Martha must carry a 45 N package up three flights of stairs. Each flight of stairs has a height of 2.3m, and the actual distance
1 answer:
Answer:
310.5 J
Explanation:
The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to
where
(mg) = 45 N is the weight of the package
is the increase in height of the package
The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is
And so, the work done by Martha is
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It's a pleasure to be helping you today with your<u> physics question!</u>
Answer:
Explanation:
We want to find the initial speed of the ball.
To do this, we have to apply the formula for the time of flight of a projectile:
where θ = angle of flight
g = acceleration due to gravity
v0 = initial speed
Therefore, substituting the given values into the formula, we have that:
⇒ 2 × ×0.8910= 9.8 × 4.2
⇒
That is the initial speed of the ball.
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By definition, the gravitational force is given by:
Where,
G: gravitational constant
m1: mass of object 1
m2: mass of object 2
r: distance between objects:
We know that the force is equal to 10N:
If the distance is double we have:
Therefore, the new force is:
2.5 newtons
Answer:
the changed force of attraction between them is:
A. 2.5 newtons
Answer:
v > 133.5 m/s
Explanation:
Let's analyze this problem a little, park run a complete circle we must know the speed of the system at the top of the circle.
Let's start by using the concepts of energy to find the velocity at the top of the circle
Initial. Top circle
Em₀ = K + U = ½ m v² + m g y
If we place the reference system at the bottom of the cycle y = 2R = L
Em₀ = ½ m v² + m g y
final. Low circle
= K = ½ m v₁²
Emo =
½ m v² + m g y = 1/2 m v₁²
v₁² = v² + (2g L)
v₁² = v² + 2 g L
The smallest value that v can have is zero, with this value the bullet + block system reaches this point and falls, with any other value exceeding it and completing the circle. Let's calculate for this minimum speed point
v₁ = √2g L
We already have the speed system at the bottom we can use the moment
Starting point before crashing
p₀ = m v₀
End point after collision at the bottom of the circle
= (m + M) v₁
The system is formed by the two bodies and therefore the forces to last before the crash are internal and the moment is conserved
p₀ =
m v₀ = (m + M) v₁
v₀ = (m + M) / m v₁
Let's replace
v₀ = (1+ M / m) √ 2g L
Let's reduce to the SI system
m = 40 g (kg / 1000g) = 0.040 kg
Let's calculate
v₀ = (1 + 1.35 / 0.040) RA (2 9.8 0.753)
v₀ = 34.75 3.8417
v₀ = 133.5 m / s
the velocity must be greater than this value
v > 133.5 m/s