Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ +
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
Answer:
Tides are very long waves that move across the oceans. They are caused by the gravitational forces exerted on the earth by the moon, and to a lesser extent, the sun. ... Because the gravitational pull of the moon is weaker on the far side of the Earth, inertia wins, the ocean bulges out and high tide occurs.
Explanation:
Answer : The specific heat of unknown sample is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
![q_1=-[q_2+q_3]](https://tex.z-dn.net/?f=q_1%3D-%5Bq_2%2Bq_3%5D)
![m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_f-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_f-T_2%29%2Bm_3%5Ctimes%20c_3%5Ctimes%20%28T_f-T_2%29%5D)
where,
= specific heat of unknown sample = ?
= specific heat of water = 
= specific heat of copper = 
= mass of unknown sample = 72.0 g = 0.072 kg
= mass of water = 203 g = 0.203 kg
= mass of copper = 187 g = 0.187 kg
= final temperature of calorimeter = 
= initial temperature of unknown sample = 
= initial temperature of water and copper = 
Now put all the given values in the above formula, we get
![0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]](https://tex.z-dn.net/?f=0.072kg%5Ctimes%20c_1%5Ctimes%20%2839.4-80.0%29%5EoC%3D-%5B%280.203kg%5Ctimes%204186J%2Fkg%5EoC%5Ctimes%20%2839.4-11.0%29%5EoC%29%2B%280.187kg%5Ctimes%20390J%2Fkg%5EoC%5Ctimes%20%2839.4-11.0%29%5EoC%29%5D)

Therefore, the specific heat of unknown sample is, 