A car traveling at 50 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 61 cm (with respect to the
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Torque can cause the angular momentum vector to rotate in UCM. This motion is called _Conservation of Angular momentum__________.
Answer:
Conservation of Angular momentum
Explanation:
The motion of an object in a circular path at constant speed is known as uniform circular motion (UCM). An object in UCM is constantly changing direction, and since velocity is a vector and has direction, you could say that an object undergoing UCM has a constantly changing velocity, even if its speed remains constant.
The law of conservation of angular momentum states that when no external torque acts on an object, no change of angular momentum will occur.
Key Points
When an object is spinning in a closed system and no external torques are applied to it, it will have no change in angular momentum.
The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation.
If the net torque is zero, then angular momentum is constant or conserved.
Angular Momentum
The conserved quantity we are investigating is called angular momentum. The symbol for angular momentum is the letter L. Just as linear momentum is conserved when there is no net external forces, angular momentum is constant or conserved when the net torque is zero. We can see this by considering Newton’s 2nd law for rotational motion:
τ→=dL→dt, where
τ is the torque. For the situation in which the net torque is zero,
dL→dt=0.
If the change in angular momentum ΔL is zero, then the angular momentum is constant; therefore,
⇒
L =constant
L=constant (when net τ=0).
This is an expression for the law of conservation of angular momentum.
Example and Implications
An example of conservation of angular momentum is seen in an ice skater executing a spin, The net torque on her is very close to zero,
because (1) there is relatively little friction between her skates and the ice, and (2) the friction is exerted very close to the pivot point.
Conservation of angular momentum is one of the key conservation laws in physics, along with the conservation laws for energy and (linear) momentum. These laws are applicable even in microscopic domains where quantum mechanics governs; they exist due to inherent symmetries present in nature.
The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
<h3>What is the specific heat capacity?</h3>The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
<h3 />Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.
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Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of , and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get
Since we have
Replacing in [2]
A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it