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sergij07 [2.7K]
2 years ago
15

EASY. PLEASE HELP. David is on a school bus stopped at a pick up point where students are getting on the bus. Once everyone is s

eated, the driver starts off quickly. What happens to the passengers on the bus as the driver accelerates?
A)The passengers do not move

B) The passengers fall forward toward the front of the bus

C) The passengers move sideways

D) The passengers are pushed backwards
against their seats.​
Physics
1 answer:
goldenfox [79]2 years ago
8 0

Answer:

d

Explanation:

because when things go forward quickly things get pushed back

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A rubber balloon is rubbed with a PVC pipe and the rubber balloon becomes positively charged. Why is this? a) Because the rubber
pentagon [3]

Answer:

d)

Explanation:

Electrons are lost or gained when the ballon is rubbed with a PVC. As the rubber ballon lost electrons, it will have more protons, hence the positive charge. (More protons than electrons in the ballon).

4 0
2 years ago
Find the magnitude of the sum<br> of these two vectors:<br><br> 101 m<br> 60.0 °<br> 85.0 m
attashe74 [19]

Answer: 161.3

I have a acellus too and got this question correct, so I hope this helps y’all out

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2 years ago
The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for
abruzzese [7]

Answer:

The answer is "\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}"

Explanation:

\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta    - \omega_{0}^{r} \\\\\to \omega^{r} = 2  (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}

5 0
3 years ago
Why is warm up important?
atroni [7]
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5 0
3 years ago
Read 2 more answers
if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and
Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62

\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99

||v|| = 85

θ = 335°

\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03

\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92

Resultant

R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15

\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0
3 years ago
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