Key concepts
Heart rate
Exercising
The heart
Cardiovascular system
Health
Introduction
As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.
Background
Your heart is continuously beating to keep blood circulating throughout your body. Its rate changes depending on your activity level; it is lower while you are asleep and at rest and higher while you exercise—to supply your muscles with enough freshly oxygenated blood to keep the functioning at a high level. Because your heart is also a muscle, exercise, in turn, helps keep it healthy. The American Heart Association recommends that a person does exercise that is vigorous enough to raise their heart rate to their target heart-rate zone—50 percent to 85 percent of their maximum heart rate, which is 220 beats per minute (bpm) minus their age for adults—for at least 30 minutes on most days, or about 150 minutes a week in total. So for a 20-year-old, the maximum heart rate would be 200 bpm, with a target heart-rate zone of 100 to 170 bpm. (For those 19 or younger, target zones can vary more than they do for adults.)
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Answer:
kick 1 has travelled 15 + 15 = 30 yards before hitting the ground
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Explanation:
1st kick travelled 15 yards to reach maximum height of 8 yards
so, it has travelled 15 + 15 = 30 yards before hitting the ground
2nd kick is given by the equation
y (x) = -0.032x(x - 50)

we know that maximum height occurs is given as


and maximum height is

y = 20
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Answer:
Magnitude of impulse, |J| = 4 kg-m/s
Explanation:
It is given that,
Mass of cart 1, 
Mass of cart 2,
Initial speed of cart 1,
Initial speed of cart 2,
(stationary)
The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

V = 1 m/s
The magnitude of the impulse exerted by one cart on the other is given by:


J = -4 kg-m/s
or
|J| = 4 kg-m/s
So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is

and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

Answer:
Explanation:
Let the magnetic field be B = B₁i + B₂j + B₃k
Force = I ( L x B ) , I is current , L is length and B is magnetic field .
In the first case
force = - 2.3 j N
L = 2.5 i
puting the values in the equation above
- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]
= - 20 B₃ j + 20 B₂ k
comparing LHS and RHS ,
20B₃ = 2.3
B₃ = .115
B₂ = 0
In the second case
L = 2.5 j
Force = I ( L x B )
2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )
= - 20 B₁ k + 20B₃ i
2.3i−5.6k = - 20 B₁ k + 20B₃ i
B₃ = .115
B₁ = .28
So magnetic field B = .28 i + .115 B₃
Part A
x component of B = .28 T
Part B
y component of B = 0
Part C
z component of B = .115 T .