Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:

Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq

Where:
I is the current (2 A)
Δt is the time (20 s)


Then, we can put this value of charge in the voltage equation.

Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
The answer is A.
Sy = 1650 x sin30.5 = 837.4 m toward south
Sx = 1650 x cos30.5 = 1421.7 m toward east
Answer:
The temperature of the metal is 
Explanation:
From the question we are told that
The mass of the metal is 
The specific heat of the metal is 
The mass of the oil is 
The temperature of the oil is 
The specific heat of oil is 
The equilibrium temperature is 
According to the law of energy conservation
Heat lost by metal = heat gained by the oil
So
The quantity of heat lost by the metal is mathematically represented as

=> 
Where
the temperature of metal before immersion
The negative sign show heat lost
The quantity of gained t by the metal is mathematically represented as

=> 
So

substituting values

=> 
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
<h3>What is velocity?</h3>
Velocity is a vector quantity that tells the distance an object has traveled over a period of time.
Displacement is a vector quality showing total length of an area traveled by a particular object.
Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?
The slope of the line is equal to zero and the object will be stationary.
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
To learn more about velocity refer to the link
brainly.com/question/18084516
#SPJ2
Answer:
(a) The range of the projectile is 31,813.18 m
(b) The maximum height of the projectile is 4,591.84 m
(c) The speed with which the projectile hits the ground is 670.82 m/s.
Explanation:
Given;
initial speed of the projectile, u = 600 m/s
angle of projection, θ = 30⁰
acceleration due to gravity, g = 9.8 m/s²
(a) The range of the projectile in meters;

(b) The maximum height of the projectile in meters;

(c) The speed with which the projectile hits the ground is;
