A car is traveling due north at 23.6 m>s.
Find the velocity of the car after 7.10 s if its
acceleration is
The acceleration is known to be: a(t) = 1.7 m/s2.
We must integrate over time to obtain the velocity function, and the results are:
v(t) = (1.7m/s^2)
*t + v0
If we suppose that we begin at 23.6 m/s, then the initial velocity is: v0 = 23.6 m/s, where v0 is the beginning velocity.
The velocity formula is then: v(t) = (1.7m/s2).
*t + 23.6 m/s
We now seek to determine the value of t such that v(t) = 27.8 m/s.
Consequently, v(t) = 27.8 m/s = (1.7 m/s2)
*t + 23.6 m/s = (1.7 m/s2) 27.8 m/s - 23.6 m/s
t = 2.5 seconds when *t 4.2 m/s = (1.7 m/s/2)
At such acceleration, 2.5 seconds are required.
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Answer:
The decay constant is 1.21×10^-4/year
Explanation:
Decay constant = 0.693/half-life
Half-life = 5730 years
Decay constant = 0.693/5730 years = 1.21×10^-4/year
Answer:
A
Explanation:
Momentum conservation will cause 0.08kg to move to the west (opposite of 0.02 kg).
and because both are at the same height above the ground, they will take the same time to reach the ground.
The speed of 0.08kg will be less than 0.02 kg, let v be the speed of 0..02kg, then speed of 0.08kg V is
0.02v - (0.08)V = 0
V = 0.02 v/ 0.08 = v/4
The speed of 0.08 kg = v/4
The speed of 0.08 kg is less than 0.02kg.
So 0.02kg strikes the ground farther from the launch point than does the 0.08 kg
Answer:
m = 15.15 kg
Explanation:
Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:
Fx = ma
F Cosθ = ma
where,
F = Magnitude of Force = 85 N
θ = Angle with horizontal = 27°
m = mass of object = ?
a = acceleration of object = 5 m/s²
Therefore,
85 N Cos 27° = m(5 m/s²)
m = 75.73 N/5 m/s²
<u>m = 15.15 kg</u>
- It helps in regulating extremes in the environment.
- It can be used as a car radiators.
<u>Explanation:</u>
- The amount of heat needed to raise a certain amount of temperature to the water is known as the specific heat of the water. For one gram of water to increase 1 degree celsius, water needs to absorb 4.184 J of heat.
Some of the uses of high specific heat of water are as follows,
- The temperature of the water will remain the same from day to night in ponds or sea. During day time, the hot air from the sea or pond rises while the cool air from the sea moves to take its place. During the night, the hot air from the land moves to take its place, giving rise to a land breeze and thus maintaining the environment.
- Another example, When water is circulated throughout an engine, heat will get absorbed. This water is pumped to a radiator and then disposed to the surrounding air.
