Answer:
1.68 s
Explanation:
From newton's equation of motion,
a = (v-u)/t.................................. Equation 1
Making t the subject of the equation
t =(v-u)g............................. Equation 2
Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.
Note: Taking upward to be negative and down ward to be positive,
Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²
t = (0-8.20)/-9.8
t = -8.20/-9.8
t = 0.84 s.
But,
T = 2t
Where T = time taken for the bowling pin to return to the juggler's hand.
T = 2(0.84)
T = 1.68 s.
T = 1.68 s
1.
m = mass of Mr. Ure = 65 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Ure is given as
F = mg
F = 65 x 9.8
F = 637 N
2.
F = force of gravity on car = 3050 N
m = mass of the car = ?
g = acceleration due to gravity = 9.8 m/s²
force of gravity on car is given as
F = mg
3050 = m (9.8)
m = 3050/9.8
m = 311.22 kg
3.
m = mass of Mr. Rees = 90 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Rees is given as
F = mg
F = 90 x 9.8
F = 882 N
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;

Therefore, the electric field strength at the mid-point between the two rings is zero.
Answer and Explanation:
If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.
- Combine 50cm and 30cm to get 80 cm.
For displacement, the answer is <u>20 cm.</u>
- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)
<em><u>#teamtrees #PAW (Plant And Water)</u></em>
<em><u>I hope this helps!</u></em>
Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.
Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is
Square root of (19.6h) meters per second.
It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.