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Monica [59]
4 years ago
14

The speed of a sound wave in air is 343m/s. If the density of the air is 1.2kg/m3, find the bulk modulus.

Physics
1 answer:
Nikolay [14]4 years ago
4 0

Answer:

141178.8

Explanation:

use : density x velocity²

1.2 x 343² = 141178.8 pa

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The graph represents the reaction 2H2 + 02 32H20 as it reaches
Alex

Answer:

C and D

Explanation:

5 0
3 years ago
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An object is thrown 16m/s straight up from a 7m tall cliff. How much time does it take to hit the ground below
harkovskaia [24]

Answer:

0,54 sec

Explanation:

t=s/v

t=7/13 s=0,54sec

5 0
4 years ago
A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
. Now assuming Anna's far point was found to be 0.9 m (i.e., her eyes can't focus on any object more than 0.9 m away), what powe
Mama L [17]

Answer:1). Distance of far point x=0.9m

Therefore, since the image is virtual

-f=-x = -0.9m

Power of the concave lenses = 1/f = 1/-0.9

= -1.11D

2 ) near point is 21cm = 0.21m

Power = 4-1/near point

= 4/0.21

= 14.2D.

7 0
4 years ago
Voltage needed to raise current to 3.75a using 20,20,200 resistor set
Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

We are given:

R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega

Using above equation, we get:

\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega

Calculating the voltage by using Ohm's law:

V=IR         .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = 9.52\Omega

Putting values in equation 1, we get:

V=3.75\times 9.52\\\\V=35.7V

Hence, the voltage needed is 35.7 V

7 0
3 years ago
Read 2 more answers
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