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3 years ago

9

A seesaw pivots as shown in below. What is the net torque about the pivot point?

1 answer:

rodikova [14]3 years ago

7 0

Answer:

2.3 Nm clockwise

Explanation:

Take counterclockwise to be positive and clockwise to be negative.

∑τ = (3 N) (2.5 m) − (7 N) (1.4 m)

∑τ = 7.5 Nm − 9.8 Nm

∑τ = -2.3 Nm

The net torque is 2.3 Nm clockwise.

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For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?

Reptile [31]

Answer: Gravitational force and drag force

Explanation:

For a snowboard jumper in the air, two forces would be acting. One in the downward direction- the gravitational pull and second in the opposite direction to the motion, the drag force due to air. If the snowboard jumper jumps in the air at a certain angle with the horizontal. The forces are written as the sum of vertical and horizontal components. Hence, for the modeling the motion, gravitational force and drag force are important,

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2 years ago

If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?

AlekseyPX

The density of the material would be
25/6 grams per cm^3.
to obtain the result above this is what we do:
density is calculated as: (the mass of the given material or object) / volume of the material
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3 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

2 years ago

An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel

Tju [1.3M]

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

S = ut + ½ gt²

∴ g = 2S/t²

Substituting the given values,

g = 2 x 144 /6²

= 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

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3 years ago

A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

SSSSS [86.1K]

I had the same question

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2 years ago