Answer:
72.53 mi/hr
Explanation:
From the question given above, the following data were obtained:
Vertical distance i.e Height (h) = 8.26 m
Horizontal distance (s) = 42.1 m
Horizontal velocity (u) =?
Next, we shall determine the time taken for the car to get to the ground.
This can be obtained as follow:
Height (h) = 8.26 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
8.26 = ½ × 9.8 × t²
8.26 = 4.9 × t²
Divide both side by 4.9
t² = 8.26 / 4.9
Take the square root of both side by
t = √(8.26 / 4.9)
t = 1.3 s
Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:
Horizontal distance (s) = 42.1 m
Time (t) = 1.3 s
Horizontal velocity (u) =?
s = ut
42.1 = u × 1.3
Divide both side by 1.3
u = 42.1 / 1.3
u = 32.38 m/s
Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:
1 m/s = 2.24 mi/hr
Therefore,
32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s
32.38 m/s = 72.53 mi/hr
Thus, the car was moving at a speed of
72.53 mi/hr.
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
Explanation:
According to the fundamental law of electrostatics, like charges which are ( only positive or only negative charges) repel while unlike charges which are ( positive and negative charges ) attract each other.
that is: ( +++++ ++++++) they repel
while ( +++++++ ———) they attract
Answer:
1) For the weight to be correct, the pointer must be below the reference (zero) or in it
2) mass greater than 300 g leave the pointer below the reference,
3) put weight on the other arms the pointer gets closer to the reference
Explanation:
1) A three-arm scale is used to determine the weight of a body by successive approximations, starting with the arm with the greatest mass (middle), when the pointer is close to zero without going over, weights are placed on the second arm (above ) weight is placed to bring the pointer closer to the reference without going over and when it is used the arm with the lowest mass is used (below) with this one, masses are placed until reaching the reference,
Taking all the jinetillos placed the total weight, the product of the position of the jinetillo by the precision of the arm and then the three values are added.
In the example given, the jinetillo is placed in position 3 of the 100 arm, so the weight is 300 gr.
For the weight to be correct, the pointer must be below the reference (zero) or in it
2) objects with a mass greater than 300 g leave the pointer below the reference, it must be completed with the other two arms to reach the correct weight
3) when you put weight on the other arms the pointer gets closer to the reference
Where’s the image?? i don’t see it
