Answer:
d = 13 miles
Explanation:
Lets say the position of court house is origin in this case
her office is located at 4 miles west and 4 miles south of court house
so here we have coordinate of the office with respect to court house is given as
now the position of her home is located at 1 miles east and 8 miles north of the court house
so the coordinates of her home is given as
now the change in the position is given as the distance between office and home
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: The smallest effort = 300N
Explanation:
Using one of the condition for the attainment of equilibrium:
Clockwise moment = anticlockwise moments
900 × 1 = 3 × M
Where M = the weight of the strong man
3M = 900
M = 900/3 = 300N
Therefore, 300N is the smallest effort that the strongman can use to lift the goat
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
<u>Where</u>:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
<u>Where:</u>
g: is gravity = 9.8 m/s²
Now, the horizontal velocity of the rock is:
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:
So, the distance is:
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!