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lions [1.4K]
3 years ago
15

If you travel at 3 m/s for 12 seconds, how far did you travel?

Physics
2 answers:
Rom4ik [11]3 years ago
8 0

Answer:

\boxed {\tt 36 \ meters }

Explanation:

We want to find how far was traveled, or the <u>distance.</u>

The formula for distance is:

d=s*t

where s is the speed and t is the time.

The speed is 3 meters per second and the time is 12 seconds.

s= 3 \ m/s \\t= 12 \ s

Substitute the values into the formula.

d= 3 \ m/s * 12 \ s

Multiply. Note that when multiply, the seconds, or s will cancel.

d= 3 \ m * 12

d= 36 \ m

d= 36 \ meters

The distance traveled was 36 meters.

ella [17]3 years ago
4 0

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

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A football kick returner catches the ball just as a player from the opposing team dives to tackle him. At the time of impact, th
pochemuha

The total momentum of the players after collision is 130 kgm/s.

The given parameters:

  • <em>Initial momentum of the returner, </em>P_i_1<em> = 0 kgm/s</em>
  • <em>The initial momentum of the diving player, </em>P_i_2<em> = 130 kgm/s</em>

The total momentum of the players after collision is determined by applying the principle of conservation of linear momentum as follows;

P_f = P_i_1 + P_i_2\\\\P_f = 0 + 130\\\\P_f = 130 \ kgm/s

Thus, the total momentum of the players after collision is 130 kgm/s.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

6 0
2 years ago
The term "exfoliation dome" is best applied to ________. the term "exfoliation dome" is best applied to ________. bryce national
BlackZzzverrR [31]
An exfoliation dome is a geological structure wherein it is primarily when the overburden of a surface gets removed by erosion, thus leading to rock relaxation. In addition, the term would be best applied to the national landmark of Yosemite National Park wherein the place has one of the best waterfalls in California.
4 0
4 years ago
The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu
Gemiola [76]

Answer:

\displaystyle w=3.478\ rad/sec

M=0.0182\ J

v=0.398\ m/s

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

(a) The angular frequency of the motion is computed as

\displaystyle w=\sqrt{\frac{g}{L}}

We have the length of the pendulum is L=0.81 meters, then we have

\displaystyle w=\sqrt{\frac{9.8}{0.81}}

\displaystyle w=3.478\ rad/sec

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

U=mgh

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

H+Y=L

And

H=L\ cos\alpha

Solving for Y

Y=L(1-cos\alpha )

Since\ \alpha=8.1^o, L=0.81\ m

Y=0.0081\ m

The potential energy is

U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)

U=0.0182\ J

The mechanical energy is, then

M=K+U=0+U=U

M=0.0182\ J

(c) The maximum speed is achieved when it passes through the lowest point (the reference for h=0), so the mechanical energy becomes all kinetic energy (K). We know

\displaystyle K=\frac{mv^2}{2}

Equating to the mechanical energy of the system (M)

\displaystyle \frac{mv^2}{2}=0.0182

Solving for v

\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}

v=0.398\ m/s

4 0
3 years ago
1. What is the potential energy of a 4-kilogram potted plant that is on a 1 meter-high plant
Yuri [45]

39.2 J

Explanation:

Step 1:

To find the potential energy the following formula is used.

Potential Energy = m × g × h

Where,

m = Mass

g = Acceleration due to gravity

h =  Height

Step 2:

Here m = 4 kg, g = 9.8 m/s², h = 1 m

Potential Energy = ( 4 × 9.8 × 1)

                             = 39.2 J

4 0
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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
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