if a 10kg object is lifted a distance of 10m from the rest what was the final velocity of the object?

A light-year is the distance that light travels in one year. The speed of light is 3.00 × 108 m/s. How many miles are there in o

Alinara [238K]

Answer:

(a) $5.88\times 10^{12}mi$

Explanation:

We have given speed of light $c=3\times 10^8m/sec$

Given time = 1 year =365 days

We know that in 1 minute = 60 sec

1 hour = 60×60 = 3600 sec

In one day = 24 hour = 24×60×60=86400 sec

So in 365 days = 365×86400 $=3.1536\times 10^7sec$

We know that distance = speed ×time $=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m$

We have given that 1 mi = 1609 m

So $9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}mi$ So option (a) is correct

8 0

3 years ago

Please I need with this

guajiro [1.7K]

I think its b im not sure im in 8th grade

8 0

2 years ago

A box-shaped metal can has dimensions 5 in. by 19 in. by 4 in. high. All of the air inside the can is removed with a vacuum pump

GuDViN [60]

Answer:

The force is $F = 1397 lb$

Explanation:

From the question we are told that

The length of the box is $l = 19 \ in$

The width of the box is $w = 5 \ in$

The height is $h = 4\ in$

The pressure experience on one of the sides is mathematically represented as

$p = \frac{F}{A}$

Where A is the area of the box which is mathematically evaluated as

$A = l * w$

substituting values

$A = 5 *19$

$A = 95 \ in^2$

This pressure is equivalent to the atmospheric pressure which has a constant value of $p = 14.7 pi$

This implies that

$14.7 = \frac{F}{95}$

=> $F = 14.7 *95$

=> $F = 1397 lb$

5 0

3 years ago

A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1

nydimaria [60]

Answer:

a) 1.06*10^-5

b) 0.00105 °C^-1

Explanation:

Given that

Length of the cylinder, L = 1.5 m

Radius of the cylinder, r = 0.25 cm

Voltage across the rod, V = 15 V

I• at Temperature T• = 20° C is 18.5 A

I at Temperature T = 90° C is 17.2 A

See attachment for calculations

5 0

2 years ago

A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?

Olenka [21]

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ towards the ground. So we can use the following suvat equation: