A. A car moving at a constant speed on a flat, straight road. B. A vehicle traveling at a steady speed on a 10-degree incline. An object operates within an inertial reference frame if there is no net force acting upon it. According to Newton's second law, this implies that the object's acceleration also equals zero. Assessing the scenarios yields: A. A car moving at a constant speed on a flat road qualifies as an inertial reference frame, since its velocity and direction remain unchanged; thus, acceleration is zero. B. A car moving steadily up a 10-degree incline still constitutes an inertial reference frame, for similar reasons. C. A car accelerating after departing a stop sign does not represent an inertial frame due to its change in speed. D. A car driving at a steady speed around a curve cannot be considered an inertial reference frame since its direction is changing, resulting in a change in velocity and thus acceleration. Therefore, options A and B are correct.
(B) (length)/(time³) Explanation: The term x = ½ at² + bt³ should meet dimensional consistency. This means that both bt³ and ½ at² must share the same units, which are length. To find the dimension of b, we rearrange the equation: [x] = [b]*[t]³ leads to length = [b]*time³, hence [b] = length/time³.
Answer:
Baseball mass: 
Explanation:
Circumference of a baseball is calculated using 2πr = 23 cm
Thus, the radius comes out to be 3.66 cm, which equals 
m
The mass density of the baseball matches that of a neutron or proton.
Proton mass = 
kg
Proton diameter = 
m
Proton radius = 
m
Volume of the baseball is 
Now by substituting all values into the mass per unit volume equation for the baseball, we get:
Therefore, the baseball mass amounts to
Answer:
11109.825 N
Explanation:
Provided Information:
mass = m = 1510 kg
initial acceleration (a) = 0.75g (where g = 9.81 m/s²)
Using the formula F=ma
= (1510)*(0.75*9.81)
= 11109.825 N
<span>We will apply the momentum-impulse theorem here. The total momentum along the x-direction is defined as p_(f) = p_(1) + p_(2) + p_(3) = 0.
Therefore, p_(1x) = m1v1 = 0.2 * 2 = 0.4. Additionally, p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3, where v3 represents the unknown speed and m3 signifies the mass of the third object, which has an unspecified velocity.
In the same way, for the particle of 235g, the y-component of the total momentum is described with p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
Thus, p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3, where m3 is the mass of the third piece.
Consequently, p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; yielding v3 = 0.4/-0.1 = - 4.
Similarly, p_(fy) = 0.3525 + 0.1v3; thus v3 = - 0.3525/0.1 = -3.525.
Therefore, the x-component of the speed of the third piece is v_3x = -4 and the y-component is v_3y = 3.525.
The overall speed is calculated as follows: resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
