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Slav-nsk [51]
3 years ago
15

What will decrease the strength of the magnetic field around a wire

Physics
1 answer:
Sav [38]3 years ago
8 0
One way to increase or decrease the strength of the magnetic field is to change the number of loops in the coil. The more loops you add, the stronger the field will become. The more loops you remove, the weaker the field will become.
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What is the biggest disadvantage of solar electricity, or at least, why is it not used much more than it currently is?
Lesechka [4]
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5 0
3 years ago
1) how will the total distance traveled by a car in 2 hours be affected if the average speed is doubled?
Elodia [21]
1) d = V*t >>>as you double the av. speed the distance become doubled.
2) after you draw the victors you will find the total displacement  = 1 meter to the left.
3) V =d/t =( 8km)/(1.25hr) = 6.4km/hr
6 0
3 years ago
A car that experiences no frictional force is started and caused to move. For the car to continue in that motion, the gas pedal
Masteriza [31]

1) Answer D not at all

The car is not experiencing any frictional force so that implies that there is no force acting on the car once it starts motion. So, according law of inertia, the car will continue to move and no other force is required.

Friction force is the resistance force that opposes the motion of any object. It arises due to the contact of surfaces.

2) Answer C none

There is no force on any spaceship moving far from any planet. So, according to law of inertia the spacecraft will continue to move at same speed.

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6 0
3 years ago
Read 2 more answers
Floating piece of wood - buoyancy question.?
sashaice [31]
No. because the weight is the same. no m9re water displacement
7 0
3 years ago
A car horn emits a frequency of 400 Hz. A car traveling at 20.0 m/s sounds the horn as it approaches a stationary pedestrian. Wh
Temka [501]

Answer:

The observed frequency by the pedestrian is 424 Hz.

Explanation:

Given;

frequency of the source, Fs = 400 Hz

speed of the car as it approaches the stationary observer, Vs = 20 m/s

Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.

The observed frequency is calculated as;

F_s = F_o [\frac{v}{v_s + v} ] \\\\

where;

F₀ is the observed frequency

v is the speed of sound in air = 340 m/s

F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\

F₀ ≅ 424 Hz.

Therefore, the observed frequency by the pedestrian is 424 Hz.

8 0
3 years ago
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