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3 years ago

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a car moving on a road passes by kilometer 218 at 10:15 am and milestone 236 at 10:30 am. determine an average scalar speed in k

m / h over this stretch.

1 answer:

Leno4ka [110]3 years ago

5 0

Answer:

v?

Explanation:

v = ∆s / ∆t

v =?

∆s = 236 - 218 = 18km

=t = 10: 30 - 10:15 = 15 minutes to hour 15/60 = 0.25h

v = 18 / 0.25

v = 72km / h

speed of 72km / h

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An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

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Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

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3 years ago