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3 years ago

what is haopening to the speed and acceleration as a car travels from rest to 72 km/hr in 4.0 seconds?

1 answer:

6 0

From the given information, we don't know. All we can tell is the net effect of all the changes during those 4 seconds ... the "averages".

The average change in speed is an increase of 72 km/hr (20 m/s) during that time.

The average acceleration is a constant 18 km/hr/sec (5 m/s^2) during the same time.

But both the speed and the acceleration may have gone up or down many times during the 4 seconds.

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neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Vikentia [17]

As per Kepler's third law we know that

$\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

now here we know that

$T_1$ = year of Neptune

$T_2$ = year of Earth

$R_1$ = distance of Neptune from Sun

$R_2$ = Distance of Earth from Sun

so now we will have

$\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}$

$T_1^2 = 27000$

$T_1 = 164.3 years$

so length of year of Neptune is 164.3 years

6 0

3 years ago

How are neutron starts formed ?

Hitman42 [59]

Neutron stars are formed when a massive star runs out of gas and collapses.

8 0

3 years ago

An experiment is designed to investigate the relation between the pressure and temperature in a tank that has a constant volume

viva [34]

Answer:

3rd order polynomial

Explanation:

Given that the increase in the order of the polynomial the error between the curve fit and measured data will decreases hence :

The polynomial order that is best to use is the 3rd order polynomial, this is because using a 3rd order polynomial will produce a less variance and a low Bias

4 0

3 years ago

Which expression correctly describes energy using SI units

mote1985 [20]

Answer:

Joule

Explanation:

energy, work, quantity of heat

m2·kg·s-2

5 0

2 years ago

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago