Answer:
A rise in temperature increases the kinetic energy and speed of particles; it does not weaken the forces between them. The particles in solids vibrate about fixed positions; even at very low temperatures. Individual particles in liquids and gases have no fixed positions and move chaotically.
Explanation:
The snapshot of light as the cart moves with constant velocity is represented by a graph with uniform displacement at each time interval.
The change in displacement with time is uniform at constant velocity. The displacement of the supplied moving item grows at the same pace.
The beginning velocity equals the ultimate velocity at constant velocity.
v₁ = v₂
The object's acceleration at constant velocity is zero since the velocity change with time is zero.
As a result, we may deduce that the graph with equal displacement at each time interval reflects a snapshot of light as the cart moves at a constant speed.
A moving object's displacement-time graph shows the distance traveled by a moving item as time passes. A vector quantity is displacement. The slope or gradient of this graph represents the velocity of the item. The displacement-time graph, also known as the position-time graph, describes an object's motion. In this graph, the displacement of the moving item is displayed on the y-axis as a dependent variable, while time is shown on the x-axis as an independent variable.
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Answer:
The direction should be in the positive x direction.
Explanation:
Given that
Electric filed E= 410 V/m (In Negative y direction)
V= 9.1 x 10 ⁵ m/s
As we know that
F= E q + q.V x B
Given that protons is moving constant speed so ,F=0
F= E q +q. V x B
0= E + V x B
E = -V x B
- 410 j = - ( 9.1 x 10 ⁵ k x B i)
The direction should be in the positive x direction.
Answer:
u₂ = 32.29 m/s
Explanation:
m₁ = 12 kg
u₁ = 0 m/s
m₂ = 417 g = 0.417 kg
d = 15 cm = 0.15 m
μ = 0.40
u₂ = ?
It is an inelastic collision. After the collision
We can apply ∑ F = m*a for the system (m₁ + m₂)
where m = m₁ + m₂ = 12 kg + 0.417 kg = 12.417 kg
then
∑ F = - Ff = m*a
if
Ff = μ*N = μ*W = μ*(m*g) = μ*m*g
⇒ - μ*m*g = m*a ⇒ a = - μ*g = - 0.40*9.8 m/s² = - 3.92 m/s²
⇒ a = - 3.92 m/s²
Since vf = 0 m/s (for the system)
we use the equation
vf² = vi²+2*a*d ⇒ 0 = vi²+2*a*d ⇒ vi = √(-2*a*d)
⇒ vi = √(-2*(- 3.92 m/s²)*0.15 m)
⇒ vi = 1.0844 m/s
we can use the equation for an inelastic collision
m₁*u₁ + m₂*u₂ = (m₁ + m₂)*vs
since m = m₁ + m₂; u₁ = 0 m/s and vs = vi
we have
m₁*0 + m₂*u₂ = m*vi
⇒ u₂ = m*vi / m₂
⇒ u₂ = 12.417 kg*1.0844 m/s / 0.417 kg
⇒ u₂ = 32.29 m/s (→)