Answer:
The answer to your question is
Explanation:
Data
mass = 0.5kg
T1 = 35
T2 = ?
Q = - 6.3 x 10⁴ J = - 63000 J
Cp = 4184 J / kg°C
Formula
Q = mCp(T2 - T1)
T2 = T1 + Q/mCp
Substitution
T2 = 35 - 63000/(0.5 x 4184)
T2 = 35 - 63000/2092
T2 = 35 - 30.1
T2 = 4.9 °C
Start by facing East. Your first displacement is the vector
<em>d</em>₁ = (225 m) <em>i</em>
Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,
<em>d</em>₂ = (350 m) <em>j</em>
Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,
<em>d</em>₃ = (125 m) (cos(60º) <em>i</em> + sin(60º) <em>j</em> )
<em>d</em>₃ ≈ (62.5 m) <em>i</em> + (108.25 m) <em>j</em>
The net displacement is
<em>d</em> = <em>d</em>₁ + <em>d</em>₂ + <em>d</em>₃
<em>d</em> ≈ (287.5 m) <em>i</em> + (458.25 m) <em>j</em>
and its magnitude is
|| <em>d</em> || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m
Answer:
yes
Explanation:
objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium
a = 3.09 m/s²
<h3>Explanation</h3>
This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.
In the <em>timeless</em> suvat equation,

where
is the acceleration of the car;
is the <em>final</em> velocity of the car;
is the <em>initial</em> velocity of the car; and
is the displacement of the car.
Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.
In this question,
Apply the <em>timeless</em> suvat equation:
.
The value of
is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.
Answer:
Angular frequency will increase
No change in the amplitude
Explanation:
At extreme end of the SHM the energy of the SHM is given by

here we know that

now at the extreme end when one of the mass is removed from it
then in that case the angular frequency will change

So angular frequency will increase
but the position of extreme end will not change as it is given here that the top block is removed without disturbing the lower block
so here no change in the amplitude