*15 points*

In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40

VARVARA [1.3K]

Answer:

Bulk modulus = 1.35 × $10^{10}$ Pa

Explanation:

given data

density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

we get here bulk modulus of the liquid that is

we know Bulk Modulus = $v^2*\rho$ ...............

here $\rho$ is density i.e 1400 kg/m³

and v is = frequency × wavelength

v = 370 × 8.40 = 3108 m/s

so here bulk modulus will be as

Bulk modulus = 3108² × 1400

Bulk modulus = 1.35 × $10^{10}$ Pa

3 0

3 years ago

A net force is acting on an object. The object is moving in a

Hoochie [10]

Answer:

The object's velocity would increase due to the change in force.

Explanation:

4 0

2 years ago

A small economy car (low mass) and a limousine (high mass) are pushed from rest across a parking lot, equal distances with equal

Studentka2010 [4]

Answer:

The car that receives more kinetic energy is the small economy car.

Explanation:

K.E = 0.5*mv²

Where;

K.E is the kinetic Energy

M is the mass of an object

V is the velocity of the moving object

But F = m(v/t), from Newton's second law of motion

If equal forces were applied to the two cars, then the velocity of each car will be calculated as follows.

v = (Ft/m)

v² = (Ft/m)²

Substitute in the value of v² into Kinetic energy equation

K.E = 0.5*mv²

K.E = 0.5*m(Ft/m)² = (0.5*F²t²)/m

Also assuming equal distance, equal force and assuming equal time for both cars.

The above equation will reduce to, K.E = k/m

Where k = 0.5*F²t², which is equal in both cars.

Thus, Kinetic energy will depend only on the mass of each car.

From the above expression, Kinetic Energy received by each car is inversely proportional to the mass of the car.

A small economy car (low mass) will receive more kinetic energy while a limousine (high mass) car will receive less kinetic energy.

Therefore, the car that receives more kinetic energy is the small economy car.

6 0

3 years ago

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

liq [111]

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

$(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]$

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

$y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]$

With this time we can calculate the horizontal distance:

$x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]$

3 0

3 years ago

A star with the mass (M=2.0×1030kg) and size (R=3.5×108m) of our sun rotates once every 30.0 days. After undergoing gravitationa

Gre4nikov [31]

Answer:

$r=97.22x10^{3} m$

Explanation:

Using the angular formulas can determine the radius using both values neutron star and the the knowing star so

$L=I*w$

$L_{1}=I_{1}*w_{1}=L_{2}=I_{2}*w_{2}$

$I_{1}*w_{1}=I_{2}*w_{2}$

I=Inertia of the star

w=angular velocity

$I=\frac{2*m*r^{2}}{5}$

$w=\frac{2\pi}{t}$

Notice the angular velocity determinate by the time and the Inertia have the radius value so

$\frac{2}{5}*m*r_{sn}^{2}*\frac{2\pi }{t_{1}}=\frac{2}{5}*m*r_{s}^{2}*\frac{2\pi }{t_{2}}$

$r_{sn}^{2}*\frac{1}{t_{1}}=r_{s}^{2}*\frac{1}{t_{2}}$

$r_{sn}^{2}=r_{s}^{2}*\frac{t_{1}}{t_{2}}$

$t_{1}=0.2s\\t_{2}=30day*\frac{24hr}{1day}*\frac{60minute}{1hr}*\frac{60seg}{1minute}=2.592x10^{6}s$

$r_{sn}=3.5x10^{8}m*\sqrt{\frac{0.2s}{2.592x^{6}s}}$

$r_{sn}=97.22x10^{3} m$

8 0

3 years ago