Answer:
Time - taken = 2.5 s
deceleration= -8 m/s²
Solution:
Given:
speed, v = 8 m/s
distance, d = 20m
To Find:
deacceleration = ?
As we know speed is defined as
v = d/t
plugging in the values
t = 20/ 8
t = 2.5s
Now from deceleration formula
a = - v/ t
a = - 20/ 2.5
a = - 8 m/s²
Thus, the time taken and acceleration is 2.5 s and -8 m/s²
respectively.
Learn more about deceleration here:
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Answer:
A.−2.1 × 10^10 N
Explanation:
Using the formula;
E = k Q1Q2/d²
Where;
E is the electrical force
k is the constant
Q1, Q2 are the two charges and
d is the distance between the two charges
Therefore;
E = (9 x 10^9) × (0.0042) × (-0.0050) / (0.0030)²
= -2.1 x 10^10 N
Therefore; electrical force acting between the two charges is -2.1 x 10^10 N.
Answer:
v_f = 6.92 x 10^(4) m/s
Explanation:
From conservation of energy,
E = (1/2)mv² - GmM/r
Where M is mass of sun
Thus,
E_i = E_f will give;
(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)
m will cancel out to give ;
(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)
Let's make v_f the subject;
v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]
G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²
Mass of sun is 1.9891 x 10^(30) kg
v_i = 2.1×10⁴ m/s
r_i = 2.5 × 10^(11) m
r_f = 4.9 × 10^(10) m
Plugging in all these values, we have;
v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12
v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]
v_f = √[(441000000) + (435.38 x 10^(7))
v_f = 6.92 x 10^(4) m/s
Answer: a) C decreases; b) Q stays the same; c) E is the same
d) ΔV increase
Explanation: In order to explain this problem we have to consider the following:
C=εoA/d where A and d are the area and the separation of the plates, respectively.
Increasing d, produces a decrease of C.
Q remain constant becasuse the plates are charges and the wire are isoloted each other.
We also know that ΔV=E*d where E is electric field between the plates.
And E= Q/εo*A ( a constant between the plates)
As we can see from above, ΔV depends directely of the d so if d increase ΔV also increase. To do that we have to do work on the system.
