An archer shot a 0.06 kg arrow at a target. The arrow accelerated at 5,000 m/s2 to reach a speed of 50.0 m/s as it left the bow.

A reactant is any substance that takes part in a chemical reaction. Conversely, a product is what emerges from the chemical transformation. A familiar example of a chemical change is rust formation. In this reaction, oxygen and iron, which serve as the reactants, react to produce a substance known as iron oxide, or rust.

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C  = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = ............................1

v =

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken =

which gives us:

time taken =

so, time taken = 19.92 × seconds = 20μs.

Thus, the required time is 20 μs.

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.

In this scenario, there exists a constant electric field produced by a large sheet. This electric field can be defined as... The force acting on the ball due to this field acts horizontally, and this force must be counterbalanced by the horizontal tension component of the string to maintain equilibrium. Similarly, the vertical tension component in the string must equal the weight of the small sphere. Hence, we can derive two equations to illustrate this.

C.) The energy stored in the capacitor becomes twice its initial value.