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OLEGan [10]
4 years ago
14

Apakah maksud rintangan?

Physics
2 answers:
Goshia [24]4 years ago
7 0
A thing that blocks one's way or prevents or hinders progress
Valentin [98]4 years ago
3 0
I’m confused???????
You might be interested in
A clarinetist, setting out for a performance, grabs his 3.230 kg3.230 kg clarinet case (including the clarinet) from the top of
SpyIntel [72]

Answer:

- 0.5 m/s²

Explanation:

m = mass of the clarinet case = 3.230 kg

W = weight of the clarinet case in downward direction

a = vertical acceleration of the case

Weight of the clarinet case is given as

W = mg

W = 3.230 x 9.8

W = 31.654 N

F = Upward force applied = 30.10 N

Force equation for the motion of the case is given as

F - W = ma

30.10 - 31.654 = 3.230 a

a = - 0.5 m/s²

8 0
3 years ago
Joan wants to test if salt lowers the temperature at which water boils.
ohaa [14]
In order to build a hypothesis, Joan should take an educated guess at what the effect of adding salt to the water will be on the water's boiling point. For example, the hypothesis may be,"The boiling point will increase". This hypothesis is one that can be verified or rejected via experimentation. The next step for Joan would be to set up an experiment to test this hypothesis. 
6 0
3 years ago
Read 2 more answers
3. Which part of the back seat of a car—the left side, the right side, or the
Nina [5.8K]

Answer:

The sides

Explanation:

Because there's a seat in front of the child to avoid him/her from flying

7 0
4 years ago
Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and was approaching at
GuDViN [60]

Answer:

a) v = 11.24 m / s ,    θ = 17.76º   b) Kf / K₀ = 0.4380

Explanation:

a) This is an exercise in collisions, therefore the conservation of the moment must be used

Let's define the system as formed by the two cars, therefore the forces during the crash are internal and the moment is conserved

Recall that moment is a vector quantity so it must be kept on each axis

X axis

initial moment. Before the crash

     p₀ₓ = m₁ v₁

where v₁ = -25.00 me / s

the negative sign is because it is moving west and m₁ = 900 kg

final moment. After the crash

      p_{x f}= (m₁ + m₂) vx

       p₀ₓ =  p_{x f}

       m₁ v₁ = (m₁ + m₂) vₓ

     vₓ = m1 / (m₁ + m₂) v₁

let's calculate

       vₓ = - 900 / (900 + 1200) 25

       vₓ = - 10.7 m / s

Axis y

initial moment

      p_{oy}= m₂ v₂

where v₂ = - 6.00 m / s

the sign indicates that it is moving to the South

final moment

     p_{fy}= (m₁ + m₂) v_{y}

     p_{oy} = p_{fy}

     m₂ v₂ = (m₁ + m₂) v_{y}

     v_{y} = m₂ / (m₁ + m₂) v₂

we calculate

    v_{y} = 1200 / (900+ 1200) 6

    v_{y}  = - 3,428 m / s

for the velocity module we use the Pythagorean theorem

      v = √ (vₓ² + v_{y}²)

      v = RA (10.7²2 + 3,428²2)

      v = 11.24 m / s

now let's use trigonometry to encode the angle measured in the west clockwise (negative of the x axis)

      tan θ = v_{y} / Vₓ

      θ = tan-1 v_{y} / vₓ)

      θ = tan -1 (3,428 / 10.7)

       θ = 17.76º

This angle is from the west to the south, that is, in the third quadrant.

b) To search for loss of the kinetic flow, calculate the kinetic enegy and then look for its relationship

      Kf = 1/2 (m1 + m2) v2

      K₀ = ½ m₁ v₁² + ½ m₂ v₂²

      Kf = ½ (900 + 1200) 11.24 2

      Kf = 1.3265 105 J

      K₀ = ½ 900 25²  + ½ 1200 6²

      K₀ = 2,8125 10⁵ + 2,16 10₅4

        K₀ = 3.0285 105J

the wasted energy is

        Kf / K₀ = 1.3265 105 / 3.0285 105

        Kf / K₀ = 0.4380

         

this is the fraction of kinetic energy that is conserved, transforming heat and transforming potential energy

5 0
3 years ago
Capacitors C1 = 5.85 µF and C2 = 2.80 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon
posledela

Answer:

Q1_new = 515.68 µC

Q2_new = 246.82 µC

Explanation:

Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.

Then it is only necessary to calculate the charge on each capacitor:

Q1 = 5.85 µF * 250 V = 1462.5 µC

Q2 = 2.8 µF * 250 V = 700 µC

Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.

When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:

1462.5 µC - 700 µC = 762.5 µC

Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.

This 762.5 µC will be divided proportionately between the two capacitors.

Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC

Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC

4 0
4 years ago
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