A long plastic rod of 20-mm diameter (k = 0.3 W/m · K and rhocp = 1040 kJ/m3 · K) is uniformly heated in an oven as preparation

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law;

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

D. Value of the centripetal force

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + a

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + (2.25 x )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

= + 2gs

= + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying = + 2gs

= + 2(9.81 x 328)

= 0 + 6435.36

v = = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.