9. True or False: Sawmills and gristmills are machines that do not need

Define the word logrolling in your own words?

Elan Coil [88]

A crazy sport thats kinda dangerous

3 0

3 years ago

A soccer player kicks a rock horizontally off a 37 m high cliff into a pool of water. If the player hears the sound of the splas

Alchen [17]

<h2>Answer:</h2>

(a). 16.741 m/s

(b). 15.75 m/s

<h2>Explanation:</h2>

In the question,

Height of the cliff, h = 37 m

Time taken to reach the sound to us = 2.92 s

Speed of the sound in air at room temperature = 343 m/s

Now,

Let us say the speed of the ball = u m/s

So,

Time taken by the ball to reach at the bottom, t is given by,

$t=\sqrt{\frac{2h}{g}}\\t=\sqrt{\frac{2\times 37}{9.8}}\\t=2.747\,s$

So,

Splash is heard after = 2.92 s

So,

Time taken by sound to travel the shortest distance along the hypotenuse of the triangle thus formed is,

t = 2.92 - 2.747

t = 0.172 s

Now,

Distance traveled by sound is given by,

$Distance=343\times 0.172\\Distance=59.02\,m$

So,

In the triangle using the Pythagoras Theorem,

Horizontal distance traveled is,

$D=\sqrt{59.02^{2}-37^{2}}\\D=45.99\,m$

So,

Speed of throwing of ball is given by,

$Distance=Speed\times Time\\45.99=u\times 2.747\\u=16.741\,m/s$

Therefore, the speed of the ball = 16.741 m/s.

(b).

If,

Speed of sound = 331 m/s

So,

Distance traveled by sound is,

$Distance=331\times 0.172\\Distance=56.932\,m$

So,

Distance traveled in the horizontal by ball is,

$Distance=\sqrt{56.932^{2}-37^{2}}\\Distance=43.269\,m$

So,

Speed of the ball thrown is given by,

$Speed=\frac{Distance}{Time}\\Speed=\frac{43.269}{2.747}\\Speed=15.75\,m/s$

Therefore, the speed of the ball = 15.75 m/s.

8 0

3 years ago

gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure

Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be ' $\text{P}_2$ '.

As we know that Boyle's law formula states that;

$P_1 \times V_1 = P_2 \times V_2$

where, $P_1$ = original pressure of gas in the container = 310 mm hg

$P_2$ = required new pressure

$V_1$ = volume of gas in the container = 185 ml

$V_2$ = desired new volume of the gas = 74 ml

So, $P_2 = \frac{P_1 \times V_1}{V_2}$

$P_2 = \frac{310 \times 185}{74}$

= 775 mm hg

Hence, the required new pressure is 775 mm hg.

7 0

3 years ago

Which statement correctly describes the location, charge, and mass of the

Alenkinab [10]

The neutrons are inside the nucleus, have no charge, and have mass.

7 0

3 years ago

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.

From the attached figure, the x and y component of forces is given by :

$T_{1x} =-T_1 cos (20)$

$T_{2x} = T_2 cos (40)$

$mg_x = 0$

$T_{1y} = T_1 sin (20)$

$T_{2y} = T_2 sin (40)$

$mg_y= -mg$

Let $R_x$ and $R_y$ is the resultant in x and y direction.

$R_x=-T_1 cos (20)+T_2 cos (40)+0$

$R_y=T_1 sin(20)+T_2 sin(40)-mg$

As the system is balanced the net force acting on it is 0. So,

$-T_1 cos (20)+T_2 cos (40)+0=0$ .............(1)

$T_1 sin(20)+T_2 sin(40)-100\times 9.8=0$ ..................(2)

On solving equation (1) and (2) we get:

$T_1=866.86\ N$ (tension on the left rope)

$T_2=1063.36\ N$ (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.

7 0

3 years ago