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3 years ago

A typical adult human has a mass of about 70 kg.

1 answer:

3 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. FM GmMmr2
= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
= 2.40 x 10-3 N

b. FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%

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At what angle two forces P + Q and (P - Q) act so that their resultant is :

stiv31 [10]

Use resultant formula

$\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}$

A be p+q and B be p-q

$\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2$

$\\ \rm\Rrightarrow 2cos\alpha=1$

$\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}$

$\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)$

$\\ \rm\Rrightarrow 2cos\beta=0$

$\\ \rm\Rrightarrow cos\beta=0$

$\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2$

$\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)$

$\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}$

$\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)$

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3 years ago

Calculate The voltage if the current passing through the wire is 3A, and has a resistance of 10 Q

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We use the formula V=IR where I is current, v is voltage, and R is resistance. This is V=(3)(10) which is 30 Volts, answer choice (c)

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2 years ago

Kristen is running a half marathon. By her tenth minute of running, her breathing rate has increased from her resting rate.

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They need more oxygen to function, so gas exchange needs to increase

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3 years ago

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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3 years ago