HELP PLEASE (Look at the picture)

If the same pressure is exerted over a greater area will more of less force result? Group of answer choices :

Zigmanuir [339]

Answer:

More force

Explanation:

Pressure and force are related by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area

We can re-arrange the equation as

$F=pA$

In this problem, the pressure is kept the same (p' = p) while the area is increased. As we can see from the previous equation, the force applied is directly proportional to the area: therefore, a greater area means also a greater force.

8 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

So

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago

A person travels by car from Tucson to Phoenix at a constant speed of 75 km/hr. They then return from Phoenix to Tucson at a con

kompoz [17]

Answer:

$v=0$

Explanation:

Knowing that the formula for average velocity is:

$v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}$

Being said that, we know that the person's displacement is zero because it returns to its starting point

$x_{2}=x_{1}$

That means $x_{2}-x_{1}=0$

$v=\frac{0}{t_{2}-t_{1}}=0$

7 0

3 years ago

A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t

jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is $9.8 \mathrm{m} / \mathrm{s}^{2}$

(a) Calculating the work done:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

$\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})$

$\text { Work }=-15.03 \mathrm{J}$

The work done by the man is -15.03J.

(b) Calculating the force:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is $\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

$\mathrm{F}=13 \times 9.8$

F = 127.4N

The force exerted on the object is 127.4N.

4 0

3 years ago

Consider a railroad bridge over a highway. A train passing over the bridge dislodges a loose bolt from the bridge, which proceed

schepotkina [342]

Answer:

12.35m

Explanation:

Hello! To solve this problem we must consider the following:

1. The car moves with constant speed, which means that the distance traveled is equal to the multiplication of time by speed.

X = VT

we solve the equation for time

$T=\frac{x}{V} =\frac{27}{17} =1.588s$

2. The bolt moves with constant acceleration, with acceleration of 9.81m / s ^ 2, so we could apply the following equation.

note=remember that "a uniformly accelerated motion", means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

$Y= VoT-\frac{1}{2}gt^{2}$

where

Vo = Initial speed

=0

T = time

=1.588s

g =gravity=9.8m/s^2

Y = bridge height

solving

$Y= \frac{1}{2}gt^{2}\\Y=(0.5)(9.8)(1.588^2)=12.35m$

4 0

3 years ago