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ra1l [238]
4 years ago
14

In a circuit a battery is used to

Physics
2 answers:
jeyben [28]4 years ago
8 0

The power of batteries in a circuit. The key functions of a battery and bulb in a circuit are explained. A battery is a source of energy which provides a push - a voltage - of energy to get the current flowing in a circuit. A bulb uses the electrical energy provided by the battery, but does not use current.

HOPE IT HELPS YOU OUT!

Gekata [30.6K]4 years ago
7 0

Answer:

convert electrical energy to some other form of energy

Explanation:

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The answer is D. Unbalanced

This is because in order to change the speed or direction of an object, unbalanced forces need to be working on it.

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You are an astronaut on a spacewalk to fix your spacecraft with a hammer. Your lifeline (tether) breaks and your jetpack on your
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Explanation:

Here we will apply the law of conservation of momentum which is one of the powerful laws of physics. As this law states that the "if no external forces are acting on the system then the net momentum of the system before and after must remain conserved. As the astronaut has a hammer, if he throws it in the direction opposite of his space craft, he will automatically move towards the space craft to conserve the momentum. That's how he can reach the space craft easily by throwing away the hammer.

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3 years ago
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In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallw
RSB [31]

Answer:

The coefficient of kinetic is

u_{k}=0.59

Explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

F_{u} =u_{k} *N\\F_{N}=25N\\N=W\\N=3.5kg*9.8\frac{m}{s^{2} }=34.3N

F_{N}-F_{u}=m*a\\F_{N}-u_{k}*N=m*a\\u_{k}*N=F_{N}-m*a\\u_{k}=\frac{F_{N}-m*a}{N}

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

v_{f} ^{2}=v_{o}^{2}+2*a(x_{f}-x_{o})\\x_{o}=0\\v_{o}=0\\v_{f} ^{2}=2*a*x_{f}\\a=\frac{v_{f} ^{2}}{2*a*x_{f}}\\ a=\frac{(1.53\frac{m}{s} )^{2}}{2*0.91m}\\a= 1.28 \frac{m}{s^{2} }

So replacing the acceleration can fin the coefficient:

u_{k}=\frac{F_{N}-m*a }{N}\\u_{k}=\frac{25N-(3.5kg*1.28\frac{m}{s^{2}} }{34.3N} \\u_{k}=0.59

5 0
3 years ago
Suppose two carts, one twice as massive as the other, fly apart when the
disa [49]

Answer:

the question this morning

Explanation:

90+70 look than

8 0
3 years ago
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
yuradex [85]

Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

  • mass of John, m_J=30\ kg
  • mass of William, m_W=30\ kg
  • length of slide, l=3\ m

(A)

height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

v_J=5.94\ m.s^{-1}

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

30\times 5.94+0=(30+20)v

v=3.564\ m.s^{-1} is the velocity with which they leave the slide.

(B)

  • frictional force due to mud, f=105\ N

<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

8 0
3 years ago
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