Answer:
3349J/kgC
Explanation:
Questions like these are properly handled having this fact in mind;
Quantity of heat = mcΔ∅
m = mass of subatance
c = specific heat capacity
Δ∅ = change in temperature
m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)
m₁ = mass of block = 500g = 0.5kg
c₁ = specific heat capacity of unknown substance
∅₂ = block initial temperature = 50oC
∅₁ = equilibrium temperature of block and water after mix= 25oC
m₂= mass of water = 2kg
c₂ = specific heat capacity of water = 4186J/kg C
∅₃ = intial temperature of water = 20oC
0.5c₁(50-25) = 2 x 4186(25-20)
And we can find c₁ which is the unknown specific heat capacity
c₁ = 
= 3348.8J/kg C≅ 3349J/kg C
Answer:
The driver's average velocity is 82.35 km/h.
Explanation:
Given:
The motion of the driver can be divided into 3 parts:
i. Displacement of the driver in 1.5 hours = 135 km
ii. Rest for 45 minutes.
iii. Displacement in next 2 hours = 215 km
The direction of motion remains same (east).
Now, total displacement of the driver is, 
km.
Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor 
hour.
So, 45 minutes in hours is equal to 
hours.
Now, total time taken for the complete journey is, 
Average velocity is given as:
Therefore, the driver's average velocity is 82.35 km/h
Answer:
SRY I CAN'T ANS :( and hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Answer:
2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]
Explanation:
2a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2032%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%201%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D32%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A1%5Bh%5D%5C%5Cx%3D32%5Bmil%7D)
2b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B420%7D%7B840%7D%5C%5C%20t%3D0.5%5Bh%5D)
3a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B35%7D%7B14%7D%5C%5C%20t%3D2.5%5Bh%5D)
3b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2074%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%202.5%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D74%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A2.5%5Bh%5D%5C%5Cx%3D185%5Bmil%7D)
Answer:
Explanation:
Parameters given:
Charge of object, q = 5 mC = 
Acceleration of object, a = 
Mass of object, m = 2.0 g
The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).
We know that Electrostatic force, F, is given in terms of Electric field, E, as:
F = qE
This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).
The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:
F = ma
Equating the two forces of the object, we get:
-qE = ma
=> 
Solving for E, we have:
The magnitude will be:
The electric field has a magnitude of 0.002 N/C.
