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3 years ago

If a 50 N force was applied to a really massive object as well as a tiny object, what can you say

Physics

1 answer:

frozen [14]3 years ago

6 0

Answer:

The tiny object's acceleration will be much greater.

Explanation:

Because of Newton's second law of motion, (which is F=ma or Force= Mass*acceleration), then if the force maintains the same (which in this case it does, because it says 50 N to both obejcts) and one mass was much greater than the other, then the ould be less for the more massive object and much greater for the lighter object.

For example: If 50 N were applied to a 500 kg object and a 50 kg object, then theformulas for each (respectively) are:

50 = 500*acceleration

and

50 = 50*acceleration

(Because of Newtons Second Law of Motion)

Then, solving for the equations, we get for equation 1:

Acceleration = .1 m/s^2

And for equation 2:

Accleration = 1 m/s^2

Thus, you can see that more massive objects (when applied he same amount of force as the smaller object) clearly have less accleration than the smaller objects.

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A typical 12-V car battery can deliver about 750,000 C of charge before dying. This is not a lot of charge. As a comparison, cal

dimaraw [331]

Answer:

3.982 kg

Explanation:

The latent heat of vaporization = 540 cal/g

= 5.4 ×10⁵ cal/kg

L = 5.4 ×10⁵ × 4.19

= 2.26 × 10⁶ J/kg

Q = 12 × 750,000

= 9, 000, 000

= 9 × 10⁶ J

the maximum number of kg of water (at 100 degrees Celsius) that could be boiled into steam (at 100 degrees Celsius) is:

= $\frac{ 9*10^6}{2.26*10^6 }$

= 3.982 kg

6 0

3 years ago

Anyone help me please?

choli [55]

Answer:

I believe that its B my apologies if its wrong/

Explanation:

3 0

3 years ago

Cho hai điện tích q1=q2=8.10^-7 C đặt cách nhau 5cm. Xác định cường độ điện trường tại điểm:

Aleksandr [31]

Answer: b

Explanation:

5 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind.

AleksAgata [21]

Answer:

a) $f_1=5.587Hz$

b) $f_{n+1}-f_n=5.587Hz$

Explanation:

The frequency of the $n^{th}$ harmonic of a vibrating string of length L, linear density $\mu$ under a tension T is given by the formula:

$f_n=\frac{n}{2L} \sqrt{\frac{T}{\mu}$

a) So for the fundamental mode (n=1) we have, substituting our values:

$f_1=\frac{1}{2(347m)} \sqrt{\frac{65.4\times10^6N}{4.35kg/m}}=5.587Hz$

b) The frequency difference between successive modes is the fundamental frequency, since:

$f_{n+1}-f_n=\frac{n+1}{2L} \sqrt{\frac{T}{\mu}}-\frac{n}{2L} \sqrt{\frac{T}{\mu}}=(n+1-n)\frac{1}{2L} \sqrt{\frac{T}{\mu}}=\frac{n}{2L} \sqrt{\frac{T}{\mu}}=f_1=5.587Hz$

3 0

3 years ago