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NISA [10]
2 years ago
15

4. List 3 detailed examples that illustrate Newton's 2nd Law.

Physics
1 answer:
NNADVOKAT [17]2 years ago
3 0

Examples of Newton's Second Law of Motion

Pushing a Car and a Truck. ...

Pushing a Shopping Cart. ...

Two People Walking Together. ...

Hitting a Ball. ...

Rocket Launch. ...

Car Crash. ...

Object thrown from a Height. ...

Karate Player Breaking Slab of Bricks.

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You drop a rock off of a cliff at exactly the edge after 25 seconds exactly you hear a splash your physics friend tells you the
Reptile [31]
The speed of sound is 340.29 meters per second.

Knowing that, we can calculate how high this cliff is by 340.29 * .4 

The cliff is 340.29 * .4 = 136.12 meters
8 0
4 years ago
Object A travels at a constant velocity of 2.0 m/s right, and object B travels in the same direction at a
pashok25 [27]

==> Object A travels for 60 seconds before Object B starts out.

==> Object A moves at 2 m/s.

==> So Object A has a lead of 120 m when Object B starts out.

==> Object B moves at 3 m/s . . . 1 m/s faster than Object A.

==> So Object B catches up on Object A by 1 m every second.

==> Object B closes up Object A's lead of 120 m in <em>120 seconds</em>.  

8 0
3 years ago
A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
4 0
3 years ago
If you could travel with speed of light (3.0e5 km/s) it would take only 2.5 minutes to reach venus. how far is venus from earth
motikmotik
D=S/T ,S=3.0e5km/s × 2.5min×60
D=4.5×e6km
7 0
4 years ago
NO ONE WILL HELP, PLEASE PLEASE HELP, I HAVE AN HOUR TO GET 4 PAGES DONE! Find the average speed of a marble that takes 6 second
weqwewe [10]

distance
speed = — ————
time
s= 30m
———
6s
the average speed of the marble is 5s/m
8 0
3 years ago
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