We need to considerate only the horizontal component of the motion of the toy car.
The formula for the distance in a decelerated motion is:
s = s₀ + v₀·t - 1/2·a·t²
where:
s₀ = initial position = 0
v₀ = initial velocity = 1.21 m/s
t = time elapsed = 0.342 s
a = deceleration = 0.131 m/s²
Plugging in numbers:
s = 0 + 1.21×0.342 - 0.5×0.141×(0.342)²
= 0.406 m
Hence, the toy car traveled a distance of about 41 cm.
Answer: 90 m/s
Explanation:
Given
mass of racecar 
velocity of racecar 
mass of still honeybadger 
after collision race car is traveling at a speed of 
conserving linear momentum
![Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]](https://tex.z-dn.net/?f=Mu%2Bm%5Ctimes0%3DMv_1%2B%20mv_2%5Cquad%5Bv_2%3D%5Ctext%7Bvelocity%20of%20honeybadger%20after%20colllision%7D%5D)


Answer:
Ro = 133 [kg/m³]
Explanation:
In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

where:
m = mass [kg]
V = volume [m³]
We will convert the units of length to meters and the mass to kilograms.
L = 15 [cm] = 0.15 [m]
t = 2 [mm] = 0.002 [m]
w = 10 [cm] = 0.1 [m]
Now we can find the volume.
![V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]](https://tex.z-dn.net/?f=V%20%3D%200.15%2A0.002%2A0.1%5C%5CV%20%3D%200.00003%20%5Bm%5E%7B3%7D%20%5D)
And the mass m = 4 [gramm] = 0.004 [kg]
![Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]](https://tex.z-dn.net/?f=Ro%20%3D%200.004%2F0.00003%5C%5CRo%20%3D%20133%20%5Bkg%2Fm%5E%7B3%7D%5D)
Answer:
The height of the building is 8,302.5 m
Explanation:
Given;
velocity of the projectile, u = 36 m/s
time of motion, t = 45 s
Let the upward direction of the bullet be negative,
The height of the building is calculated as;

Answer:
y = 80.2 mille
Explanation:
The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening
θ = 1.22 λ/ d
in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m
θ = 1.22 550 10⁻⁹ / 0.002
θ = 3.355 10⁻⁴ rad
Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi
tan θ = y / L
y = L tan θ
y = 2,389 10⁵ tan 3,355 10⁻⁴
y = 8.02 10¹ mi
y = 80.2 mille
This is the smallest size of an object seen directly by the eye