Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m
Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
Answer:
4.7m
Explanation:
Given parameters:
Mass of the book = 1kg
Gravitational potential energy = 46J
Unknown:
Height of the shelf = ?
Solution:
The potential energy is due to the position of a body above the ground.
Gravitational potential energy = mgh
m is the mass,
g is the acceleration due gravity = 9.8m/s²
h is the height which is unknown
46 = 1 x 9.8 x h
h = 4.7m
Answer:
Explanation:
Mechanical Advantage is the ratio of the distance of the input load (Li)from the pivot to the output load applied to the pivot(Lo)
MA = Li/Le
Given;
Li = 45cm
Lo = 1.8cm
MA = 45/1.8
MA = 25
Hence the mechanical advantage is 25
Also MA is expressed in terms of the force ratio which is the ratio of the Load to the effort applied.
MA = Load/Effort
Given
Load = 1250N
MA = 25
Effort = ?
Substitute
25 = 1250/Effort
Effort = 1250/25
Effort = 50N
Hence the minimum force exerted on the load is 50N
Answer:
202.8m
Explanation:
Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.
First calculate the total time travelled by using the second equation of motion
h = Ut + 1/2gt^2
Let assume that u = 0
And h = 3.5
Substitute all the parameters into the formula
3.5 = 1/2 × 9.8 × t^2
3.5 = 4.9t^2
t^2 = 3.5/4.9
t^2 = 0.7
t = 0.845s
To know how far the cannonball travel, let's use the equation
S = UT + 1/2at^2
But acceleration a = 0
T = 2t
T = 1.69s
S = 120 × 1.69
S = 202.834 m
Therefore, the distance travelled by the cannon ball is approximately 202.8m.