Answer:
250N
Explanation:
According to newton second law,
\sumF = ma
Fm - Ff = ma
Since the velocity is constant, a = 0m/s
Frictional force Ff = 250N
Substitute
Fm - 250 = m(0)
Fm - 250 = 0
Fm = 250N
Hence the force to keep the box sliding at constant speed is 250N
Answer:
F/2
Explanation:
In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant
Hence;
F= KQ1Q2/r^2 ------(1)
Where the charge on Q1 is doubled and the distance separating the charges is also doubled;
F= K2Q1 Q2/(2r)^2
F2= 2KQ1Q2/4r^2 ----(2)
F2= F/2
Comparing (1) and (2)
The magnitude of force acting on each of the two particles is;
F= F/2
This is a uniform rectilinear motion (MRU) exercise.
To start solving this exercise, we obtain the following data:
<h3><u>
Data:</u></h3>
- v = 4.6 m/s
- d = ¿?
- t = 10 sec
To calculate distance, speed is multiplied by time.
We apply the following formula: d = v * t.
We substitute the data in the formula: the <u>speed is equal to 4.6 m/s,</u> the <u>time is equal to 10 s</u>, which is left as follows:
Therefore, the speed at 10 seconds is 46 meters.
Answer: 4.77m/s
Explanation:
According to the law of conservation of momentum which states that the sum total of momentum of bodies before collision is equal to the sum of their momentum after collision. Note that the two bodies will move at a common velocity after colliding.
Let m1 and m2 be the mass of the first and second railroad cars
u1 and u2 be the velocities of the railroad cars
v be the common velocity
Using the formula
m1u1 + m2u2 = (m1 +m2)
m1 = 1.20×10⁴kg
m2 = 1.20×10⁴kg (body of same mass)
u1 = 7.70m/s
u2 = 1.84m/s
v = ?
(1.20×10⁴×7.7) + (1.20×10⁴×1.84) = (1.20×10⁴ + 1.20× 10⁴)v
9.24×10⁴ + 2.21×10⁴ = 2.4×10⁴v
11.45×10⁴ = 2.4×10⁴v
v = 11.45×10⁴/2.4×10⁴
v = 4.77m/s
The velocity of the cars after collision will be 4.77m/s
Elliptical, as shown by most projections
