In the horizontal direction, the forces acting on the person are
• friction with magnitude <em>f</em>, opposing motion, and
• the horizontal component of the pulling force (itself with mag. <em>p</em> ) with mag. <em>p</em> cos(30º), in the direction of motion.
There is no friction in the vertical direction, so we omit any discussion of the vertical forces.
By Newton's second law, we then have
<em>p</em> cos(30º) - <em>f</em> = <em>m</em> <em>a</em> cos(30º)
where <em>m</em> is the person's mass, and <em>a</em> is their acceleration so that <em>a</em> cos(30º) is the magnitude of the horizontal component of acceleration. The person is pulled by a force of <em>p</em> = 401 N, so solve for <em>f</em> :
(401 N) cos(30º) - <em>f</em> = (53 kg) (0.59 m/s²) cos(30º)
<em>f</em> ≈ 320 N
Answer:
322 kJ
Explanation:
The work is the energy that a force produces when realizes a displacement. So, for a gas, it occurs when it expands or when it compress.
When the gas expands it realizes work, so the work is positive, when it compress, it's suffering work, so the work is negative.
For a constant pressure, the work can be calcutated by:
W = pxΔV, where W is the work, p is the pressure, and ΔV is the volume variation. To find the work in Joules, the pressure must be in Pascal (1 atm = 101325 Pa), and the volume in m³ (1 L = 0.001 m³), so:
p = 60 atm = 6.08x10⁶ Pa
ΔV = 82.0 - 29.0 = 53 L = 0.053 m³
W = 6.08x10⁶x0.053
W = 322x10³ J
W = 322 kJ
Answer:
June Mar Fajardo
plus he plays for Phillipines men's national basketball team
and baseball is my favorite team sport because everyone has a role no matter how big or small it is
