<span>The fluid in a graduated cylinder should be read at the BOTTOM of the meniscus.</span>
<h2>Potential energy lost by 10 N rock will be greater</h2>
Explanation:
Two rocks of 5N and 10N falls from the same height . Thus they will loose the potential energy.
The potential energy lost = mass x acceleration due to gravity x height
The potential energy lost by first 5 N rock = 5 h
Because weight of rock m g = 5 N
Similarly , the potential energy lost by 10 N Rock = 10 h
here weight of rock m g = 10 N
Thus comparing these two , the potential energy lost by 10 N rock is greater than that of 5 N rock .
Answer:
Explanation:
Given:
Specific heat of gold = 0.031cal/°C
Specific heat of silver = 0.057cal/°C
To know the metals that will heat up faster, we must understand the meaning of specific heat capacity.
It is the amount of heat required to raise the temperature of 1g of a substance by 1°C.
Now,
The higher the specific heat capacity the more energy it is required to heat up the substance.
So, Gold with a specific heat capacity of 0.031cal/°C will heat up faster.
Answer:
1.5F
Explanation:
Using
E= F/q
Where F= force
E= electric field
q=charge
F= Eq
So if qis tripled and E is halved we have
F= (E/2)3q
F= 1.5Eq=>> 1.5F
Answer:
The observed frequency by the pedestrian is 424 Hz.
Explanation:
Given;
frequency of the source, Fs = 400 Hz
speed of the car as it approaches the stationary observer, Vs = 20 m/s
Based on Doppler effect, as the car the approaches the stationary observer, the observed frequency will be higher than the transmitted (source) frequency because of decrease in distance between the car and the observer.
The observed frequency is calculated as;
![F_s = F_o [\frac{v}{v_s + v} ] \\\\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C)
where;
F₀ is the observed frequency
v is the speed of sound in air = 340 m/s
![F_s = F_o [\frac{v}{v_s + v} ] \\\\400 = F_o [\frac{340}{20 + 340} ] \\\\400 = F_o (0.9444) \\\\F_o = \frac{400}{0.9444} \\\\F_o = 423.55 \ Hz \\](https://tex.z-dn.net/?f=F_s%20%3D%20F_o%20%5B%5Cfrac%7Bv%7D%7Bv_s%20%2B%20v%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%5B%5Cfrac%7B340%7D%7B20%20%2B%20340%7D%20%5D%20%5C%5C%5C%5C400%20%3D%20F_o%20%280.9444%29%20%5C%5C%5C%5CF_o%20%3D%20%5Cfrac%7B400%7D%7B0.9444%7D%20%5C%5C%5C%5CF_o%20%3D%20423.55%20%5C%20Hz%20%5C%5C)
F₀ ≅ 424 Hz.
Therefore, the observed frequency by the pedestrian is 424 Hz.
