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2 years ago

11

Which items or activities should a pregnant woman avoid to maintain her health and the health of her fetus? Check all that apply

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1 answer:

Fynjy0 [20]2 years ago

6 0

Answer:

Alcohol and drugs are a no no. you should put the answers in so people can answer correctly.

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A car traveling at 23 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration?

kykrilka [37]

We could determine the acceleration using this formula
$\boxed{a= \dfrac{v_{1}-v_{0}}{t} }$

Given from the question v₀ = 23 m/s, v₁ = 0 (the car stops), t = 5 s
plug in the numbers
$a= \dfrac{v_{1}-v_{0}}{t}$
$a= \dfrac{0-23}{5}$
$a= \dfrac{-23}{5}$
a = -4.6
The acceleration is -4.6 m/s²

8 0

3 years ago

Refraction occurs when __________.

ser-zykov [4K]

Refraction occurs when a wave enters a new medium and changes its speed.

i hope this helps

4 0

3 years ago

Read 2 more answers

As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/

zhuklara [117]

Acceleration = (change in speed) / (time for the change)

- 4.1 m/s² = (-9 m/s) / (time for the change)

Time for the change = (-9 m/s) / (-4.1 m/s²) = 2.2 seconds

3 0

4 years ago

what can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis ? Also draw

MaRussiya [10]

Answer:

It says about the motion and the graph of the object is stationary, basically travelling at the same speed at any time of the graph. It will never change.

Explanation:

To draw a diagram:

1. Draw an object and represent the speed as stationary and constant at any time.

5 0

3 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

6 0

4 years ago