<span>If the Earth rotated more slowly about its axis, your apparent weight would
A) increase.
B) decrease.
C) stay the same.
D) be zero.
</span>A) increase.
Answer:
The answer is A. on edgen.
Explanation:
A. adding in the boxes an arrow that points from Qh to Qc
Answer:
the ball travelled approximately 60 m towards north before stopping
Explanation:
Given the data in the question;
First course :
= 0.75 m/s²,
= 20 m,
= 10 m/s
now, form the third equation of motion;
v² = u² + 2as
we substitute
² = (10)² + (2 × 0.75 × 20)
² = 100 + 30
² = 130
= √130
= 11.4 m/s
for the Second Course:
= 11.4 m/s,
= -1.15 m/s²,
= 0
Also, form the third equation of motion;
v² = u² + 2as
we substitute
0² = (11.4)² + (2 × (-1.15) ×
)
0 = 129.96 - 2.3
2.3
= 129.96
= 129.96 / 2.3
= 56.5 m
so;
|d| = √(
² +
² )
we substitute
|d| = √( (20)² + (56.5)² )
|d| = √( 400 + 3192.25 )
|d| = √( 3592.25 )
|d| = 59.9 m ≈ 60 m
Therefore, the ball travelled approximately 60 m towards north before stopping
A magnet has a South Pole and a North Pole. South Pole and South Pole can't connect to her other, same as North and North. The same poles push each other away.
South Pole and North Pole connect.
Answer:
14.7 m/s.
Explanation:
From the question given above, the following data were obtained:
Time (t) = 1.5 s
Acceleration due to gravity (g) = 9.8 m/s².
Height = 11.025 m
Final velocity (v) = 0 m/s
Initial velocity (u) =?
We, can obtain the initial velocity of the penny as follow:
H = ½(v + u) t
11.025 = ½ (0 + u) × 1.5
11.025 = ½ × u × 1.5
11.025 = u × 0.75
Divide both side by 0.75
u = 11.025/0.75
u = 14.7 m/s
Therefore, the penny was travelling at 14.7 m/s before hitting the ground.