Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2
20N/4 = 5N
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Answer:
The image distance is 30 cm
image height = - 5 cm
Explanation:
The formula for calculating the image distance is expressed as
1/f = 1/u + 1/v
where
f is the focal length
u is the object distance
v is the image distance
From the information given,
u = 30
f = 15
By substituting these values into the formula,
1/15 = 1/30 + 1/v
1/v = 1/15 - 1/30 = (2 - 1)/30 = 1/30
Taking the reciprocal of both sides,
v = 30
The image distance is 30 cm
magnification = image height/object height = - v/u
Given that object height = 5 cm, then
image height/5 = - 30/30 = - 1
image height = - 5 * 1
image height = - 5 cm
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
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So, the answer is B.
