Fnet =ma
1560)(1.3102)
the answer is b
Answer:
The runner's speed at the following times would remain 8.64 m/s.
Explanation:
Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.
In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.
First equation of motion: Vf = Vi + a×t
Vf stands for final velocity
Vi stands for initial velocity
a stands for acceleration
t stands for time
In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.
The acceleration (a) is given as 2.4 m/s²
The time (t) is given as 3.6 s
Now put the values of Vi, a and t in first equation of motion
Vf = Vi + a×t
Vf = 0 + 2.4×3.6
Vf = 2.4×3.6
Vf = 8.64 m/s
So,the runner's speed at the following times would remain 8.64 m/s.
Answer:
Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion. There are 2 types of friction: Static friction: If the two surfaces in contact do not move relative to each other, one has static friction.
Answer:
7.36 × 10^22 kg
Explanation:
Mass of the man = 90kg
Weight on the moon = 146N
radius of the moon =1.74×10^6
Weight =mg
g= weight/mass
g= 146/90 = 1.62m/s^2
From the law of gravitational force
g = GM/r^2
Where G = 6.67 ×10^-11
M = gr^2/G
M= 1.62 × (1.74×10^6)^2/6.67×10^-11
= 4.904×10^12/6.67×10^-11
=0.735×10^23
M= 7.35×10^22kg. (approximately) with option c
(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
