x 1.6x10^-19 x (0.18)^2 / ( 1.6 x 10 ^ 3 V/m ) = 4.05 x 10 ^ -25 kg

8 0
2 years ago
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