Answer:
Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).
Explanation:
The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):
José


Txomin


Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).
Answer:

Explanation:
The equation that relates heat Q with the temperature change
of a substance of mass <em>m </em>and specific heat <em>c </em>is
.
We want to calculate the final temperature <em>T, </em>so we have:

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add
to our temperature in
to have it in
as it must be):

Complete question is;
A ski jumper travels down a slope and leaves the ski track moving in the horizontal direction with a speed of 24 m/s. The landing incline below her falls off with a slope of θ = 59◦ . The acceleration of gravity is 9.8 m/s².
What is the magnitude of the relative angle φ with which the ski jumper hits the slope? Answer in units of ◦
Answer:
14.08°
Explanation:
The time covered will be given by the formula;
t = (2V_x•tan θ)/g
t = (2 × 24 × tan 59)/9.8
t = 8.152 s
Now, the slope of the flight path at the point of impact will be given by the formula;
tan α = V_y/V_x
We are given V_x = 24 m/s
V_y will be gotten from the formula;
v = gt
Thus;
V_y = gt
V_y = 9.8 × (8.152) = 78.89 m/s
Thus;
tan α = 78.89/24
tan α = 3.2871
α = tan^(-1) 3.2871
α = 73.08°
Thus ;
Relative angle φ = α - θ = 73.08 - 59 = 14.08°
Answer:
The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>
Explanation:
As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.
<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>
Where,
<em>R = Orbital radius (in this case) = 30.1 AU</em>
<em />
Plug the value of R in the equation (A):
<em>(A) => The circumference of the circle = 2*π*(30.1)</em>
<em> The circumference of the circle = </em><em>60.2π</em>
Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>
Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .