Answer:
r = 0.0173 m = 1.73 cm
Explanation:
Here, the centripetal force of the block will be providing the required breaking tension in the string:
[tex]Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\[/tex]
where,
r = radius = ?
m = mass of block = 0.13 kg
v = tangential spee of block = 4 m/s
T = Breaking Strength = 30 N
Therefore,
[tex]r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}[/tex]
r = 0.0173 m = 1.73 cm
When finding the radius of the string at the point it breaks, the tangential
velocity is assumed to be constant.
The radius when the string breaks is [tex]\underline{6.9 . \overline 3 \times 10^{-3}} \ m[/tex]Reasons:
The mass of the small block, m = 0.130 kg
Initial radius of the circle of rotation = 0.800 m
Tangential velocity, v = 4.00 m/s
The radius of the path of rotation is reduced as the string is pulled
Breaking strength of the string = 30.0 N
Required:
The radius of the circle when the string brakes
Solution:
[tex]Centripetal \ force = \dfrac{m \cdot v^2}{r}[/tex]
Where;
r = The radius of the circle of rotation
When the string brakes, w have;
Centripetal force = Breaking strength of the string = 30.0 N
Which gives;
[tex]\displaystyle r = \mathbf{\dfrac{m \cdot v^2}{Centrifugal \ force}} = \frac{0.130 \times 4^2}{30} =6.9\overline 3 \times 10^{-2}[/tex]
The radius of the circle when, the string breaks r = [tex]\underline{6.9\overline 3 \times 10^{-2}} \ m[/tex]
Learn more here:
https://brainly.com/question/20905151
A wave has a frequency of 30Hz and wave length of 40cm. What is the velocity of the wave?
Answer:
12m/s
Explanation:
v=fλ
30×(40÷100)=
12m/s
Before we make measurements, let's make sure we understand the circuit. 1. Select all of the following that correctly describe what a volt meter and ammeter measure. Select all that apply: A volt meter measures the potential difference (or voltage) across a circuit element. A volt meter measures the potential difference (or voltage) passing through a circuit element. A ammeter measures the electric current passing through a circuit element. A ammeter measures the electric current across a circuit element.
Answer:
the correct answers are a and c
Explanation:
In an electrical circuit there are two important quantities to measure, such as voltage and current.
Voltage is the potential difference between two points in a circuit
current is the number of electrons you pass through a given point per unit of time.
Now let's analyze each answer
a) true. The potential difference across an element
b) False. The potential difference is u field there is no physical entity that moves
c) True. The current is electrons in motion and these pass through the given element
d) False. There is a physical quantity that passes through the point
the correct answers are a and c
A 5kg block rests on a 30° incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding. a) up the incline, b) down the incline ?
Answer:
Hope It Help
Explanation:
That's all I know
Why do astronomers use frequencies other than the visible ones when they are
investigating the universe?
1 kg block slides down a frictionless inclined plane that makes an angle of 300 with respect to the ground. The total length of the plane is 2 m, but midway down it collides with a second block, weighing 0.5 kg. The two blocks stick together and travel as one unit the rest of the way down the ramp. What is the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane
Answer:
the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J
Explanation:
Given that the data in the question;
angle of inclination with respect to the ground [tex]\theta[/tex] = 30°
length of plane d = 2m
m₁ = 1 kg
m₂ = 0.5 kg
now, velocity of the first block at midpoint;
[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]
[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]
v² = gsin[tex]\theta[/tex]d
v = √( gsin[tex]\theta[/tex]d)
g is 9.8 m/s
so we substitute
v = √( 9.8 × sin30° × 2)
v = √( 19.6 )
v = 3.13 m/s
Now, velocity just after collision of the blocks will be;
(m₁ + m₂)v₂ = m₁v
v₂ = m₁v / (m₁ + m₂)
we substitute
v₂ = (1 × 3.13) / (1 + 0.5)
v₂ = 3.13 / 1.5
v₂ = 2.0866 m/s
now, final kinetic energy will be;
[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy
[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²
we substitute
[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]
[tex]KE_f[/tex] = 7.35 + 3.2654
[tex]KE_f[/tex] = 10.62 J
Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J
1.A body of mass 10kg falling freely was found to be falling at a rat of 20m/s what force will stop the body in 2second?
Answer:
50N
Explanation:
force it is falling with can be found by mass into acceleration and then devide by half to find force that could stop it in 2 sec