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3 years ago

Someone help please by providing work and answers please :)

Physics

1 answer:

ohaa [14]3 years ago

7 0

Mass is a property of the object. It doesn't depend on anything else,
and it doesn't change when the object goes to different places.
Its mass is always the same.

If it's 20,000 kilograms on Earth, then it's 20,000 kilograms on the
Moon, on Mars, in school, in bed, in love, in trouble, in water,
on Pluto, on Halley's comet, or in free fall in outer space.

That's why choice-B is the correct choice.

Now, if the question was talking about WEIGHT, then we would have
a lot to talk about, because WEIGHT depends on where you are, and
what else is around you. But mass doesn't.

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Natasha_Volkova [10]

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3 years ago

Read 2 more answers

The velocity of a an object in linear motion changes from +25 meters per second to +15 meters per second in 2.0 seconds.

kodGreya [7K]

Answer:

$-5.0 m/s^2$

Explanation:

Missing question:

What is the object's acceleration?

Solution:

The acceleration of an object is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the change in velocity

For the object in this problem,

u = +25 m/s

v = +15 m/s

t = 2.0 s

Substituting,

$a=\frac{15-25}{2}=-5.0 m/s^2$

And the acceleration is negative because its direction is opposite to that of the velocity.

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3 years ago

Scientists have invented the equipment to harness the power of lighting for energy use. True or false

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I think it’s false .
Let me know if it’s incorrect .

8 0

3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

2 years ago