Ask question

Ask question

All categories

mrs_skeptik [129]

3 years ago

9

A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter st

ick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively.
A. 57.6 cm
B. 57.4 cm
C. 57.2 cm
D. 57.8 cm

1 answer:

devlian [24]3 years ago

4 0

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

You might be interested in

If a machine has an efficiency of 50% and an input of 3 J, what is the output?

uranmaximum [27]

Answer:

150J

Explanation:

work output/work input=100%

so just make work output the subject

6 0

3 years ago

A container with volume 1.83 L is initially evacuated. Then it is filled with 0.246 g of N2. Assume that the pressure of the gas

alisha [4.7K]

Answer:

The pressure is $P = 1652 \ Pa$

Explanation:

From the question we are told that

The volume of the container is $V = 1.83 \ L = 1.83 *10^{-3 } \ m^3$

The mass of $N_2$ is $m_n = 0.246 \ g = 0.246 *10^{-3} \ kg$

The root-mean-square velocity is $v = 192 \ m/s$

The root -mean square velocity is mathematically represented as

$v = \sqrt{ \frac{3 RT}{M_n } }$

Now the ideal gas law is mathematically represented as

$PV = nRT$

=> $RT = \frac{PV}{n }$

Where n is the number of moles which is mathematically represented as

$n = \frac{ m_n }{M }$

Where M is the molar mass of $N_2$

So

$RT = \frac{PVM_n }{m _n }$

=> $v = \sqrt{ \frac{3 \frac{P* V * M_n }{m_n } }{M_n } }$

=> $v = \sqrt{ \frac{ 3 * P* V }{m_n } } }$

=> $P = \frac{v^2 * m_n}{3 * V }$

substituting values

=> $P = \frac{( 192)^2 * 0.246 *10^{-3}}{3 * 1.83 *10^{-3} }$

=> $P = 1652 \ Pa$

3 0

3 years ago

Which type of energy is a result of motion

Stells [14]

Kinetic energy I think.

6 0

4 years ago

Read 2 more answers

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$

So, the y-component of the resultant force is

$F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N$

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}$

Where in this problem, we have:

$F_x=-7.14 N$ is the x-component

$F_y=2.70 N$ is the y-component

And substituting, we find:

$F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N$

6 0

4 years ago

Read 2 more answers

What causes friction between two solid forces? After the roughness of the surfaces, what most affects the frictional force exper

Sveta_85 [38]

Friction is caused by two solids rubbing together quickly.

7 0

3 years ago