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ololo11 [35]
3 years ago
6

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

Physics
1 answer:
Anastasy [175]3 years ago
5 0
I think it will be 100 kg
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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}miSo 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:

v=u+at

where

v is the  final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

a=g=9.8 m/s^2 is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

v=0+(9.8)(0.75)=7.35 m/s

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

8 0
2 years ago
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