Peter often gets sore muscles in the back of his lower legs from jogging what flexibility exercise could help him

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

2 years ago

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A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W

7nadin3 [17]

For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>

7 0

2 years ago

A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this

noname [10]

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

$W=Fdcos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

The displacement of the object is d = 3.00 m, while $\theta=180^{\circ}$ , because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

$W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J$

And the work done is negative, because it is done against the motion of the object.

6 0

2 years ago

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1. A diffraction grating with 5.000 x 103 lines/cm is used to examine the sodium

Nuetrik [128]

Answer:

0.0002°, 0.1691°, 0.338°

Explanation:

Difference between the two line = 5.97 * 10-⁸m

d = 1 / N

N = 5.0 * 10³

d = 2.0 * 10⁴m

nL = Nsin¤

For first order

588.995 * 10-⁹ = 2.0 * 10-⁴ sin ¤

Sin¤ = 2.944*10-³

¤ = sin-¹ 0.002944

¤ = 0.1687°

First order ¤ =

Sin-¹(589.592*-⁹ / 2.0 * 10-⁴)

Sin-¹ (0.002947) = 0.1689°

Angular separation = 0.1689 - 0.1687 = 0.0002°

Second order ¤ = sin-¹ [2 (589.59*10-⁹ / 2.0*10-⁴)] = sin-¹ (0.005895)

Second order ¤ = 0.3378°

Angular difference = 0.3378° - 0.1687° = 0.1691°

Third order ¤ = sin-¹ [3(589.59*10-⁹ /2.0*10-⁴] = 0.5067°

Angular difference = 0.5067° - 0.1687° = 0.338°

7 1

2 years ago

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If in a vernier callipers 10 VSD coincides with 8 MSD then the least count of varnier calliper is

ratelena [41]

Answer:0.2mm

Explanation:

The length of one VSD=8/10=0.8mm

The least count of the instrument is the difference between the length of one MSD and length of one VSD

The length of inebriated MSD=1mm

Therefore,

The least count=1-0.8=0.2mm

3 0

2 years ago

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