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2 years ago

Do we include signs if we calculate Electric field strength?

Physics

1 answer:

expeople1 [14]2 years ago

6 0

Yes because if not people wouldn't understand how did you calculate electric field strength.

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A car engine produces a force of 2000N while moving a distance of 200m in a time of 10s. What is the power developed by the engi

andreyandreev [35.5K]

Power = work/time
Work = force * distance

(2000 N * 200 m) / 10 s = 40,000 Joules

3 0

2 years ago

__________ is when a person becomes less sensitive to a drug. A. Tolerance B. Craving C. Addiction D. Dependence

Luden [163]

A) Tolerance

Tolerance is developed after using a drug repeatedly, so the body adapts to it. Because of that, people who develop a tolerance would then need to use more of that drug to get the same effect.

4 0

2 years ago

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

3 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0

2 years ago

A force of 8.0 N is along x direction, another force of 6.0 N is along +y direction. If both forces are acting on a point object

Darya [45]

Answer:

Resultant force, R = 10 N

Explanation:

It is given that,

Force acting along +x direction, $F_x=8\ N$

Force acting along +y direction, $F_y=6\ N$

Both the forces are acting on a point object located at the origin. Let the resultant force of the object is given by R. So,

$R=\sqrt{F_x^2+F_y^2+F_xF_y\ cos\theta}$

Here $\theta=90^{\circ}$

$R=\sqrt{F_x^2+F_y^2}$

$R=\sqrt{8^2+6^2}$

R = 10 N

So, the resultant force on the object is 10 N. Hence, this is the required solution.

6 0

3 years ago