Answer:
The value is 
Explanation:
From the question we are told that
The speed of the rope with hook is 
The angle is 
The speed at which it hits top of the wall is 
Generally from kinematic equation we have that
Here h is the height of the wall so
![[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h](https://tex.z-dn.net/?f=%5B16.3%20sin%20%2865%29%5D%5E2%20%3D%20%20%5B24.1%20sin%20%2865%29%5D%20%5E2%2B%20%20%202%20%28-9.8%29%2A%20h)
=>
Answer:
0.25m/s
Explanation:
Given parameters
m₁ = 5kg
v₁ = 1.0m/s
m₂ = 15kg
v₂ = 0m/s
Unknown:
velocity after collision = ?
Solution:
Momentum before collision and after collision will be the same. For inelastic collision;
m₁v₁ + m₂v₂ = v(m₁ + m₂)
Insert parameters and solve for v;
5 x 1 + 15 x 0 = v (5 + 15 )
5 = 20v
v = 
= 0.25m/s
Answer:
Kinematics is the study of motion, without any reference to the forces that cause the motion. It basically means studying how things are moving, not why they're moving. It includes concepts such as distance or displacement, speed or velocity, and acceleration, and it looks at how those values vary over time.
Answer:
the <em>ratio F1/F2 = 1/2</em>
the <em>ratio a1/a2 = 1</em>
Explanation:
The force that both satellites experience is:
F1 = G M_e m1 / r² and
F2 = G M_e m2 / r²
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- r is the orbital radius
- M_e is the mass of Earth
Therefore,
F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]
F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]
F1/F2 = m1/m2
F1/F2 = 1000/2000
<em>F1/F2 = 1/2</em>
The other force that the two satellites experience is the centripetal force. Therefore,
F1c = m1 v² / r and
F2c = m2 v² / r
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- v is the orbital velocity
- r is the orbital velocity
Thus,
a1 = v² / r ⇒ v² = r a1 and
a2 = v² / r ⇒ v² = r a2
Therefore,
F1c = m1 a1 r / r = m1 a1
F2c = m2 a2 r / r = m2 a2
In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,
F1 = F1c
G M_e m1 / r² = m1 a1
a1 = G M_e / r²
also
a2 = G M_e / r²
Thus,
a1/a2 = [G M_e / r²] / [G M_e / r²]
<em>a1/a2 = 1</em>
Answer:
6.136 mm
Explanation:
given,
frequency emitted by the bat = 5.59 x 10⁴ Hz
speed of sound = 343 m/s
smallest insect bat can hear will be equal to the wavelength of the sound the bat make.
λ = 6.136 mm
so, the smallest size of insect that bat can hear is equal to 6.136 mm
