A block is suspended from a scale and then lowered into a bucket of water. The density of the water is 1 gm/cm3. The initial rea

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.

An apple strikes the ground at a velocity of 16.2 m/s.

The angle between the velocity of the apple and a line normal to the inclined surface is 20 degrees.

The parallel and perpendicular components of its velocity concerning the surface are as follows:

This gives us:

The velocity along the inclined plane measures 5.5 m/s.

Response: a. The mirrors and eyepiece of a large telescope are designed with spring-loaded components to quickly return to a predetermined position.

Justification:

Adaptive optics refers to a technique employed by various astronomical observatories to compensate in real-time for the atmospheric turbulence that impacts astronomical imaging.

This is executed by integrating advanced deformable mirrors into the telescope's optical pathway, operated by a set of computer-controlled actuators. This allows for obtaining clearer images despite the atmospheric fluctuations that create distortions.

However, locating such stars is not always feasible, prompting the use of a strong laser beam directed at the upper atmosphere to create artificial stars.

Answer:

17 m/s south

Explanation:

Mass of the dog = 10 kg

Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

Velocity of the dog = 1 m/s

Velocity of skateboard

In this scenario, linear momentum is conserved

The speed of the skateboard post-dog jump will be 17 m/s south, considering north as the positive direction

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.