If you hold your arm outstretched with palm upward, the force to keep your arm from falling comes from your deltoid muscle. assu

. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be

Softa [3030]

A 40 kg child throws a 0.5 kg stone at a velocity of 5 m/s. To find the recoil, we apply the conservation of momentum formula: m1•v1 + m2•v2 = 0, where m1 is the mass of the child, and v1 is the child's recoil velocity. Applying the known values results in 40•v1 = -0.5 × 5, leading to v1 = -2.5 / 40, which simplifies to v1 = -0.0625 m/s. Thus, the child's recoil speed is 0.0625 m/s.

6 0

2 months ago

A tennis player who is recovering from an ankle injury and is not allowed to change directions can maintain her cardio fitness l

Softa [3030]

A tennis player recovering from an ankle injury, restricted from pivoting, can keep her cardiovascular fitness by using the rowing machine, pedaling on a stationary bike with one leg, or swimming. These exercises do not necessitate directional changes and are safe for her injury.

7 0

2 months ago

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A rigid tank is divided into two equal parts by a partition. Initially one side of the tank contains 5 kg of water at 200 kPa an

Ostrovityanka [3204]

In the system analyzed, there is no heat transfer as the temperature remains constant at 25°C before and after the interaction. The heat exchange can be expressed by the equation Q = mCΔT, indicating that if ΔT equals zero, there is evidently no heat gain or loss. Considering the vacuum side of the tank, where m equals 0, we find Q equals 0 KJ, signifying that the system experienced no heat exchange.

8 0

2 months ago

A crow drops a 0.11kg clam onto a rocky beach from a height of 9.8m. What is the kinetic energy of the clam when it is 5.0m abov

kicyunya [3294]

Solution:

The kinetic energy of the clam at an elevation of 5.0 m is 5.19 J and the velocity of the clam at that height is 9.71 m/s.

Explanation:

Throughout its motion, mechanical energy remains constant. We understand that mechanical energy is the summation of potential energy and kinetic energy. Potential energy = $m \times g \times h$ , Kinetic energy = $\frac{1}{2} \times m \times v^{2}$ and Mechanical energy = $m \times g \times h+\frac{1}{2} \times m \times v^{2}$ Initial kinetic energy is zero. At a height of 9.8 m, the mechanical energy of the clam with a mass of 0.11 kg and g=9.81 $\frac{m}{s^{2}}$ is calculated as follows: 0.11×9.81×9.8 = 10.58 J.

Mechanical energy of the clam at a height of 5.0 m = $0.11 \times 9.81 \times 5+\frac{1}{2} \times m \times v^{2}$ = $5.39+\frac{1}{2} \times m \times v^{2}$ . Given that mechanical energy is conserved, we can state that the mechanical energy of the clam at a height of 9.8 m is equal to that at 5.0 m. The representation is as follows:

10.58 = $5.39+\frac{1}{2} \times m \times v^{2}$ 10.58 – 5.39 = $\frac{1}{2} \times m \times v^{2}$ 5.19 = $\frac{1}{2} \times m \times v^{2}$ the clam's kinetic energy measures 5.19 J.

Lastly, the speed of the clam at 5.0 m is computed; thus, 5.19 = $\frac{1}{2} \times 0.11 \times v^{2} \frac{5.19 \times 2}{0.11}=v^{2}$ 94.36 = $v^{2} \sqrt{94.36}=v \quad v$ = 9.71 m/s. The clam's speed is determined to be 9.71 m/s.

6 0

2 months ago