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JulijaS [17]
3 years ago
13

What is the momentum of a 12 kg condor flying at 6 m/s?

Physics
1 answer:
rusak2 [61]3 years ago
5 0

Answer:

72

Explanation:

Formula

p=mv\\=12*6\\=72

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Which four equations can be used to solve for acceleration
emmainna [20.7K]

The four equations for acceleration are obtained from the three equations of motion and from second law of motion.

Explanation:

Acceleration is defined as the rate of change of velocity with respect to time. So the change in velocity with respect to time can be determined using the three equations of motions.

So from the first equation of motion, v = u + at , we can determine the value of acceleration if time taken, final and initial velocity is known. The equation can be re-written as a = \frac{v-u}{t}

Similarly, from the second equation of motion, s = ut + 1/2 at², we can determine the equation for acceleration as a = 2*\frac{s-ut}{t^{2} }

So this is second equation for acceleration.

Then from the third equation of motion, v^{2}- u^{2} = 2* a *s

the acceleration equation is determined as a = \frac{v^{2}-u^{2}  }{2s}

In addition to these three equation, another equation is present to determine the acceleration with respect to force from the Newton's second law of motion. F = Mass × acceleration. From this, acceleration = Force/mass.

So, these are the four equations for acceleration.

8 0
3 years ago
Water raises a boat once every 3.0 seconds. What is the frequency (f) of the waves passing the boat?
Goshia [24]

Answer:

I think its A

Explanation:

if its every 3 seconds wouldnt it be 3.0 Hz

6 0
3 years ago
In order to identify them, animal fossils and other fossils nearby can be: A. compared B. piled C. seen
tigry1 [53]

Answer:

A.compared

Explanation:

Fossils help figure out the time that organisms lived. If you know one of the fossils, it can be used as a reference for others around.

5 0
3 years ago
The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the
8090 [49]
Compute first for the vertical motion, the formula is:

y = gt²/2 

0.810 m = (9.81 m/s²)(t)²/2 

t = 0.4064 s 


whereas the horizontal motion is computed by: 

x = (vx)t 

4.65 m = (vx)(0.4064 s) 

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed. 

(vy) = gt 

(vy) = (9.81 m/s²)(0.4064 s) 

(vy) = 3.99 m/s 


speed with which it hit the ground: 

v = sqrt[(vx)² + (vy)²] 

v = sqrt[(11.44 m/s)² + (3.99 m/s)²] 

v = 12.12 m / s
6 0
3 years ago
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
3 years ago
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