Answer:
The frictional force acted on mass, m = 7.5 kg
Explanation:
Given data,
The acceleration of the box, a = 2 m/s²
The force acting on the box, F = 15 N
The mass of the box, m = ?
The force is defined as the product of mass and acceleration.
F = m x a newton
Therefore,
m = F / a
= 15 / 2
= 7.5 kg
Hence, the frictional force acted on mass, m = 7.5 kg
Answer:
conditions have changed significantly
Explanation:
This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
You are given the mass of the book with an initial
height of 0.77m. You are also given the final height of the book, 2m above the
floor. The problem requires you to find the work done on the book. It is said
that work is equal to force times distance. And that force is equal to mass
times acceleration due to gravity.
W=Fd
W=mgd=mgh
W=(1.7kg)(9.8m/s^2)(2m-0.77m)
W=20.49J
<span>The work is equal to 20.49J.</span>