In case of an object sitting at rest on another base, there are two equal and opposite forces – Normal force and the gravity.
Answer: Option A
<u>Explanation:
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When an object is placed at rest position on another object, there is a force exerted by the surfaces of the two contact objects. This force is denoted as Normal Force.
When an object such as a box is placed on a shelf, its surface exerts a contact force on the base of the shelf- The Normal force directed upward. Meanwhile, the gravity stays at its action and tries to pull the box towards itself.
Both of these forces however are equal and opposite and therefore, there is zero net force on the box. That's why it remains at rest, holding on Newton's third law.
Answer:
The evolutionary success of bats is accredited to their ability, as the only mammals, to fly and navigate in darkness by echolocation, thus filling a niche exploited by few other predators. Over 90% of all bat species use echolocation to localize obstacles in their environment by comparing their own high frequency sound pulses with returning echoes. The ability to localize and identify objects without the use of vision allows bats to forage for airborne nocturnal insects, but also for a diverse range of other food types including motionless perched prey or non-animal food items.
The agility and precision with which bats navigate and forage in total darkness, is in large part due to the accuracy and flexibility of their echolocation system. The echolocation clicks of the few echolocating Pteropodidae (Rousettus) are fundamentally different from the echolocation sounds produced in the larynx that we focus on here, and thus not part of this review. Many studies have shown that bats adapt their echolocation calls to a variety of conditions, changing duration and bandwidth of each call and the rate at which calls are emitted in response to changing perceptual demands . In recent years the intensity and directionality of echolocation signals has received increasing research attention and it is becoming evident that these parameters also play a major role in how bats successfully navigate and forage. To perceive an object in its surroundings, a bat must ensonify the object with enough energy to return an audible echo. Hence, the intensity and duration of the emitted signal act together to determine how far away a bat can echolocate an object. Equally important is signal directionality. Bat echolocation calls are directional, i.e., more call energy is focused in the forward direction than to the sides (Simmons, 1969; Shimozawa et al., 1974; Mogensen and Møhl, 1979; Hartley and Suthers, 1987, 1989; Henze and O'Neill, 1991). An object detectable at 2 m directly in front of the bat may not be detected if it is located at the same distance but off to the side. Consequently, at any given echolocation frequency and duration, it is the combination of signal intensity and signal directionality that defines the search volume, i.e., the volume in space where the bat can detect an object.
The aim of this review is to summarize current knowledge about intensity and directionality of bat echolocation calls, and show how both are adapted to habitat and behavioral context. Finally, we discuss the importance of active motor-control to dynamically adjust both signal intensity and directionality to solve the different tasks faced by echolocating bats.
Explanation:
Answer:
1. 1, 2, 4 all show some form of refraction as the bending of a light ray when passing from one media to another.
Explanation:
Number 4 is the most accurate as it also shows some light being reflected and the bending of the refracted light ray in the correct direction for going from a medium of low refractive index (air) into a higher refractive index material (crown glass)
Answer:
Explanation:
<u>Frictional Force
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When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:
The centripetal acceleration a_c is computed as
Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one
For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as
The normal force N is equal to the weight of the car, thus
Equating both forces
Simplifying
Substituting the values
Answer:
1500 mph
Explanation:
Take east to be +x and north to be +y.
The x component of the velocity is:
vₓ = 889 cos 0° + 830 cos 59°
vₓ = 1316.5 mph
The y component of the velocity is:
vᵧ = 889 sin 0° + 830 sin 59°
vᵧ = 711.4 mph
The speed is found with Pythagorean theorem:
v² = vₓ² + vᵧ²
v² = (1316.5 mph)² + (711.4 mph)²
v = 1496 mph
Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.
