The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.
If it accelerates at 20 m/s² during the hit, then
Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .
Answer:
The ratio of T2 to T1 is 1.0
Explanation:
The gravitational force exerted on each sphere by the sun is inversely proporational to the square of the distance between the sun and each of the spheres.
Provided that the two spheres have the same radius r, the pressure of solar radiation too, is inversely proportional to the square of the distance of each sphere from the sun.
Let F₁ and F₂ = gravitational force of the sun on the first and second sphere respectively
P₁ and P₂ = Pressure of solar radiation on the first and second sphere respectively
M = mass of the Sun
m = mass of the spheres, equal masses.
For the first sphere that is distance R from the sun.
F₁ = (GmM/R²)
P₁ = (k/R²)
T₁ = (F₁/P₁) = (GmM/k)
For the second sphere that is at a distance 2R from the sun
F₂ = [GmM/(2R)²] = (GmM/4R²)
P₂ = [k/(2R)²] = (k/4R²)
T₂ = (F₂/P₂) = (GmM/k)
(T₁/T₂) = (GmM/k) ÷ (GmM/k) = 1.0
Hope this Helps!!!
Answer:
1. Energy = 2880 Joules.
2. Energy = 60 Joules.
3. Quantity of charge = 120 Coulombs.
Explanation:
Given the following data;
1. Voltage = 12 Volts
Current = 0.5 Amps
Time, t = 8 mins to seconds = 8 * 60 = 480 seconds
To find the energy;
Power = current * voltage
Power = 12 * 0.5
Power = 6 Watts
Next, we find the energy transferred;
Energy = power * time
Energy = 6 * 480
Energy = 2880 Joules
2. Charge, Q = 4 coulombs
Potential difference, p.d = 15V
To find the total energy transferred;
Energy = Q * p.d
Energy = 4 * 15
Energy = 60 Joules
3. Voltage = 6 Volts
Current = 1 Amps
Time = 2 minutes to seconds = 2 * 60 = 120 seconds
To find the quantity of charge;
Quantity of charge = current * time
Quantity of charge = 1 * 120
Quantity of charge = 120 Coulombs
The object’s resultant angle of motion with the +x-axis after the collision is 47°
<span>From object A:
1) x-momentum is 5.7 × 10^4 kilogram meters/second,
2) y-momentum is 6.2 × 10^4 kilogram meters/second.
Now, we know, tan</span>Ф =

⇒tanФ =

⇒tanФ = 1.088
⇒ Ф =

1.088
= 47.4 ≈ 47
Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So

As we find acceleration.Now we need to find time
So

Now for distance
So
