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Margarita [4]
3 years ago
12

My question is An object may be acted on by several forces. What name is given to the single force that has the same effect as t

hese forces ?
Physics
2 answers:
12345 [234]3 years ago
8 0
Dat wuld be the Resultant Force
wariber [46]3 years ago
6 0
Newtons Law of motion
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by an echo meter

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5 0
3 years ago
When does a ball that is thrown vertically upwards reach its maximum speed?
ladessa [460]
As soon as you let go of it it is at its max speed because gravity is constantly pulling it down
4 0
3 years ago
assume the suns total energy output is 4.0 * 10^26 watts, and 1 watt is 1 joule/second. assume 4.3 * 10^-12 J is released from e
Dmitry [639]

Answer:

9.3\cdot 10^{37}

Explanation:

Power is defined as the energy produced (E) per unit of time (t):

P= \frac{E}{t}

This means that the energy produced in the Sun each second (1 s), given the power P=4.0\cdot 10^{26}W, is

E=Pt=(4.0\cdot 10^{26}W)(1s )=4.0\cdot 10^{26} J

Each p-p chain reaction produces an amount of energy of

E_1 = 4.3\cdot 10^{-12} J

in order to get the total number of p-p chain reactions per second, we need to divide the total energy produced per second by the energy produced by each reaction:

n=\frac{E}{E_1}=\frac{4.0\cdot 10^{26} J}{4.3\cdot 10^{-12} J}=9.3\cdot 10^{37}

3 0
3 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
A ball is tossed int the air and rises to a height of 12m. How long is the ball in the air?
Eddi Din [679]
1.5 seconds that its in the air
7 0
3 years ago
Read 2 more answers
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