Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
where 
is the mass of the truck, 
is velocity of the truck, 
is the common velocity of moving and standing truck after collision and 
is the mass of the standing truck
Making the subject we obtain
Substituting as 25000 Kg, as 22.3 m/s, as 2000 Kg we obtain
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s
<h2>Question:</h2>
In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?
Answer:
9.1Ω
Explanation:
The circuit diagram has been attached to this response.
(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e
=> 
------------(i)
From the question;
R1 = 3Ω,
R2 = 7Ω
Substitute these values into equation (i) as follows;
Ω
(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.
Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e
R = Rₓ + R3
Rₓ = 2.1Ω
R3 = 7Ω
=> R = 2.1Ω + 7Ω = 9.1Ω
Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω
Assume the motion when you are in the car or in the school bus to go to the school.
To describe the motion the first thing you need is a point of reference. Assume this is your house.
This should be a description:
- When you are sitting and the car has not started to move you are at rest.
- The car starts moving from rest, gaining speed, accelerating. You start to move away from your house, with a positive velocity (from you house to your school) and positive acceleration (velocity increases).
- The car reaches a limit speed of 40mph, and then moves at constant speed. The motion is uniform, the velocity is constant, positive, since you move in the same direction), and the acceleration is zero.
- When the car approaches the school, the driver starts to slow down. Then, you speed is lower but yet the velocity is positive, as you are going in the same direction. The acceleration is negative because it is in the opposite direction of the motion.
- When the car stops, you are again at rest: zero velocity and zero acceleration.
- In all the path your velocity was positive, constant at times (zero acceleration) and variable at others (accelerating or decelerating).
- When you comeback home, then you can start to compute negative velocities, as you will be decreasing the distance from your point of reference (your house).
I would choose B. exosphere. Astronomers study outer space, planets and ect.
Answer:
magnitude=34.45 m
direction=
Explanation:
Assuming the initial point P1 of this vector is at the origin:
P1=(X1,Y1)=(0,0)
And knowing the other point is P2=(X2,Y2)=(19.5,28.4)
We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.
For the magnitude we will use the formula to calculate the distance 
between two points:
(1)
(2)
(3)
(4) This is the magnitude of the vector
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
(5)
(6)
(7)
Finding 
:
(8)
(9) This is the direction of the vector
