Let's check the relationship
So
- Raindrops will fall faster . .
- Also walking on ground would become more difficult as g increases.
Option C is wrong by now .Let's check D once
- So time period of simple pendulum would decrease.
car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw
ard towards car 2 at a velocity of 1.5 m/s. calculate the force of Car 3 and Car 2. dunno if I phrased this right so I included a screenshot too. pls help asap will give brainliest
1 answer:
Answer:
The force of car 3 on car 2 ≈ 1810.82 N
Explanation:
The equation for the change in momentum of the two cars are;
Conservation of linear momentum
150( 2.2 - v) = 265(1.5-v)
150 × 2.2 - 265×1.5 = (150+265)v
150 × 2.2 - 265×1.5 = -67.5 = 415×v
∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East
The impulse of the net force is the amount of momentum change experienced given by the equation;
Impulse force = -
Where;
= The final velocity
= The initial velocity
For the the 265 kg mass, we have;
= 0.1627 m/s
= 1.5 m/s
Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s
The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J
Whereby the distance moved in one second is 0.1627 m, we have;
Work done = Force × Distance = Force × 0.1627 = 294.62
Force = 294.62/0.1627 = 1810.82 N.
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Answer:
Your friend has to wait 0.26 s after you throw the ball to start running.
Explanation:
The equation that gives the position vector of the ball is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t ·sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal positon
v0 = initial velocity
t = time
α = throwing angle
y0 = initial vertical position
g = acceleration due to gravity
The equation of displacement of your friend is as follows:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of your friend at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Please, see the attached figure for a description of the situation. Notice that the frame of reference is located at the throwing point.
Let´s find the time of flight of the ball. We know that at the final time, the y-component of the vector r has to be -6.00 m (1 m above the ground). Then:
y = y0 + v0 · t ·sin α + 1/2 · g · t²
-6.00 m = 0 m + 9.00 m/s · t · sin 33.0° - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 9.00 m/s · sin 33.0° · t + 6.00 m
Solving the quadratic equation:
t = 1.71 s
Now that we have the time of flight, we can calculate the x-component of the vector r (the horizontal distance traveled by the ball):
x= x0 + v0 · t · cos α
x = 0m + 9.00 m/s · 1.71 s · cos 33°
x = 12.9 m
Then, your friend will have to run (12.9 m - 11.0 m) 1.9 m to catch the ball 1 m above the ground.
Let´s see, how much time it takes your friend to run that distance:
x = x0 + v0 · t + 1/2 · a · t² (x0 = 0, v0 = 0)
x = 1/2 · a · t²
1.9 m = 1/2 · 1.80 m/s² · t²
Solving for t
t = 1.45 s
Then, since the time of flight of the ball is 1.71 s, your friend has to wait
1.71 s - 1.45 s = 0.26 s after you throw the ball to start running.
