Answer:
The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is <u>10°.</u>
Explanation:
Given:
Mass of the driver is, 
Radius of circular turn is, 
Linear speed of the car is, 
Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.
The forces are:
1. Weight = 
2. Centripetal force, 'F', which is given as:
Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,
°
Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.
Its a waste of time, you have to not only write it down, but study it after too . other than that notes are great.
Answer:
The answer is given below
Explanation:
u is the initial velocity, v is the final velocity. Given that:
a)
The final velocity of cart 1 after collision is given as:
The final velocity of cart 2 after collision is given as:
b) Using the law of conservation of energy:
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
A point charge is located at the origin of a coordinate system. A positive charge is brought in from infinity to a point. The charges are at distance for given electrical potential energy is 3.34 x 10⁷ m.
<h3>What is electric potential energy?</h3>
The electric potential energy is the work done by a test charge to bring it from infinity to a particular location.
The electric potential energy is given by the relation,
V = kQ/r
where k = 9 x 10⁹ J.m/C ,Q = 3 x 10⁻⁹ C, V =8.09 × 10⁻⁷ J.
Substitute the values into the expression to get the distance between the charges.
8.09 × 10⁻⁷ = 9 x 10⁹ x 3 x 10⁻⁹ / r
r =3.34 x 10⁷ m
Thus, the distance between the charges will be 3.34 x 10⁷ m.
Learn more about electric potential energy.
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