The impulse imparted by the wall is -2 kg·m/s. The negative sign indicates a change in direction due to the rebound of the ball.
To determine the impulse imparted by the wall, we can use the principle of conservation of momentum. The impulse is equal to the change in momentum experienced by the ball.
The momentum of an object is given by the product of its mass and velocity:
Momentum = mass × velocity
Given:
Mass of the ball (m) = 0.10 kg
Initial velocity of the ball (v₁) = 10 m/s
Final velocity of the ball (v₂) = -10 m/s (negative sign indicates a change in direction)
The initial momentum of the ball is:
Initial momentum = m × v₁ = 0.10 kg × 10 m/s = 1 kg·m/s
The final momentum of the ball is:
Final momentum = m × v₂ = 0.10 kg × (-10 m/s) = -1 kg·m/s
The change in momentum is the difference between the final and initial momentum:
Change in momentum = Final momentum - Initial momentum = (-1 kg·m/s) - (1 kg·m/s) = -2 kg·m/s
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A projectile is launched horizontally from a height of 8. 0 m. The projectile travels 6. 5 m before hitting the ground. The velocity of the projectile the moment it was launched, rounded to the nearest hundredth, is m/s.
The initial velocity of a projectile launched horizontally can be calculated using the equation of distance covered horizontally (x) = Initial velocity (u) Time of flight (t). The horizontal component of the initial velocity can be determined by x = u t, t = 1.63 s, x = 6.5 mu = x / t = 6.5 m / 1.63 su = 3.99 m/s 4.00 m/s.
The initial velocity of the projectile that was launched horizontally can be calculated using the equation below: Distance covered horizontally (x) = Initial velocity (u) × Time of flight (t) where, Time of flight (t) can be found using the formula below: t = [2 × vertical height (h)] / g where ,g is the acceleration due to gravity = 9.8 m/s².The vertical height (h) of the projectile is 8.0 m. So the time of flight of the projectile will bet = [2 × 8.0 m] / 9.8 m/s²t = 1.63 s Therefore, the horizontal component of the projectile’s initial velocity can be determined by: x = u × tt = 1.63 s, x = 6.5 mu = x / t = 6.5 m / 1.63 su = 3.99 m/s ≈ 4.00 m/s. So, the projectile was launched horizontally with a velocity of 4.00 m/s (rounded to the nearest hundredth).Content loaded: The term “content loaded” is used to indicate that the contents of a webpage or app have finished loading and are ready for viewing or use.
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A stevedore slides a crate along a dock with a 50 kg horizontal force of 175 N. The opposing force of friction is 120 N. If started from rest, what is the crates’s final velocity after 0.5s?
To determine the crate's final velocity after 0.5 seconds, we can use the concept of Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
In this scenario, the stevedore applies a horizontal force of 175 N to move the crate along the dock. However, there is also an opposing force of friction acting in the opposite direction, which has a magnitude of 120 N. The net force is the difference between these two forces, so we can calculate it as follows:
Net force = Applied force - Frictional force
Net force = 175 N - 120 N
Net force = 55 N
Now, using Newton's second law of motion, we can determine the acceleration of the crate. Rearranging the equation, we have:
Net force = mass * acceleration
55 N = 50 kg * acceleration
Solving for acceleration:
acceleration = 55 N / 50 kg
acceleration = 1.1 m/s²
Since we know the initial velocity of the crate is zero (as it starts from rest), and we want to find the final velocity after 0.5 seconds, we can use the equation of motion:
final velocity = initial velocity + (acceleration * time)
Plugging in the values:
final velocity = 0 + (1.1 m/s² * 0.5 s)
final velocity = 0.55 m/s
Therefore, the crate's final velocity after 0.5 seconds is 0.55 m/s. This means that after being subjected to a 175 N force and experiencing 120 N of friction, the crate gains a velocity of 0.55 m/s in the direction of the applied force.
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Steam burns are pretty dangerous because there's often a lot of
thermal energy in the steam. When the steam hits you, it is going to
transfer some of that energy to you, cooling the steam. If a 6.4 kg cloud of
steam at 150 degrees, hits you and cools to 100 degrees, What is the loss
of heat energy? (no scientific notation)
The loss of heat energy when a 6.4 kg cloud of steam at 150 degrees Celsius hits you and cools to 100 degrees Celsius is 13,440,000 Joules.
To calculate the heat energy loss, we can use the formula:
Q = mcΔT
Where Q represents heat energy, m is the mass of the steam cloud (6.4 kg), c is the specific heat capacity of water (4,186 J/kg°C), and ΔT is the change in temperature (150°C - 100°C = 50°C).
Plugging in the values, we have:
Q = (6.4 kg) × (4,186 J/kg°C) × (50°C)
Q = 13,440,000 Joules
Therefore, the loss of heat energy when the steam cools from 150°C to 100°C is 13,440,000 Joules.
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