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3 years ago

A 28 kg student stand on the surface of the Earth. What is the magnitude of the

Physics

1 answer:

Contact [7]3 years ago

4 0

Answer:

F = 274.68[N]

Explanation:

The gravitational force is equal to the weight of a body, or this case that of a person. Weight can be calculated by means of the product of mass by gravitational acceleration. In this way we have the following equation:

$F=m*g$

where:

F = force or weight [N]

m = mass = 28 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

$F=28*9.81\\F=274.68[N]$

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 780 N on him. The

stich3 [128]

Answer:

$f = 556.4 N$

Explanation:

Mass of the losing player with its all equipment is given as

M = 86 kg

Net force applied on him by another player is given as

F = 780 N

also we know that acceleration of the losing player is given as

$a = 2.6 m/s^2$

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3 years ago

The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general

Sever21 [200]

Answer:

The moment is 81.102 k N-m in clockwise.

Explanation:

Given that,

Force = 260 N

Side = 580 mm

Distance h = 370 mm

According to figure,

Position of each point

$O=(0,0)$

$A=(0,-b)$

$B=(h,0)$

We need to calculate the position vector of AB

$\bar{AB}=(h-0)i+(0-(-b))j$

$\bar{AB}=hi+bj$

We need to calculate the unit vector along AB

$u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}$

$u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}$

We need to calculate the force acting along the edge

$\hat{F}=F(u_{AB})$

$\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})$

We need to calculate the net moment

$\hat{M}=\hat{F}\times OA$

Put the value into the formula

$\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))$

$\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})$

$\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}$

Put the value into the formula

$\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}$

$\hat{M}=-81.102\ \hat{k}\ N-m$

Negative sign shows the moment is in clockwise.

Hence, The moment is 81.102 k N-m in clockwise.

5 0

3 years ago