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3 years ago

Ok these are my last points say hi if u want brainliest

Physics

2 answers:

lawyer [7]3 years ago

5 0

Answer:

Explanation:

il63 [147K]3 years ago

3 0

Hiii! i'd love brainliest cuz i'm new here and i've been helping people all day!

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Technician A says that hill assist and hill descent controls are added features to some electronic stability control systems. Te

Pani-rosa [81]

Answer:

Both technicians A and B

Explanation:

Both trailer sway control, hill assist and hill descent controls are additional featires that enhance stability of electronics and their control systems. Majorly, these features track and reduce skidding in electronics, therfore, enhancing electronic system stability. During the process, these newly added features help to automatically apply brakes and direct the sytem where the controller wants to take it.

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3 years ago

Which of these is a benefit of social networking ?

Alisiya [41]

Answer:

Staying connected to friends

Explanation:

hope this helps

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3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,

laila [671]

Answer:

1.45544 J

Explanation:

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3 years ago

A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.

Mekhanik [1.2K]

Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

vf^2 = 1900

sqrt(vf^2) = sqrt(1900)

vf = 43.59 m/s

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3 years ago

Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

5 0

3 years ago

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