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Art [367]
3 years ago
7

What does a machine change to make work easier? Check all that apply. the amount of work done the amount of force applied the ti

me needed to do the work the direction of a force the distance over which a force is applied
Physics
2 answers:
raketka [301]3 years ago
7 0

The correct options are:

2.) the amount of force applied

4.) the direction of a force

5.) the distance over which a force is applied


Please mark as brainliest! :)

natulia [17]3 years ago
4 0

A machine can change the amount of force required to do the work and makes the work easier because work is the dot product of force and displacement.

A machine can change the distance required to do the work and makes the work easier because work is the dot product of force and displacement.

A machine can change the direction of force to do the work and makes the work easier because work is the dot product of force and displacement.

You might be interested in
A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly
Mnenie [13.5K]

Answer:

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The acceleration of the tortoise is 0.28 m/s²

Explanation:

The equations that describe the position and velocity of the hare and the tortoise are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

To find the acceleration of the hare once it begins to slow down, we have to find how much time the hare traveled during the deceleration and what was its initial speed.

First, the hare moves with constant acceleration for 4.7 s. Then, its velocity at  t = 4.7 s will be:

v = v0 + a · t    (v0 = 0 because the hare starts form rest)

v = a · t = 0.9 m/s² · 4.7 s = <u>4.2 m/s</u>

<u />

The distance traveled by the hare while accelerating can be calculated using the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²      (x0 = 0 and v0 = 0)

x = 1/2 · a · t² = 1/2 · 0.9 m/s² · (4.7)² = <u>9.9 m</u>

<u />

Then, the hare runs at a constant speed of 4.2 m/s for 11.7 s. The distance traveled at constant speed will be:

x =  v · t

x = 4.2 m/s · 11.7 s = <u>49.1 m</u>

<u />

Then, the distance traveled by the hare while slowing down was:

Distance traveled while slowing down = 72 m - 49.1 m - 9.9 m = 13 m

Let´s find how much time it took the hare to come to stop, so we can calculate the acceleration. We know that when the position is 13 m, the velocity is 0.

v = v0 + a · t

0 = 4.2 m/s + a · t

-4.2 m/s / t = a

Replacing in the equation of the position:

x = v0 · t + 1/2 · a · t²      (considering x0 as the point at which the hare started to slow down)

13 m = 4.2 m/s · t - 1/2 · 4.2 m/s / t · t²

13 m = 4.2 m/s · t - 2.1 m/s · t

13 m = 2.1 m/s · t

t = 13 m / 2.1 m/s

t = 6.2 s

Then, the acceleration of the hare while slowing down will be:

-v0/t = a

-4.2 m/s / 6.2 s = a

a = -0.68 m/s²

The acceleration of the hare once it begins to slow down is -0.68 m/s²

The hare traveled 72 m in (6.2 s + 11.7 s + 4.7 s) 22.6 s. The tortoise reaches the final position of the hare at the same time, so, using the equation of the position we can calculate the acceleration of the tortoise:

x = x0 + v0 · t + 1/2 · a · t²     (x0 = 0 and v0 = 0)

x = 1/2 · a · t²

72 m = 1/2 · a · (22.6 s)²

144 m / (22.6 s)² = a

a = 0.28 m/s²

The acceleration of the tortoise is 0.28 m/s²

6 0
3 years ago
If distance between two charges increased 5 times then force between them
Marrrta [24]

decreased 5 times

Explanation: if the force increases 5 times between them would decrease 5 times

4 0
3 years ago
A current of 1.8 A delivers 2.5 C of charge. How much time was required? 0.70 s 0.72 s 1.4 s 4.5 s
Natalka [10]

Answer:

1.4 s

Explanation:

Given the following data;

Quantity of charge, Q = 2.5 C

Current = 1.8 A

To find the time required;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge, Q = current * time

Substituting into the formula, we have;

2.5 = 1.8 * time

Time = 2.5/1.8

Time = 1.4 s

5 0
3 years ago
Read 2 more answers
Can someone solve this problem and explain to me how you got it​
Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

F=k\frac{q_1 q_2}{r^2}

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

q_1' = 8q_1

q_2' = 8q_2

Substituting into the equation, we find the new force:

F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
For maximum radiation protection the suggested distance between array fan-beam scanner source and the operator is:
krok68 [10]

For maximum radiation protection the suggested distance between array fan-beam scanner source and the operator is 2m.

The Fan beam 5 position reference system (PRS) uses accurate time-of-flight laser technology to determine vessel position relative to custom reflectors.

A fan beam allows only the measurement of the azimuth angle. A fan beam, one with a narrow beam width in azimuth and a broad beam width in elevation, can be obtained by illuminating an asymmetrical section of the paraboloid.

The operators’ desk should be positioned at least 1m away from a pencil beam, and at least 2m from a fan-beam system. Some older models, that are not now common, require a distance of 3.5 m.

To learn more about scanner here

brainly.com/question/28174696

#SPJ4

4 0
2 years ago
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