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ELEN [110]
2 years ago
5

Wan and Nurul sat on a see-saw. The see-saw is imbalanced because Wan’s mass is 45 kg while Nurul’s mass is only 30 kg. Suggest

how Nurul can balance the see-saw.
Physics
1 answer:
Alex Ar [27]2 years ago
5 0

We have that the see-saw will be balance if a weight of x=15kg is added to Nural's side of the see-saw

x=15kg

From the Question we are told that

Wan’s mass is M_w=45 kg

Nurul’s mass is M_n= 30 kg.

Generally

The Will be balance when the weight on both sides of the see-saw are equal

M_w=M_n+x

45=30+x

x=45-30

x=15kg

In conclusion

The see-saw will be balance if a weight of x=15kg is added to Nural's side of the see-saw

x=15kg

For more information on this visit

brainly.com/question/22255610

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the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of
nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
3 0
3 years ago
What do we call changes between solid liquid and gaseous forms of a substance
const2013 [10]

<u>We call changes between solid liquid and gaseous forms of a substance as  phase change or change of state.</u>

<u>Explanation:</u>

To change a substance from one state to another, extreme temperatures or pressures are required. Sometimes when a substance doesn't change states we should use all the ideas when that happens. To create a solid, we should decrease the temperature by a huge amount and then add pressure. For example, oxygen will solidify at -361.8 degrees Fahrenheit at standard pressure. However, it will freeze at warmer temperatures when the pressure is increased.

Phase changes happen when a substance reach some special points. Sometimes when a liquid becomes a solid a freezing point or melting point is used to measure the temperature at which a liquid changes into a solid. Some of the phase changes are: Condensation, Freezing, Melting.

5 0
3 years ago
Read 2 more answers
You normally drive a 12-h trip at an average speed of 100 km/h . Today you are in a hurry. During the first two-thirds of the di
kherson [118]

Answer:

78 km/h

Explanation:

If I normally drive a 12 hour trip at an average speed of 100 km/h, my destination has a total distance of:

  • 100 km/h · 12 h = 1,200 km

Today, I drive the first 2/3 of the distance at 116 km/h. Let's first calculate what 2/3 of the normal distance is.

  • 1,200 * 2/3 = 800 km

I've driven 800 km already. I need to drive 400 km more to reach my final destination. I need to figure out my average speed during this last 1/3 of the distance.

To do this, I first need to calculate how much time I spent driving 116 km/h for the past 800 km.

  • 116 km/1 h = 800 km/? h
  • 800 = 116 · ?
  • ? = 800/116
  • ? = 6.89655172

I spent 6.89655172 hours driving during the first 2/3 of the distance.

Now, I need to subtract this value from 12 hours to find the remaining time I have left.

  • 12 h - 6.89655172 h = 5.10344828 h  

Using this remaining time and my remaining distance, I can calculate my average speed.

  • ? km/1 hr = 400 km/5.10344828 h
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My average speed during the last third of the distance is around 78 km/h.

8 0
2 years ago
Here's a question from ~ [ AIEEE 2002 ]
lyudmila [28]
  • r=150m
  • coefficient of friction=\mu=0.6

As car is avoid skidding

\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg

  • Cancel m

\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g

\\ \sf\hookrightarrow v^2=\mu rg

\\ \sf\hookrightarrow v^2=0.6(10)(150)

\\ \sf\hookrightarrow v^2=60(150)

\\ \sf\hookrightarrow v^2=900

\\ \sf\hookrightarrow v=30ms^{-1}

Done

7 0
2 years ago
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1. A ball with a mass of 2.3 kg travels with a velocity of 10 m/s. What is it's kinetic energy?
monitta

Answer:

KE=½mv²

KE=½2.3kg×(10m/s)²

KE=½2.3kg×100m²/s²

KE=2.3kg×50m²/s²

KE=115joules

7 0
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