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daser333 [38]
3 years ago
14

Use the values from PRACTICE IT to help you work this exercise. If the current in each wire is doubled, how far apart should the

wires be placed if the magnitudes of the gravitational and magnetic forces on the upper wire are to be equal
Physics
1 answer:
melisa1 [442]3 years ago
8 0
Which behavior would best describe someone who has good communication skills with customers ? a) Following up with some customers b) Talking to customers more than listening to them c) Repeating back what the customer says d) Interrupting customers frequently
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A machine is currently set to a feed rate of 5.921 inches per minute (IPM). Te machinist changes this setting to 6.088 IPM. By h
lukranit [14]

Answer:

By 16.7% or 0.167 IPM

Explanation:

Substracting the final IPM (6.088) to the initial IPM (5.921) gives us the net difference, which is how much did it increase in IPM. Multiplying this number by 100 gives us the percentual increase in the feed rate.

4 0
2 years ago
Read 2 more answers
A mass of gas under constant pressure occupies a volume of 0.5 m3 at a temperature of 20°C. Using the formula for cubic expansio
Kryger [21]
No cubic expansion given
6 0
3 years ago
Sheila weighs 60 kg and is riding a bike. Her momentum on the bike is 340 kg • m/s. The bike hits a rock, which stops it complet
Vikki [24]

Answer:

v₂ = 5.7 m/s

Explanation:

We will apply the law of conservation of momentum here:

Total\ Initial\ Momentum = m_{1}v_{1} + m_{2}v_{2}\\

where,

Total Initial Momentum = 340 kg.m/s

m₁ = mass of bike

v₁ = final speed of bike = 0 m/s

m₂ = mass of Sheila = 60 kg

v₂ = final speed of Sheila = ?

Therefore,

340\ kg.m/s = m_{1}(0\ m/s) + (60\ kg)v_{2}\\v_{2} = \frac{340\ kg.m/s}{60\ kg}\\\\

<u>v₂ = 5.7 m/s </u>

6 0
3 years ago
1. Define weight
andrezito [222]

1. Weight is the gravitational pull with which the earth attracts the body towards the center of the earth. The S.I unit is Newton (N)

2. The weight of an object is related to its mass with the below equation.

W = mg

Where W = weight, m = mass and g = acceleration due to gravity

3. The mass m = 50 kg and g = 9.8 m/s^{2}

Substitute the parameters in the equation above.

W = 50 x 9.8

W = 490 N

4. An object is accelerating when it speeds up. If the object slows down, it means it is decelerating. The correct answer is option A

5. Newton's second law of motion is:

F = ma

Where F = force applied, m = mass and a = acceleration

Therefore, Newton's second law of motion relates an object's acceleration to its net force acting on it. The correct answer is option C

6. According to Newton's second law of motion which relates an object's acceleration to its mass, doubling the net force acting on an object a doubles its acceleration. Because mass is always constant.

7. Since the weight of an object is related to its mass with the equation.

W = mg,  If the mass of an object doubles, its weight will also doubles. Option A is the correct answer.

8. If you know the mass of an object, you can calculate its weight with the formula F = mX when X = 9.8m/s^{2}

9. Force is expressed in unit as Newton (N)

10. The parameters given are :

mass m = 20kg

Force F = 40N

To calculate acceleration, use the formula F = ma

Substitute all the parameters into the equation.

40 = 20a

a = 40/20

a = 2m/s^{2}

The correct option is D

Learn more here : brainly.com/question/18835375

8 0
3 years ago
A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal
Karolina [17]

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2

Let d be the breaking distance. It can be calculated using third equation of motion as :

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m

So, the required distance covered by the car is 44.64 m.

4 0
3 years ago
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