Answer:
1.35 kJ
Explanation:
KE = ½mv² = ½ × 0.030 kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ
Unscrambling
1. resting heart rate
2. overload
3. workout
4. specificity
5. cool-down
6. progression
7. warm-up
8. the last one can only be instance, but there was a typo on the paper.
The one fact that needs to be mentioned but isn't given anywhere on or around the graph is: The distance, on the vertical axis, is the distance FROM home. So any point on the graph where the distance is zero ... the point is in the x-axis ... is a point AT home.
Segment D ...
Walking AWAY from home; distance increases as time increases.
Segment B ...
Not walking; distance doesn't change as time increases.
Segment C ...
Walking away from home, but slower than before; distance increases as time increases, but not as fast. Slope is less than segment-D.
Segment A ...
Going home; distance is DEcreasing as time increases. Walking pretty fast ... the slope of the line is steep.
Answer:
Friction is useful in some cases like walking and cycling ..
but it is unwanted in machines as it create unwanted sounds and heat .,due to which we loss energy
Explanation:
mark me as brainliest ❤️
Answer:
a) 4 289.8 J
b) 4 289.8 J
c) 6 620.1 N
d) 411 186.3 m/s^2
e) 6 620.1 N
Explanation:
Hi:
a)
The kinetic energy of the bullet is given by the following formula:
K = (1/2) m * v^2
With
m = 16.1 g = 1.61 x 10^-2 kg
v = 730 m/s
K = 4 289.8 J
b)
the work-kinetic energy theorem states that the work done on a system is the same as the differnce in kinetic energy of the same. Since the initial state of the bullet was at zero velocity (it was at rest) Ki = 0, therefore:
W = ΔK = Kf - Ki = 4 289.8 J
c)
The work done by a force is given by the line intergarl of the force along the trayectory of the system (in this case the bullet).
If we consider a constant force (and average net force) directed along the trayectory of the bullet, the work and the force will be realted by:
W = F * L
Where F is the net force and L is the length of the barrel, that is:
F = (4 289.8 J) / (64.8 cm) = (4 289.8 Nm) / (0.648 m) = 6620.1 N
d)
The acceleration can be found dividing the force by the mass:
a = F/m = (6620.1 N) /(16.1 g) = 411 186.3 m/s^2
e)
The force will have a magnitude equal to c) and direction along the barrel towards the exit
