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Solnce55 [7]
3 years ago
8

Which statement describes what most likely occurs when a compass is placed next to a simple circuit made from a battery, a light

bulb, and a wire?
A. A magnetic field created by the compass increases the current in the electrical circuit.
B. A magnetic field created by the compass causes the light bulb to stop working.
C. A magnetic field created by the electric current places negative charges on the compass.
D. A magnetic field created by the electric current causes the compass needle to move.

Physics
2 answers:
Nastasia [14]3 years ago
8 0
<span>When a compass is placed next to a simple circuit made from a
battery, a light bulb, and a wire, the magnetic field created by the
electric current in the wire causes the compass needle to move. (D)</span>
aleksley [76]3 years ago
6 0
D. A magnetic field created by the electric current causes the compass needle to move.
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The car travels 28m in 7s. What is the speed of the car?
vovikov84 [41]

The Speed of the car is 4m/s.

What is the speed?

The speed of an object can be defined as the total distance traveled by it in a particular interval of time. This can be determined by dividing the total distance moved by the object by the time taken to move.

It is the ratio of distance in m and time taken by the object to travel in seconds.

Hence: Distance D=28m

Time       T=7s

Speed S=D/T                            

S=28m/7S                          

S=4m/S

Therefore car's  speed is 4m/S

Learn more on speed from

brainly.com/question/28626511

#SPJ1

6 0
1 year ago
Read 2 more answers
A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilomete
HACTEHA [7]

Explanation :

Speed of car, v_1=50\ km/h=13.8\ m/s

Kinetic energy of the car, KE_1=400\ J

Speed of car, v_2=100\ km/h=27.7\ m/s

Let KE_2 is the kinetic energy of the car when it is moving with 100 km/h.

\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}

\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}

KE_2=KE_1(\dfrac{v_2}{v_1})^2

KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2

KE_2=1611.6\ J  

Initial velocity of both cars are 0. Using third equation of motion :

So, S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}

and S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}

t is same. So,

\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}

\dfrac{S_1}{S_2}=\dfrac{1}{2}

S_2=2\ S_1

So, the braking distance at the faster speed is twice the braking distance at the slower speed.

4 0
4 years ago
Read 2 more answers
A 1,500 kg car’s speed changes from 30 m/s to 15 m/s after the brakes are applied. Calculate the work done onto the car from the
HACTEHA [7]

The work done onto the car is 506,250 J

The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.

From the parameters given:

  • The mass of the car = 1500 kg
  • The initial speed = 30 m/s
  • The final speed = 15 m/s

The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)

\mathbf{=\dfrac{1}{2} mv_1^2 -\dfrac{1}{2} mv_2^2}

\mathbf{=\dfrac{1}{2} m(v_1^2 - v_2^2)}

\mathbf{=\dfrac{1}{2} \times 1500 \times (30^2 - 15^2)}

= 506,250 J

Therefore, we can conclude that the work done on the car is 506,250 J

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7 0
3 years ago
PHYSICS. <br> ……………………………………………….
belka [17]

Answer:

Work, W = F * d, and

Work = change in kinetic energy, so W=deltaKE.

Hence,

deltaKE=F * d

(1/2)*m*v^2 =F * d

d=[(1/2)*m*v^2]/F

d=[(1/2)*0.6*20^2]/5

d=24 m.

Explanation:

Work = change in kinetic energy, so W=deltaKE.

5 0
2 years ago
A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W
7nadin3 [17]
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
7 0
4 years ago
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