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katen-ka-za [31]
3 years ago
15

Plz help me with my PE class

Physics
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

The answer to the first one is, group 1.

Explanation:

Sorry if this didn’t help if I find the answers to the others I’ll comment them to myself!

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a tuck at rest starts to move and accelerated by 4m/s in 25 secondd. what is the velocity of the truck at the end offf the time?
inysia [295]

Answer:

100m/s

Explanation:

4 0
3 years ago
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When observing a group of children at a daycare center, Emily made the following observations: Five year old children played in
konstantin123 [22]
This question is not about physics science.

The answer is: option <span>a. Five-year-old children have longer attention spans than three-year-old children.

It is the attention ability what let the older children to stay longer in one location instead of being moving between different activities. The younger children who cannot keep their attention long time in a same activity entertain themselves by changing activities.
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3 years ago
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. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in
kupik [55]

Answer:

(c) 16 m/s²

Explanation:

The position is r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}.

The velocity is the first time-derivative of <em>r(t).</em>

<em />v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}<em />

The acceleration is the first time-derivative of the velocity.

a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}

Since <em>a(t)</em> does not have the variable <em>t</em>, it is constant. Hence, at any time,

a = -16\hat{j}

Its magnitude is 16 m/s².

4 0
3 years ago
Can power be negative in physics? I am doing a lab in which I run down the stairs as quickly as possible. They are 5.4m tall and
RUDIKE [14]

Answer:

Yes.

Explanation:

A negative power would just represent a loss of power. So in your case it lost -1252.16 W

8 0
3 years ago
A 13.4-mH inductor carries a current i = <img src="https://tex.z-dn.net/?f=I_%7Bmax%7D" id="TexFormula1" title="I_{max}" alt="I_
Digiron [165]

The voltage across an inductor ' L ' is

V = L · dI/dt .

I(t) = I(max) sin(ωt)

dI/dt = I(max) ω cos(ωt)

V = L · ω · I(max) cos(ωt)

L = 1.34 x 10⁻² H

ω = 2π · 60 = 377 /sec

I(max) = 4.80 A

V = L · ω · I(max) cos(ωt)

V = (1.34 x 10⁻² H) · (377 / sec) · (4.8 A) · cos(377 t)

<em>V = 24.25 cos(377 t)</em>

V is an AC voltage with peak value of 24.25 volts and frequency = 60 Hz.

6 0
3 years ago
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