Answer:
a = +0.56 m/s² (due west)
Explanation:
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m= 4900 kg
Fd= 4600 N : drive force
Fw = 670 N : wind force
Fr = 1170 N : water resistive force
Problem development
We apply the formula (1)
∑Fx = m*a
Fd-Fw-Fr = m*a
4600 - 670 - 1170 = 4900 *a
2760 = 4900 *a
a= (2760) / ( 4900 )
a = +0.56 m/s² (due west)
Answer:
0.7593 seconds
Explanation:
A student waiting at a stoplight notices that her turn signal, which has a period of 0.80 s, makes one blink exactly in sync with the turn signal of the car in front of her. The blinker of the car ahead then starts to get ahead, but 14 s later the two are exactly in sync again. What is the period of the blinker of the other car?
Speed:
If a car is moving on a track and car covers x meter distance in y second time. Now the car driver want to know the distance covered by the car in unit time. So the distance covered in unit time will be calculated by dividing x meters by y seconds. This quantity is known as the speed.
Given:
The period of the blinker of the student car is t1 = 0.80
The period after which both blinks sync is t2 = 15seconds
Let the period of the blinker of the other car be t2
The blinker of the student car is 1 second and can be calculated as:
Student car blink/ sec = 1/0.80
The blink of the other car is in 1 second and can be calculated as:
Other car blink/sec = 1/t2
According to the condition,
Other car blink/sec - students car blink /sec = 1/15
1/t2 - 1/0.80 = 1/15
1/t2 = 1/15 - 1/0.80
t2 = 0.7593
t2 = 0.7593second
Do what?? i know how to do a lot of things to answer but there’s nothing to answer.♀️
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