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3 years ago

When we draw a diagram of the forces acting on an extended object, the tail of the force vector for the weight should be at?

1 answer:

6 0

Whenever we represent forces using vectors, we ensure that the vector begins at the point of force exertion. In the case of weight, the point of exertion of the force is known as the center of gravity, which is the point through which the weight of an object can be said to act. Therefore, the downward arrow representing the weight would begin at the center of gravity of the object.

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What happens when humans change the flow of water?

REY [17]

Disaster. (like my life) It'll affect the animals and there will be no way to get their food without water.Hope this helped :)

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3 years ago

Read 2 more answers

A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a

ss7ja [257]

Answer:

P = 1471500 [Pa]

Explanation:

We must remember that pressure is defined as the relationship between Force over the area.

$P=F/A$

where:

P = pressure [Pa] (units of pascals)

F = force [N] (units of Newtons)

A = area of contact = 4 [cm²]

But first we must convert from cm² to m²

$A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }$

A = 0.0004 [m²]

Also, the weight should be calculated as follows:

$w = m*g$

where:

m = mass = 60 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

$w = 60*9.81\\w = 588.6[N]$

And the pressure:

$P=588.6/0.0004\\P=1471500 [Pa]$

Because 1 [Pa] = 1 [N/m²]

8 0

3 years ago

Find the absolute value of the change of the gravitational potential energy (GPE) of the puck-Earth system from the moment the p

jekas [21]

Answer:

0.16joules

Explanation:

Using the relation for The gravitational potential energy

E= Mgh

Where,

E= Potential energy

h = Vertical Height

M = mass

g = Gravitational Field Strength

To find the vertical component of angle of launch Where the angle is 22°

h= sin theta

So E = mghsintheta

= 0.18 x 0.98 x 0.253 sin22

=0.16joules

Explanation:

7 0

3 years ago

A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from

s2008m [1.1K]

E = kQ 

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = 9e9 * 300e^-6 

(8e^-2 + 16e^-2)²

E = 4687500N/C

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2 years ago

Match the half life and time information to the percentage of radioactive isotope left.

Kazeer [188]

Answer:

Explanation:

can i have brainliest pls im new

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3 years ago