Answer:
1.11 x 10^6 m
Explanation:
mass of Pluto, M = 1.1 x 10^22 kg
acceleration due to gravity on Pluto, g = 0.59 m/s²
let R be the radius of the Pluto and G is the universal gravitational constant.
G = 6.67 x 10^-11 Nm²/kg²
use the formula for the acceleration due to gravity
R = 1.11 x 10^6 m
Thus, the radius of the Pluto is 1.11 x 10^6 m.
A) 50 cm
B) 10000 cm/s
Explanation
Step 1
A)
If you know the distance between nodes and antinodes then use this equation:
then, let
now, replace to find the wavelength
so, the wavelength is
A) 50 cm
Step 2
The speed of a wave can be found using the equation
or velocity = wavelength x frequency,
then,let
replace and evaluate
so
B) 10000 cm/s
I hope this helps you
Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where
is the mass of the truck,
is velocity of the truck,
is the common velocity of moving and standing truck after collision and
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg,
as 22.3 m/s,
as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s