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Tresset [83]
3 years ago
10

65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight

of the body in air.​
Physics
1 answer:
Anna35 [415]3 years ago
5 0

Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

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A spring gun is made by compressing a spring in a tube and then latching the spring at
inn [45]

Answer:

a)v=13.2171\,m.s^{-1}

b)H=8.9605\,m

Explanation:

Given:

mass of bullet, m=4.97\times 10^{-3}\,kg

compression of the spring, \Delta x=0.0476\,m

force required for the given compression, F=9.12 \,N

(a)

We know

F=m.a

where:

a= acceleration

9.12=4.97\times 10^{-3}\times a

a\approx 1835\,m.s^{-2}\\

we have:

initial velocity,u=0\,m.s^{-1}

Using the eq. of motion:

v^2=u^2+2a.\Delta x

where:

v= final velocity after the separation of spring with the bullet.

v^2= 0^2+2\times 1835\times 0.0476

v=13.2171\,m.s^{-1}

(b)

Now, in vertical direction we take the above velocity as the initial velocity "u"

so,

u=13.2171\,m.s^{-1}

∵At maximum height the final velocity will be zero

v=0\,m.s^{-1}

Using the equation of motion:

v^2=u^2-2g.h

where:

h= height

g= acceleration due to gravity

0^2=13.2171^2-2\times 9.8\times h

h=8.9129\,m

is the height from the release position of the spring.

So, the height from the latched position be:

H=h+\Delta x

H=8.9129+0.0476

H=8.9605\,m

4 0
3 years ago
A car of mass 1470 kg is on an icy driveway
Schach [20]

Answer:

  a = 7.5 m / s²

Explanation:

For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane

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X axis

       W sin θ = m a

 

       mg sin θ = m a

        a = g sin θ

let's calculate

        a = 9.8 cos 40

        a = 7.5 m / s²

8 0
3 years ago
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katrin2010 [14]
Energy is transferred between the ground and the atmosphere via conduction.
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7 0
3 years ago
How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t
Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

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so it is given by

W = U_f - U_i

we know that

U_f = -\frac{GMm}{6r}

U_i = -\frac{GMm}{r}

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W = -\frac{GMm}{6r} + \frac{GMm}{r}

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7 0
3 years ago
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Andru [333]
Truee hope this helps 
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3 years ago
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