Answer:
1. What is not considered a cardiovascular workout?
A. Jogging
B. Sit ups
C. Jump Ropes
D. Burpees
Explanation:
I am sorry, I am in middle school and I only have one question.
Answer:
The force of the impact would be smaller
Explanation: Examples:
If the force is big then the time would be small (2500N of Force = 10 seconds)
If the force is small then the time would be big (250N of Force = 50 seconds)
Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum
Answer:
a = 1.764m/s^2
Explanation:
By Newton's second law, the net force is F = ma.
The equation for friction is F(k) = F(n) * μ.
In this case, the normal force is simply F(n) = mg due to no other external forces being specified
F(n) = mg = 15kg * 9.8 m/s^2 = 147N.
F(k) = F(n) * μ = 147N * 0.18 = 26.46N.
Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.
Thus, F(net) = F(k) = ma
26.46N = 15kg * a
a = 1.764m/s^2
Answer:
1.98 atm
Explanation:
Given that:
Temperature = 28.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28 + 273.15) K = 301.15 K
n = 1
V = 0.500 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K
⇒P (ideal) = 49.45 atm
Using Van der Waal's equation
R = 0.0821 L atm/ K mol
Where, a and b are constants.
For Ar, given that:
So, a = 1.345 atm L² / mol²
b = 0.03219 L / mol
So,
⇒P (real) = 47.47 atm
Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm
Answer:
The potential energy stored in the spring is 0.018 J.
Explanation:
Given;
spring constant, k = 90 N/m
extension of the spring, x = 2 cm = 0.02 m
The potential energy stored in the spring is calculated as;
U = ¹/₂kx²
where;
U is the potential energy stored in the spring
Substitute the given values in the equation above;
U = ¹/₂ x 90 N/m x (0.02 m)²
U = 0.018 J
Therefore, the potential energy stored in the spring is 0.018 J.
