Answer:
(a) Maximum current through resistor is 1.43 A
(b) Maximum charge capacitor receives is
.
Explanation:
(a)
In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by

Here,
is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).
The maximum current is given by

<em>V</em> is the emf of the battery and
is the effective resistance.
In this question,
= 10.0 Ω + 25.0 Ω = 35.0 Ω

(b) The maximum charge is given
<em>Q</em> = <em>CV</em>
where <em>C</em> is the capacitance of the capacitor

Answer:
Final speed of car = 12 m/s
Explanation:
We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.
a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s
v = ?
u = 0 m/s
a = 4.0 m/s²
t = 5 s
v = u + at = 0 + 4 x 5 = 20 m/s
b) Then maintains that velocity for 10 s
v = ?
u = 20 m/s
a = 0 m/s²
t = 10 s
v = u + at = 20 + 0 x 10 = 20 m/s
c) Then decelerates at the rate of 2.0 m/s² for 4.0 s
v = ?
u = 20 m/s
a = -2.0 m/s²
t = 4 s
v = u + at = 20 + -2 x 4 = 12 m/s
Final speed of car = 12 m/s
Answer:
<em>B</em><em>.</em><em> </em><em>Kinetic</em><em> </em><em>friction</em><em> </em>
Explanation:
This is definitely the correct answer because kinetic friction acts when an object is in motion and it allows the object to move without slipping, etc
<em>ALSO</em><em>,</em><em> </em><em>PLEASE DO</em><em> </em><em>MARK</em><em> </em><em>ME AS</em><em> </em><em>BRAINLIEST UWU</em><em> </em>
<em>Bonne</em><em> </em><em>journée</em><em> </em><em>;</em><em>)</em><em> </em>
Answer:
d). The value of y should be -32m
Vx=0.92 m/s
Explanation:
Using equation of motion uniform to y motion

So to find t that is the same time for all the motion

The value of Xf=-3.2m because the g is negative from the axis
Now in the axis 'x' to find Vx

<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.