All categories

2 years ago

Big Bee has a mass of 5 kg and hits a newly cleaned windshield accelerating at a rate of 2 m/s2.

Physics

1 answer:

Juli2301 [7.4K]2 years ago

8 0

Answer

The Little Bee will have a greater force of 12N

Explanation

As we know that Force is equal to the product of mass and acceleration (F=ma)

The Force with which the Big Bee will hit the windshield = (5kg x 2m/s^2)

Force of Big Bee= 10 Newton (N)

The Force with which the Little Bee will hit the windshield = (3kg x 4m/s^2)

Force of Little Bee = 12N

Hence, the little bee will hit the windshield with a greater force.

You might be interested in

It is August 1st and you are at a Science Camp in Florida. During an outdoor science quiz, you are asked to identify the tempera

KonstantinChe [14]

Fahrenheit because the boiling point of water is 100 degrees Celsius which is 212 Fahrenheit which is very hot, and that would be about 200 Kelvin so therefore the answer is that the temperature was recorded in Fahrenheit not Kelvin or Celsius

3 0

2 years ago

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

Nadya [2.5K]

Answer: v = 2.24 m/s

Explanation: The Law of Conservation of Energy states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called Elastic Potential Energy (PE), transforms into the movement of the plastic sphere, called Kinetic Energy (KE). Since total energy must be constant:

$KE_{i}+PE_{i}=KE_{f}+PE_{f}$

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

$\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}$

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$

$\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}$

$v^{2}=\frac{50(0.1)^{2}}{0.1}$

$v=\sqrt{50(0.1)}$

$v=\sqrt{5}$

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

3 0

2 years ago

How many elements are in the mixture pictured? A. 7 B. 4 C. 3 D. 2

alex41 [277]

Answer:

C: 3

Explanation:

Cause of the colors

5 0

1 year ago

Identify and name the global wind belts. A. B. C. D. E. F. G.

Yuri [45]

Answer:

Explanation:

That is is the equator, prolly

6 0

2 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

2 years ago