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3 years ago

5

The BMI of someone that weights 121 pounds and is 64 inches tall:

2 answers:

nordsb [41]3 years ago

8 0

Answer:

B

;)

Explanation:

Ainat [17]3 years ago

7 0

Answer:

B - 32.52

Explanation:

Hope this Helps!

:)

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Which of the following is true of a parasitic relationship?

NISA [10]

Answer:

when one organism benefits but harms the other

6 0

3 years ago

What is the temperature of a gas in which the atoms are moving very very slowly?

pickupchik [31]

Cold.
Example: A frozen object (such as ice) is solid. Its molecules move extremely slow because the object itself isn't moving... It's like the skin on your bod... Imagine that wiggling and writhing on your bones. It's not like a gas, that can move freely and has no specific form or hold.
So again, the answer is Cold.

5 0

3 years ago

PLZ HELP THIS IS FOR AN UNIT

LiRa [457]

Answer:

5,000+25,000=

Explanation:

<h2>You need to plus it answer it now go girl #CarryOnLearning</h2>

6 0

3 years ago

1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche

Tatiana [17]

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2$

$A^2=\dfrac{mv^2}{k}$

Put the value into the relation

$A^2=\dfrac{1.60\times13.0^2}{315}$

$A=\sqrt{0.858}$

$A=0.926\ m$

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

$a=A\omega^2$

$a=A\times\dfrac{k}{m}$

Put the value into the formula

$a=0.926\times\dfrac{315}{1.60}$

$a=182.31\ m/s^2$

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

$F=ma$

Put the value into the formula

$F= 1.60\times182.31$

$F=291.69\ N$

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

7 0

3 years ago

A 3.00 × 10^-6 kg ant rides the second hand of an enormous vertical wall clock. The ant smoothly sweeps a circle of radius 50.0

kotykmax [81]

Answer:

1.65×10⁻⁶ N

Explanation:

m = Mass of ant = $3\times 10^{-6}\ kg$

r = Radius = 50 m

t = Time taken to complete on rotation = 60 seconds

Angular velocity

$\omega=\frac{2\pi}{t}\\\Rightarrow \omega=\frac{2\pi}{60}$

Centripetal acceleration is force that is acting outward

$F_c=m\omega^2r\\\Rightarrow F_c=3\times 10^{-6}\times \left(\frac{2\pi}{60}\right)^2\times 50\\\Rightarrow F_c=1.65\times 10^{-6}\ N$

The magnitude of the upward force felt by the ant due to the second hand would be 1.65×10⁻⁶ N

4 0

3 years ago