Answer:
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Explanation:
The distance travelled on the rough ice is equal to the width of the rough ice.
distance d = 5.0 m
Initial speed u = 9.2 m/s
Final speed v = 5.8 m/s
The time taken to move through the rough ice can be calculated using the equation of motion;
d = 0.5(u+v)t
time t = 2d/(u+v)
Substituting the given values;
t = 2(5)/(9.2+5.8)
t = 2/3 = 0.66667 second
The acceleration is the change in velocity per unit time;
acceleration a = ∆v/t
a = (v-u)/t
Substituting the values;
a = (5.8-9.2)/0.66667
a = -5.099974500127
a = -5.10 m/s^2
her acceleration on the rough ice is -5.10 m/s^2
Answer:
- Decreasing the resistance
- Using a shorter length
- Using a smaller area wire
Explanation:
Formula for conductance in wires is;
G = 1/R
Where;
G is conductance
R is resistance
This means that increasing the resistance leads to a larger denominator and thus a smaller conductance but to decrease the denominator means larger conductance.
Thus, to increase the conductance, we have to decrease the resistance.
Resistance here has a formula of;
R = ρL/A
Where;
ρ is resistivity
L is length of wire
A is area
Thus, to decrease the resistance, we will have to use a shorter length and smaller area of wire.
Answer:
(4) 8.5 m/s
Explanation:
You add both the meters together and both the seconds together and then divide them both.
Answer:
thanks for da 5points hoi
Explanation: thanks dawg
