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3 years ago

In the example of jumping off a chair, what is the impulse that will stop your fall?

Physics

1 answer:

liq [111]3 years ago

5 0

Answer:

You will reach both your arms out to break your fall and save your head.

Explanation:

It common sense you don't want your head injured. Do you?

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The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

now here we can say

$m_1 = 10 g$

$v_{1i} = 0$

$m_2 = 550 g$

$v_{2i} = 3.5 m/s$

now here we can say

$10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}$

$192.5 = v_{1f} + 55 v_{2f}$

now by coefficient of restitution

for elastic collision we know that e = 1

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

$v_{2f} - v_{1f} = 0 - 3.5$

now by solving the two equation

$56v_{2f} = 189$

$v_{2f} = 3.375 m/s$

also we know that

$v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s$

so final speed of the nail is 6.875 m/s

6 0

3 years ago

Read 2 more answers

Two bodies are falling with negligible air resistance, side by side, above a horizontal plane. If one of the bodies is given an

tankabanditka [31]

Answer:4-strikes the plane at same time as the other body

Explanation:

Given

If both bodies is falling on a horizontal plane and second body is given an acceleration in horizontal direction then it does not change the time to reach the Horizontal Plate as there is no change in vertical direction.

Horizontal acceleration will give only horizontal range and horizontal velocity.

8 0

3 years ago

Read 2 more answers

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

cupoosta [38]

Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0

3 years ago

Describe the differences among ultraviolet waves, visible light waves, and infrared waves. how are these waves alike?

sergeinik [125]

Our eyes are detectors which are designed to detect visible light waves (or visible radiation). ... The electromagnetic spectrum includes gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, and radio waves. The only difference between these different types of radiation is their wavelength or frequency.

4 0

3 years ago

Two masses, each weighing 1.0 × 103 kilograms and moving with the same speed of 12.5 meters/second, are approaching each other.

juin [17]

A perfectly elastic<span> collision is defined as one in which there is no loss of </span>kinetic energy<span> in the collision. Therefore, we just add the kinetic energies of each system. We calculate as follows:

KE = 0.5(</span>1.0 × 10^3)(12.5 )^2 + 0.5(1.0 × 10^3)(12.5 )^2
KE = 156250 J = 1.6 x 10^5 J -------> OPTION A

5 0

3 years ago