Answer:
28.7 m at 46.9°
Explanation:
The x component of the displacement is:
x = 6 m cos 0° + 25 m cos 57°
x = 19.6 m
The y component of the displacement is:
y = 6 m sin 0° + 25 m sin 57°
y = 21.0 m
The total displacement is found with Pythagorean theorem:
d = √(x² + y²)
d = 28.7 m
And the direction is found with trig:
θ = tan⁻¹(y/x)
θ = 46.9°
Answer:
a) T = 608.22 N
b) T = 608.22 N
c) T = 682.62 N
d) T = 533.82 N
Explanation:
Given that the mass of gymnast is m = 62.0 kg
Acceleration due to gravity is g = 9.81 m/s²
Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.
So;
To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;
T = mg
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs the rope at a constant rate tension in the string is
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs up the rope with an upward acceleration of magnitude
a = 1.2 m/s²
the tension in the string is T - mg = ma (Since acceleration a is upwards)
T = ma + mg
= m (a + g )
= (62.0 kg)(9.81 m/s² + 1.2 m/s²)
= (62.0 kg) (11.01 m/s²)
= 682.62 N
When the gymnast climbs up the rope with an downward acceleration of magnitude
a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)
T = mg - ma
= m (g - a )
= (62.0 kg)(9.81 m/s² - 1.2 m/s²)
= (62.0 kg)(8.61 m/s²)
= 533.82 N
Answer with Explanation:
We are given that
Mass of superman=m=78 kg
Mass of train=m'=17863 kg
Speed of train=u'=75 km/h=
Let initial speed of superman=u
Momentum=mv
Using the formula
Average horizontal force=0.58
Deceleration 
Final speed of train=v'=0
Using the formula
Using the formula
Answer:
15.45 m/s
Explanation:
F = 1.2 x 10^3 N, Fk = 615 N, u = 0, s = 25 m
Let the speed after traveling 25 m is v.
mass = F / g = 1200 / 9.8 = 122.45 kg
Net force, Fnet = 1.2 x 1000 - 615 = 1200 - 615 = 585 N
acceleration = Fnet / mass = 585 / 122.45 = 4.78 m/s^2
Use third equation of motion
v^2 = u^2 + 2 a s
v^2 = 0 + 2 x 4.78 x 25 = 238.88
v = 15.45 m/s
