Time is being measured in days here so if you want to calculate the rate of increase after one day, substitute 1 for t in your rate equation:
dB/dt = 0.677598 cos(1.232(1))
dB/dt = 0.22
The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².
<h3>What is acceleration?</h3>
The rate of velocity change concerning time is known as acceleration.
Given data;
Initial velocity, u= 0 m/s
Final velocity, v= 4.2 m/s
Time elapsed, t = 7.3 seconds.
To find ;
Acceleration, a
The acceleration when the change in velocity is observed by the formula as:
a= (v-u)/(t)
Substitute the given values:
a= (4.2-0)/(7.3)
a=(4.2)/(7.3)
a= 0.575 m/s²
Hence, the car's acceleration will be 0.575 m/s².
To learn more about acceleration, refer to the link brainly.com/question/2437624
#SPJ1
Use the following formula for the electric field strength between two parallel plates:
E = V/d
where,
V: potential difference = 25V
d: distance between plates = 5 cm = 0.05 m
Replace the previous values of the parameters into the formula for E:

Hence, the electric field strength is 500V/m
Answer:
V=533.33 V
Explanation:
Given that
A= 30 cm²
d= 0.059 cm
Q= 0.0240 μ C
We know that capacitance given as

Now by putting the values


Voltage difference given as
Q= V .C
V=Q/C

V=533.33 V
Doesn't seem like we know much here, but we can answer it. Let's talk about what we know. We know it takes 3.24 s for the ball to go up and drop back down again. We know that gravity is the only force acting after the ball leaves the hand, so a = 9.8 m/s^2 (we'll say it's negative in our equations because down being negative is intuitive). We also know that it stops moving for a brief moment at the top of the arc, where v = 0 m/s. Because gravity is the only force, and it slows it down on the way up at the same rate it speeds it up on the way down and the distance covered in upward and downward motion is the same, we can confidently say that it will reach the top of its arc (where v = 0 and it turns around) in half the total time it is in the air, so it takes 1.62 s to reach the peak. Now we can use a kinematics equation, let's use vf = vi + a*t, where vf is final velocity and is 0, vi is initial velocity and is some unknown v we need to solve for, a is acceleration and is -9.8 m/s^2 and t is time and since this is just to the top of the arc, we'll use half the time so 1.62 s. We can solve for vi and plug stuff in like so: v = -a*t = -(-9.8m/s^2)*(1.62s) = 15.876 m/s.