If we decrease the distance an object moves we will

Answers

Answer 1

Answer: ... without changing the speed of the object, then
we will not have to wait so long for the object to get
where it is going.

Answer 2

Answer: If we decrease the distance an object moves, we will decrease the amount of work done.

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Brianna fires a gun horizontally from a height of 500 m and with a muzzle velocity of 240 m/s. How fast does it travel horizontally after one second?

Answers

In this kind of projectile problem, we always ignore air resistance. (At least
through high school Physics.) If we tried to include air resistance, that would
make the problem hopelessly complicated, when all we're really trying to do is
work with the effects of gravity ... separating out what gravity does and what it doesn't do.

Ignoring the effects of air resistance, there is no horizontal force acting on
the bullet after it leaves the gun, so its horizontal acceleration is zero, and
its horizontal velocity doesn't change. If it left the muzzle with a horizontal
velocity of 240m/s, then its horizontal velocity is that same number, from
then until it hits something.

(By the way . . . your question says that Brianna fires a gun horizontally, and
then it asks how fast does it travel after one second. I took the liberty of
addressing the horizontal velocity of the bullet instead of the gun.)

What is the density of water?
gram/mL
(fill in the number)

Answers

Answer:

1.00 gram per milliliter (1 g/mL). In other words, 1 milliliter of water has a mass of 1 gram. A drop of water is 0.05 mL of water, so its mass would be 0.05 grams.

Explanation:

on a high way a car is driven 80 km the first 1 he of travel, 50km during the next 0.5 he, and 40 km in the final 0.5 hr. What is the cars average speed for the entire trip?

Answers

Average speed = (total distance covered) / (total time to cover the distance) .

Total distance = (80 + 50 + 40) = 170 km

Total time = (1 + 0.5 + 0.5) = 2 hours

Average speed = (170 km) / (2 hrs)= 85 km/hr .

Water _____. 1 has a pH value of 0
2 dissolves in most substances
3 is made of helium and oxygen
4 becomes less dense when it freezes

Answers

The answer is statement 4. This is the most unique property of water since most materials becomes more dense when it undergoes transition from liquid to solid. However, this is not the case of water. When water freezes it becomes less dense because the hydrogen bonds causes the molecule to push each other farther apart.

Answer:

The answer is D:becomes less dense when it freezes.

Explanation:

D,because water becomes less dense when it freezes.Please rate and thank me and give me points.I am a newbie.(:

The position of an object moving along a line at any time t 2 0 is given by the function s(t) = t3 – 9t2 + 18, where s is measured in feet and t is measured in seconds. A) Find the velocity of the particle at t = 4.

Answers

Answer:

S = t^3 - 9 t^2 + 18 position of particle

V = dS / dt = 3 t^2 - 18 t speed of particle at time t

V = 3 * 4^2 - 18 * 4 = 48 - 72 = - 24 ft/sec

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on the other to be LaTeX: 1.35\times10^{-4}N1.35 × 10 − 4 N when they are 20 cm apart. Accidentally, one the the experimenters causes the balls to collide and then repositions them 20 cm apart . Now the repulsive force is found to be LaTeX: 1.406\times10^{-4}N1.406 × 10 − 4 N. What are the initial charges on the two metal balls?

Answers

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=(1)/(4\pi\spsilon_0)(q_1q_2)/(d^2)-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$(1)/(4\pi\spsilon_0)=9*10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35*10^(-4)$ N

So, from equation (i)

$1.35*10^(-4)=9*10^9(q_1q_2)/((0.2)^2)$

$\Rightarrow q_1q_2=6*10^(-16)\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$(q_1+q_2)/(2)$

d=20cm=0.2m, and $F_c=1.406*10^(-4)$ N

So, from equation (i)

$1.406*10^(-4)=9*10^9(\left((q_1+q_2)/(2)\right)^2)/((0.2)^2)$

$\Rightarrow (q_1+q_2)^2=2.50*10^(-15)$

$\Rightarrow q_1+q_2=\pm5* 10^(-8)$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5* 10^(-8)\;\cdots(iii)$