An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following pre

Answer:

Let the Volvo's speed upon braking be v, and its mass be m1.

The speed of the Volvo just before the crash is represented as v1.

After applying the brakes, the skidding distance is noted as 30 m.

The equation governing this scenario is v1^2- v^2 = - 2 * a * s, where a= represents deceleration and s= denotes skidding distance.

Assuming the deceleration to be -0.3 g = - 2.94 m/s^2 (as provided by the Volvo website for the car's emergency system).

The equation can be modified to solve for v: v^2 = v1^2 + (2 * a * s )= v1^2 + 176.4.

After the collision, both vehicles merge and proceed together (m1+m2) at an initial speed of V (let's say).

They then move 12.25 m and eventually come to a stop.

Using the equation (0)^2 -(V)^2 = - 2 * a' * s', where a' signifies the combined deceleration and s' denotes the distance traveled by the two cars.

Let's assume the combined system’s deceleration is notably higher at (- g).

Utilizing the format V^2 = 2 * 9.8 * 12.25.

This results in V = 15.5 m/s.

Using momentum conservation in the X direction (East): m1 * v1 = (m1+ m2) * V * cos 15.

Solving for v1 gives: v1 = (1650+ 975) Kg * 15.5 m/s * cos 15 /1650 Kg = 23.8 m/s.

Hence, v^2 = V1^2 + 176.4 = (23.8)^2 + 176.4.

This results in v = 27.3 m/s = 61.1 mph (indicating that he exceeded the speed limit when applying the brakes).

In the Y-direction momentum conservation shows: m2 * v2 = (m1+m2) * V * sin 15.

Solving for v2 results in: v2 = (m1+m2) * V * sin 15/m2.

We conclude with v2 = 41.7 m/s (93.3 m/hr) indicating he also exceeded the speed limit.

Answer:

The first experiment measures inertial mass, while the second experiment measures gravitational mass.

Explanation:

A student conducts two different experiments to observe resistance to changes in motion, both when at rest and in motion.

In the initial experiment, an object is forcefully pushed against a flat surface while its speed is tracked by a sensor. This setup involves work done against the object's inertia, identifying the mass as inertial mass.

Conversely, in the subsequent experiment, the object is lifted or thrown upward with an applied force and the speed is recorded. Here, the mass refers to gravitational mass, as the work performed combats gravity or the object's weight.

To tackle this question we will apply the kinematic equations that describe the motion of a projectile, where both maximum height and distance traveled are defined. This scenario demonstrates a lion achieving a height (H) of 3m and covering a horizontal distance (R) of 10m. The equations governing this kind of motion are expressed as follows:

By dividing the two equations, we determine:

Plugging in the values for H and R yields:

After substituting \theta into the relevant equation, we find:

In conclusion, the mountain lion's launch speed upon takeoff is approximately 9.98m/s at an angle of 50.2°.

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

T1 = 525 K

T2 = 275 K

It is known that

n and v are constant on both sides. Therefore we have

..............1

let the final pressure be P and the temperature

..................2

similarly

.............3

divide equation (2) by equation (3)

thus, the left side temperature equals 1.48 times the right side temperature