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katrin [286]
3 years ago
11

A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the

upthrust on the body. (iii) volume of the body.​
Physics
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

difference  \: in \: weight = 150n - 100n = 50n

Now,buyantant force

difference \: in \: weight \: = volume(body) \times density \: of \: water \:  \times g

so;

50 =  {v}^{b}  \times 1 \times  {10}^{3}  \times 9.8m {s}^{2}

{v}^{b}  =  \frac{50}{1000 } \times 9.8

=  \frac{50}{9800}

= 0.0051

Now,

mass \: in \: air \:  = 150n =  \frac{150}{9.8kg}

density =  \frac{weght}{volume}

=  \frac{150}{0.0051}  \times 9.8 \\ x = 3000

And now,

specific \: density \:  =  \frac{density of \: the \: body}{density \: of \: water}

=  \frac{3000}{1000}

= 3

Hence that,specific density of a given body is 3

please mark me as brainliest, please

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Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

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3 years ago
The ballistic pendulum is a device used to measure the speed of a projectile such as a bullet. The projectile of mass m is fired
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Answer:

Relation between initial speed of bullet and height h is given as

v = \frac{m + M}{m}\sqrt{2gh}

Explanation:

As we know that system of block and bullet swings up to height h after collision

So we have

(m + M)gh = \frac{1}{2}(m + M)v_1^2

so we have

v_1 = \sqrt{2gh}

so speed of the block + bullet just after the impact is given by above equation

Now we also know that there is no force on the system of bullet + block in the direction of motion

So we can use momentum conservation

mv = (m + M)v_1

now we have

v = \frac{m + M}{m}\sqrt{2gh}

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A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s
djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, u=32 m/s

Acceleration of the driver is, a=-1.5 m/s²

Time taken to reach final velocity is, t=10.8 s.

The final velocity is given using the Newton's equations of motion as:

v=u+at, where, v is the final velocity.

Now, plug in the given values and solve for v.

v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}

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Answer:

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