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Dmitry_Shevchenko [17]
3 years ago
11

Which statement are true about moving the compass around the wire? Check all that apply

Physics
2 answers:
hoa [83]3 years ago
7 0

With no current, the compass needle will not move.

If the light bulb is on, the compass needle will move.

Changing the battery terminals will change the current flow and the compass will point differently

yanalaym [24]3 years ago
4 0

The following apply among the choices;

1. With no current, the compass needle will not move

2. If the light bulb is on, the compass needle will move

3. Changing the battery terminals will change the current flow and the compass will point differently






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When the car was stopped by the tree, its change in velocity during the collision was-6 meters/second. This change in
HACTEHA [7]

Answer: -3 meters/second^2.

Explanation:

3 0
3 years ago
A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i
Effectus [21]

Answer:

s =  1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore  

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}  

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}  

• s = \sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}

s = \sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}

s =  1.7 m

7 0
3 years ago
A 3.0-kg brick rests on a perfectly smooth ramp inclined at 34° above the horizontal. The brick is kept from sliding down the pl
Firdavs [7]

Answer:

d=0.137 m ⇒13.7 cm

Explanation:

Given data

m (Mass)=3.0 kg

α(incline) =34°

Spring Constant (force constant)=120 N/m

d (distance)=?

Solution

F=mg

F=(3.0)(9.8)

F=29.4 N

As we also know that

Force parallel to the incline=FSinα

F=29.4×Sin(34)

F=16.44 N

d(distance)=F/Spring Constant

d(distance)=16.44/120

d(distance)=0.137 m ⇒13.7 cm

4 0
3 years ago
A car with a mass of 2200 kg is travelling at a rate of 55 m's. What is the cars momentum? *
Varvara68 [4.7K]
  • Mass=2200kg
  • Velocity=55m/s

\\ \sf\bull\dashrightarrow Momentum=Mass\times Velocity

\\ \sf\bull\dashrightarrow Momentum=2200(55)

\\ \sf\bull\dashrightarrow Momentum=121000kgm/s

7 0
2 years ago
Read 2 more answers
HELP PLS
melisa1 [442]

Answer:

red I think

Explanation:

it's on red so I googled some of it and the closest was red

4 0
3 years ago
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