<span>Th find the average speed of a trip we need to dived the total distance by the total time.
Let's find the total distance d.
d = (300 mi/h)(2.00 h) + 750 miles
d = 600 miles + 750 miles
d = 1350 miles
The total distance is 1350 miles
Let's find the total time t.
t = 2.00 hours + (750 mi / 250 mi/h)
t = 2.00 hours + 3.00 hours
t = 5.00 hours
The total time of the trip is 5.00 hours.
We can find the average speed.
d / t = 1350 miles / 5.00 hours
d / t = 270 miles/ hour
The average speed of the trip is 270 mi/h
(Note that the direction does not matter when we find the average speed.)</span>
Answer: 114 km/h
Explanation:
The formula for determining average speed is expressed as
Average speed = total distance/total time
The car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. It means that the total distance that the car travels is
85 + 200 = 285 km
The total time spent by the car is
0.5 + 2 = 2.5 hours
Therefore,
Average speed = 285/2.5 = 114 km/h
Answer:
12.7m/s
Explanation:
Given parameters:
Mass of diver = 77kg
Height of jump = 8.18m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we apply the motion equation below:
v² = u² + 2gH
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
H is the height
Now insert the parameters and solve;
v² = 0² + 2 x 9.8 x 8.18
v = 12.7m/s
Answer:
-611.32 N/C
0.43723 m
Explanation:
k = Coulomb constant = 
q = Charge = -4.25 nC
r = Distance from particle = 0.25 m
Electric field is given by
The magnitude is 611.32 N/C
The electric field will point straight down as the sign is negative towards the particle.
The distance from the electric field is 1.71436 m
Answer:
The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Explanation:
Given;
intensity of the sound level, dB = 60 dB
The intensity of the sound in W/m² is calculated as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C)
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
I is intensity of the sound in W/m²
Substitute the given values and for I;
![dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C60%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C6%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C10%5E6%20%3D%20%5Cfrac%7BI%7D%7BI_o%7D%20%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%20I_o%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%201%5E%7B-12%7D%20%5C%20W%2Fm%5E2%20%5C%5C%5C%5CI%20%3D%201%5C%20%5Ctimes%20%5C%2010%5E%7B-6%7D%20%5C%20W%2Fm%5E2)
Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
