1. What coefficients would balance the following equation?

Physics

2 answers:

Dovator [93]3 years ago

4 0

Answer:

Explanation:

Stolb23 [73]3 years ago

3 0

Answer:

no question matee

Explanation:

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Difference between freefall and weightlessness.

Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>

When a body falls only under the influence of gravity, it is called free fall.
Freefall is not possible in absence of gravity.
A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>

Weightlessness is a condition at which the apparent weight of body becomes zero.
Weightlessness is possible in absence of gravity.
A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0

3 years ago

A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction,

olchik [2.2K]

<span>The velocity would be 54.2 m/s We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>

5 0

4 years ago

Read 2 more answers

Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B

WARRIOR [948]

$\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

Explanation:

Given:

$\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}$

$\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}$

The cross product $\textbf{A}×\textbf{B}$ is given by

$\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|$

$= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}$