Physics: Two cars are traveling along perpendicular roads, car a at 40 mi/hr, car b at 60 mi/hr. at noon, when car a reaches the intersection, car b is 90 mi away, and moving toward it. at 1 p.m. the distance between the cars is changing, in miles per hour, at the rate of:

Two cars are traveling along perpendicular roads, - id4235857, xxxharveyweinsteinxx

yyyyyy290

13.01.2022

$\frac{dD}{dt} = -4 miles/hour$

negative sign indicates that the distance is decreasing with time

Explanation:

Let at any time t after noon that is 12 p.m.

distance traveled by car A = 40t

distance traveled by car B = 90-60t

then distance between the two cars at time t

D^2= (40t)^2+(90-60t)^2 ............1

also, at time 1 p.m.

distance $D^2= (40\times1)^2+(90-60\times1)^2$

D=50 Km

differentiating equation 1 w.r.t. t we get

$2D\frac{dD}{dt}= 2\times40t\times40+2(90-60t)(-60)$

put t= 1 and D= 50 we get

$2\times50\frac{dD}{dt}= 3200\times1-3600\times1$

$\frac{dD}{dt} = -4 miles/hour$

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thao6570

12.09.2020

Unit of G = m³/kg s²

Explanation:

The force of gravitational attraction is represented by the equation.

$F=G\dfrac{m_1m_2}{r^2}$

F is the magnitude of the gravitational attraction on either body

m₁ and m₂ are the masses of the bodies, r is the distance between them. G is the gravitational constant.

We need to find the unit of G.

Rearranging the above formula for G.

$G=\dfrac{Fr^2}{m_1m_2}$

Putting units of F,m and r such that,

$G=\dfrac{N\times m^2}{kg^2}\\\\\because N=kgm/s^2\\\\G=\dfrac{kg\times m\times m^2}{kg^2\times s^2}\\\\G=m^3/kgs^2$

Hence, the correct option is (c) m³/kg s².

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