So we know F=ma
m means mass and a means acceleration
so Force= ma
so F=1300X1.07=1391N
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Fitness can be measured during the actual performance.
Which means that there are a lot of factors that can affect agility, speed, and coordination beside the physical factors itself. Meanwhile, flexibility solely depends on the physical Factors.
The answer is flexibility .
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Answer:
the maximum height reached by the object from where it was launched is 0.4591 m
Explanation:
initial speed of the object, u= 3 m/s
The velocity at the maximum height will always be 0.
Therefore, final velocity, v= 0 m/s
Using the Newton's equation of motion,
v^2 - u^2 = 2*g*h(max)
0 - u^2= 2*g*h(max)
h(max) = -u^2 /2* g
where g is the gravitational acceleration.
g= - 9.8 m/s^2
substituting the values in equation,
h(max)= - (3*3) / 2*(-9.8)
h(max) = 0.4591 m
the maximum height reached by the object from where it was launched is 0.4591 m
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Answer:

Explanation:
Given data
Mass m=1.2 g=0.0012 kg
diameter d=0.76 m
Friction Force F=3.6 N
To find
Velocity v
Solution
From the Centripetal force we know that

Where m is mass
v is velocity
r is radius
Substitute the given values to find velocity v
So

Answer:
Explanation:
Given:
- Acceleration a_dry = 0.4*a_wet
- initial velocity V_i
Find:
By how much does time taken to stop of dry snow differ from time taken to stop on wet snow?
Solution:
- Time taken to stop on wet snow t_wet:
V_f = V_i + a_wet * t_wet
t_wet = - a_wet / V_i
- Time taken to stop on wet snow t_wet:
V_f = V_i + a_dry * t_dry
t_dry = - a_dry / V_i
- Difference in t_wet and t_dry:
t_dry - t_wet = - a_dry / V_i + a_wet / V_i
dt = (a_wet - a_dry) / V_i
dt = (a_wet - 0.4*a_wet) V_i
dt = 0.6*a_wet / V_i