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o-na [289]
3 years ago
6

Identify a situation in which you would want to have a high

Physics
2 answers:
Andreyy893 years ago
7 0

Answer: A voltmeter must have a high resistance where as an ammeter must have a low resistance.

Explanation:

A voltmeter is a device which is connected in parallel to the component across which voltage needs to be measured. In a parallel circuit voltage drop is same at the nodes. The parallel connection must not offer easier path for current to divert from the main circuit and travel. Thus, a voltmeter must have high resistance.

On the other hand, an ammeter which is used to measure current in the circuit must have low resistance as it is connected in series. It should not offer resistance as it would reduce the actual current and measurement would be inaccurate.

Allushta [10]3 years ago
7 0
Higher resistance - on heater (need water temperature ASAP)
Lower resistance  - any electricity wire on the street (energy supply to the households, not to loose much on the way)
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Why is the amount of mechanical energy produced less than the amount of electrical energy used?
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A car is traveling at a speed of 54 km/h. Breaks are applied so as to produce a uniform acceleration of -0.5 m/s2. Find how far
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Explanation:

u=54 km/h

54*5/18=15 m/s

v=0m/s

t=?

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we know that a=v-u/t

so,

t=v-u/a

t=15-0/0.5

=15/0.5

=30

therefore, the time is 30 second

Hope this answer helps you..

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The multiple reflection of a single sound wave is a/an
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A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0
3 years ago
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