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Leto [7]
3 years ago
10

PLEASE HELP ME WITH THIISSSSS UGGHHH

Physics
1 answer:
grigory [225]3 years ago
5 0

(b) is the answer because when the cat jumps on the fridge the gravity is pulling it down so it's less force but when the cat jumps down for the fridge the gravity also pull it down so it go down fast than it go up so it's twice as great (b)

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Basketball= What is the half of the court that is farthest from the offensive basket?
klasskru [66]
The answer would be half court I think
6 0
3 years ago
Marking as brainliest last attempt
arsen [322]

Answer:

Bullying in the schools has negative effects on individual students and on the school climate as a whole. Bullying can cause long-term problems for both the victims of bullying and the bullies themselves. To explore the effects of bullying on adolescents, we will define bullying, identify the characteristics of bullies and victims, outline the extent and consequences of bullying, and present resources for further information and assistance.

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Characteristics of Bullies and Victims

There are specific behaviors that bullies tend to exhibit. The bullies often need to feel powerful and in control. They may feel no remorse when they inflict injury and suffering on others. Bullies tend to defy authority and are likely to break school rules. They seem to have little anxiety and appear to possess high self-esteem. Students who come from homes characterized by physical punishment tend to be more likely to exhibit these types of behaviors.

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3 0
2 years ago
A child with a weight of 110 N swings on a playground swing attached to 2.00 m long chains. What is the gravitational potential
Ierofanga [76]

Answer

given,

Weight of the child = 110 N

length of the swing,L = 2 m

now, calculating the potential energy when the string is horizontal

  Potential energy = m g h

 now, h = L (1 - cos θ)   where θ is the angle made by the string with the vertical.

     PE = m g L (1 - cos θ)

   when rope is horizontal θ = 90°

     PE = 110 x 2 (1 - cos 90°)

    PE = 220 J

now, calculating potential energy when string made 25° with horizontal

PE = m g L (1 - cos θ)

   when rope is horizontal θ = 25°

     PE = 110 x 2 (1 - cos 25°)

     PE = 20.61 J

5 0
4 years ago
What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2
Margarita [4]

Answer:

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

Explanation:

  • We have to make a hollow sphere of inner  radius r_1 and outer radius r_2.

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

V_t=\frac{4}{3} \pi.r_2^3

And the volume of the hollow space in the sphere:

V_h=\frac{4}{3} \pi.r_1^3

Therefore the net volume of material required to make the sphere:

V=V_t-V_h

V=\frac{4}{3} \pi(r_2^3-r_1^3)

  • Now let the density of the of the material be \rho.

<u>Then the mass of the material used is:</u>

m=\rho.V

m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3  )

4 0
4 years ago
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

7 0
4 years ago
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