The portion of a light ray that falls on a surface is a/an A. reflected ray.
B. diffuse reflection.
C. specular reflection.
D. incident ray.
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Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16
vB = 15.4 m/s : speed of the cart at B
Answer:
the guy above me is correct
Explanation:
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Answer:
Explanation:
As we know that the combination is maintained at rest position
So we will take net torque on the system to be ZERO
so we know that
here we will have
so we have
so we have
Final answer:
The concept of torques and equilibrium is used to calculate the pulling force on the larger flywheel, which is found to be approximately 29.55 Newtons. This force will balance the system and prevent it from rotating.
Explanation:
To solve this problem, we need to understand the concept of torque and equilibrium. We know that torque (τ) is the rotational equivalent of linear force. It's calculated by the formula τ = force × radius. Thus, for the system to stay at equilibrium (not rotate), the torques need to balance each other out.
On the smaller flywheel, the torque τ₁ is given by the pulling force (F₁ = 50 N) and the radius (r₁ = 13 cm, or 0.13 m), hence τ₁ = F₁ × r₁ = 50 N x 0.13 m = 6.5 N.m.
In order for the system to stay at equilibrium, the same amount of torque needs to be applied to the larger flywheel. We already know the radius of the larger flywheel (r₂ = 22 cm, or 0.22 m). To keep the system at equilibrium, the pulling force F₂ on the larger flywheel should be such that the torque τ₂ = τ₁ = 6.5 N.m. From the formula τ = F × r, we can solve for F₂ as follows: F₂ = τ₂ / r₂ = 6.5 N.m / 0.22 m = 29.55 N, approximately. Therefore, a pulling force of about 29.55 N should be applied to the cord connected to the larger flywheel to prevent the system from rotating.
Learn more about Torques and Equilibrium here:
#SPJ3
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Answer:
4.1. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In other words, when one object exerts a force on another object, the second object exerts an equal force in the opposite direction on the first object.
4.2. Here's a labeled free-body diagram for Block A:
```
T (tension in the string)
↑
│
│
│
│
│
F (applied force)
──→ (direction of motion)
```
In this diagram, "T" represents the tension in the string, and "F" represents the applied force at an angle of 30° to the horizontal. The arrow indicates the direction of motion.
4.3. To find the frictional force acting on block A as it accelerates, we can use Newton's Second Law:
\[F_{\text{net, A}} = m_A \cdot a\]
Where:
- \(F_{\text{net, A}}\) is the net force acting on block A.
- \(m_A\) is the mass of block A (given as 15 kg).
- \(a\) is the acceleration (given as 2.08 m/s²).
Rearranging the equation to solve for \(F_{\text{net, A}}\):
\[F_{\text{net, A}} = 15 kg \cdot 2.08 m/s² = 31.2 N\]
Now, we need to consider the frictional force, which opposes the motion and acts in the direction opposite to the applied force. So, the frictional force is 31.2 N in the opposite direction of motion, making it:
Frictional force on block A = -31.2 N
However, since you want it in magnitude, it's 31.2 N.
4.4. To calculate the mass of block B, we can use the fact that block A and block B are connected by a string, so they experience the same acceleration. Therefore, we can use the following equation:
\[F_{\text{net, B}} = m_B \cdot a\]
Where:
- \(F_{\text{net, B}}\) is the net force acting on block B, which is the tension in the string.
- \(m_B\) is the mass of block B (unknown).
- \(a\) is the acceleration (given as 2.08 m/s²).
We already calculated that the tension in the string is 31.2 N. Plugging in the values:
\[31.2 N = m_B \cdot 2.08 m/s²\]
Now, solving for \(m_B\):
\[m_B = \frac{31.2 N}{2.08 m/s²} \approx 15 kg\]
So, the mass of block B is approximately 15 kg.
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When the formation of carbon dioxide is negative, it means that it releases heat to the surroundings. When it releases heat to the surroundings, the reaction is exothermic.