Answer:
Fr = 48 [N] forward.
Explanation:
Suppose the movement is on the X axis, in this way we have the force of the engine that produces the movement to the right, while the force produced by the brake causes the vehicle to decrease its speed in this way the sign must be negative.
∑F = Fr
![F_{engine}-F_{brake} =F_{r}\\F_{r}=79-31\\F_{r}=48[N]](https://tex.z-dn.net/?f=F_%7Bengine%7D-F_%7Bbrake%7D%20%3DF_%7Br%7D%5C%5CF_%7Br%7D%3D79-31%5C%5CF_%7Br%7D%3D48%5BN%5D)
The movement remains forward, since the force produced by the movement is greater than the braking force.
The British physicist Joseph John (J. J.) Thomson (1856–1940) performed a series of experiments in 1897 designed to study the nature of electric discharge in a high-vacuum cathode-ray tube, an area being investigated by many scientists at the time. Thomson's model showed the atom as a positively charged ball of matter with negatively changed electrons floating freely around inside of it. This model showed the atom having no structure. There are also no protons and neutrons in this model. Thomson knew that the atom had positively and negatively charges particles in it he just didn't know how they were arranged. <span>Today's model gives us a much clearer picture of the atom. There is a positively charged center of the atom that is denser than the rest of it called the nucelus. This dense center is made up of positively charged protons and neutrally charged neutrons. Around the outside of the nucleus the electrons are organized on rings. These electrons are arranged in a certain pattern that is the same for all atoms.</span>
The first question would be A
The second would be either A or D
The ability of singers to shatter glass is due to the control of frequency in their voices. The sound waves will vibrate the air particles that surround the glass at its resonant frequency. The glass will break if a singer sings too loudly.
Answer:
28.3 m/s
Explanation:
From the question given above, the following data were obtained:
Angle of projection (θ) = 30°
Maximum height (H) = 10 m
Acceleration due to gravity (g) = 10 m/s²
Initial velocity (u) =?
Thus, we can obtain the minimum velocity cannon ball by using the following formula:
H = u²Sine² θ / 2g
10 = u² × (Sine 30)² / 2× 10
10 = u² × (0.5)² / 20
10 = u² × 0.25 / 20
10 = u² × 0.0125
Divide both side by 0.0125
u² = 10/ 0.0125
u² = 800
Take the square root of both side
u = √800
u = 28.3 m/s
Therefore, the minimum speed of the cannon ball is 28.3 m/s
