Answer:
weight =m x g =2kg × 10m/s 2 = 20N
<h3>Formulas:</h3>
weight = m × gravity
=unit for weight is newtons
Mass=weight ÷ gravity
= unit for mass is kilograms
Gravity=weight ÷ mass
=unit for gravity is 10m/sec 2
Answer:
B, to predict star count
Explanation:
the rest can be done with rulers, Einstein cans and public surveys, we can't predict how many stars there will be, but we can take pictures of them instead
Answer: a) v = ω /k, b) v = - ωAcos( kx −ωt)
Explanation:
y(x,t)=Asin(kx−ωt) defines the wave equation.
a)
We are asked to find wave speed (v)
Recall that v = fλ
From the wave equation above,
k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k
ω = 2πf where f is the frequency and ω is the angular frequency.
f = ω/ 2π.
By substituting for λ and ω into the wave speed formulae, we have that
v =( ω/ 2π) × (2π /k)
v = ω/k
b)
y(x,t)=Asin(kx−ωt)
The first derivative of y with respect to x give the velocity (vy)
By using chain rule, we have that
v = dy/dt = A cos( kx −ωt) × (−ω)
v = - ωAcos( kx −ωt)
Answer:
15.67 m/s
Explanation:
The ball has a projectile motion, with a horizontal uniform motion with constant speed and a vertical accelerated motion with constant acceleration g=9.8 m/s^2 downward.
Let's consider the vertical motion only first: the vertical distance covered by the ball, which is S=50 m, is given by
where t is the time of the fall. Substituting S=50 m and re-arranging the equation, we can find t:
Now we now that the ball must cover a distance of 50 meters horizontally during this time, in order to fall inside the carriage; therefore, the velocity of the carriage should be:
<span> For any body to move in a circle it requires the centripetal force (mv^2)/r.
In this case a ball is moving in a vertical circle swung by a mass less cord.
At the top of its arc if we draw its free body diagram and equate the forces in radial
direction to the centripetal force we get it as T +mg =(mv^2)/r
T is tension in cord
m is mass of ball
r is length of cord (radius of the vertical circle)
To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So
minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s
In the second case the speed of ball at top = (2*3.285) =6.57 m/s
Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom
we get velocity at bottom as 9.3m/s.
Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force
T-mg=(mv^2)/r
We get tension in cord T=13.27 N</span>
