Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
Answer:
A. attract each other
Explanation:
When the neutral sphere is brought close to the negatively charged, as charges can move freely in a conductor, some electrons will be repelled, leaving a net positive charge (equal in magnitude ot the charge in the other sphere) on the surface of the neutral sphere, which will cause that both spheres attract each other.
Answer:
1) as far as I remember
Let's take 20 as vf (final velocity) and 11 as (initial velocity) and 4 as time
So we would use this formula a=vf-vi/t
So 20-11/4
Asnwer 2.25
Answer:
a = 0.16 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity = 3.2 [m/s]
Vo = initial velocity = 1.1 [m/s]
t = time = 13 [s]
a = acceleration [m/s²]
Now replacing:
![3.2=1.1+a*13\\2.1=13*a\\a=0.16[m/s^{2} ]](https://tex.z-dn.net/?f=3.2%3D1.1%2Ba%2A13%5C%5C2.1%3D13%2Aa%5C%5Ca%3D0.16%5Bm%2Fs%5E%7B2%7D%20%5D)
