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iVinArrow [24]
3 years ago
12

10m= (5.0) + (.5)(9.8)(5.0)

Physics
1 answer:
Norma-Jean [14]3 years ago
8 0
M=2.45 because you multiply out the equation on the right and divide by 10
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to digitize sound, a process called _____ is used to measure the amplitude of a sound wave thousands of times per second.
Leona [35]

Answer: Sound recording and production

Explanation: Digital audio is also the name for the entire technology of sound recording and reproduction using audio signals that have been encoded in digital form.

7 0
3 years ago
A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?
Genrish500 [490]
A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps 
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3 years ago
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Explain why a moving object cannot come to a stop instantaneously (in zero seconds). Hint: Think about the acceleration that wou
gizmo_the_mogwai [7]
To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:F=ma<span>So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, </span><span><span>m<span>v⃗ </span></span><span>m<span>v→</span></span></span>, we can use the following equation to calculate force:F=p/h<span>However, because the force is nowhere close to infinity, time will never tend to zero either, which means that you cannot come to an instantaneous stop.</span>
7 0
3 years ago
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
3 years ago
How many days does it take for a free to grow?
vovikov84 [41]
Idk what is growing but if it’s a free than c
7 0
3 years ago
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