The work done to stop the car is -208.33 kJ
From work-kinetic energy principles, the change in kinetic energy of the car ,ΔK equals the work done to stop the car, W.
W = ΔK = 1/2m(v'² - v²) where
- m = mass of car = 1500 kg,
- v = initial velocity of car = 60 km/h = 60 × 1000 m/3600 s = 16.67 m/s and
- v' = final velocity of car = 0 m/s (since the car stops).
<h3 /><h3>Calculating the work done</h3>
Substituting the values of the variables into the equation, we have
W = 1/2m(v'² - v²)
W = 1/2 × 1500 kg(v(0 m/s)² - (16.67 m/s)²)
W = 750 kg(0 (m/s)² - 277.78 (m/s)²)
W = 750 kg(- 277.78 (m/s)²)
W = -208333.33 J
W = -208.33333 kJ
W ≅ -208.33 kJ
So, the work done to stop the car is -208.33 kJ
Learn more about work done here:
brainly.com/question/9821607
Answer:
c
Explanation:
Solution) It is option (c)
It will move to bottom of the screen
We know that the direction of electric field is the direction of in which test positive charged is moved. Therefore, electron having negative charged is moved in the direction opposite to electric field.
Hence, it will move to the bottom of the screen
Answer:
i. -4m
ii. 20m
Explanation:
The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)
Total distance = 8m going east + 8m back to origin + 4m west = 20m
Answer:
1.76m/s²
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 65m/s
Distance traveled = 1200m
Unknown:
Acceleration = ?
Solution:
This is linear velocity and we apply the appropriate motion equation to solve this problem.
V² = U² + 2as
S is the distance
u is the initial velocity
V is the final velocity
a is the acceleration
Now, insert the parameters and solve;
65² = 0² + 2 x a x 1200
4225 = 2400a
a = 1.76m/s²
Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.
