To determin the density of an empty plastic jug what would you measure

Tyler has learned that potential energy is energy stored. Tyler's teacher asks the students to come up with a demonstration of p

Advocard [28]

D.A child at the top of a slide

7 0

2 years ago

Read 2 more answers

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0

2 years ago

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

2 years ago

A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin

Dovator [93]

Answer:

fcosθ + Fbcosθ =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ =Wtanθ

7 0

2 years ago

A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu

Pepsi [2]

Answer:

$v = 10.89\ m/s$

Explanation:

given,

radius of loop = 12.1 m

to find the minimum speed transverse by the rider to not to fall out upside down

centripetal force = $\dfrac{mv^2}{r}$

gravitational force = m g

computing both the equation]

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{rg}$

$v = \sqrt{12.1 \times 9.8}$

$v = \sqrt{118.58}$

$v = 10.89\ m/s$

5 0

2 years ago