hey you look nice (pic).
According to Newton’s first law, if no force is applied to a ball, it will continue moving at the same speed and direction as it did before. When we put the ball on the grass it stays in its place, namely it stays in zero motion since no force is applied to it. However, after we kick the ball, it will continue moving in the direction we kicked it. Its speed will drop gradually, due to friction (a force applied on the ball in the opposite direction to its motion), but the direction of its motion will remain the same.
According to Newton’s second law, a force applied to an object changes that object’s acceleration – namely, the rate at which the speed of the object changes. When we kick the ball, the force we apply to it causes it to accelerate from a speed of 0 to a speed of dozens of kilometers per hour. When the ball is released from the foot, it begins to decelerate (negative acceleration) due to the force of friction that is exerted upon it (as we observed in the previous example). If we were to kick a ball in outer space, where there is no friction, it would accelerate during the kick, and then continue moving at a constant speed in the direction that we kicked at, until it hits some other object or another force is applied to it.
A. It is a virtual image.
B. It is upside down.
C. It is larger than the actual bird.
D. It is capable of being projected.
The most likely answer is A. It is a virtual Image (on Edge2020)
Answer:
499.73 m
Explanation:
We'll begin by calculating the change in the temperature. This can be obtained as follow:
Initial temperature (T₁) = 25 °C
Final temperature (T₂) = –20° C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = –20 – 25
ΔT = –45 °C
Finally, we shall determine the length on a cold day at –20° C. This can be obtained as follow:
Length at 25°C (L₁) = 500 m
Coefficient of linear expansion (α) = 12×10¯⁶ /°C
Change in temperature (ΔT) = –45 °C
Length at –20 °C (L₂) =?
α = L₂ – L₁ / L₁ΔT
12×10¯⁶ = L₂ – 500 / (500 × –45)
12×10¯⁶ = L₂ – 500 / –22500
Cross multiply
L₂ – 500 = 12×10¯⁶ × –22500
L₂ – 500 = –0.27
Collect like terms
L₂ = –0.27 + 500
L₂ = 499.73 m
Therefore, the length on a cold day at –20° C is 499.73 m.
A mass balance is used to measure the mass in an object.
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated