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PilotLPTM [1.2K]
3 years ago
13

A woman wearing athletic clothing outside. The woman is stretching her legs in a lunge position. Based on the nonverbal messages

you see in this photograph, which of the following most likely describes this person’s situation? a. This woman is a successful and confident business owner. b. This woman is an overworked and overly stressed medical doctor. c. This woman is a stay-at-home mom who is very conscious of her health. d. There are not enough nonverbal messages to say for sure. Please select the best answer from the choices provided A B C D
Physics
1 answer:
Inessa05 [86]3 years ago
4 0

Answer:

D

Explanation:

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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow
ale4655 [162]

Answer: The riders are subjected to 11.5 revolutions per minute

Explanation: Please see the attachments below

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3 years ago
Compare and contrast repetition and replication in a science investigation
Bas_tet [7]
Repetition occurs when multiple sets of measurements are made during one scientific investigation. Replication occurs when a scientific investigation is <span>reproduced by another person.

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3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
3 years ago
a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. If a 19,600 N car is parked 8 meters from one end.
LUCKY_DIMON [66]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

Mg(r - L/2) = mg(L/2 - 8)

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

400,000(r - 10) = 19,600 (10 - 8)

r - 10 = 0.098

r = 10.098 m

so pillar is at distance 10.098 m from one end of the slab

7 0
3 years ago
When is a hypothesis developed in the scientific method?
ivanzaharov [21]

Answer:

aren't there pics ?

Explanation:

7 0
3 years ago
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