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2 years ago

In order to be _______________ forces, their effects must cancel each other out and not cause a change in the object's motion

Physics

1 answer:

Tju [1.3M]2 years ago

4 0

Balancing.

When the same force is applied from both sides,the forces cancel out each other.

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Atoms containing radioactive nuclei are called

miss Akunina [59]

The correct answer is radioisotopes.

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2 years ago

Please help me with this 10 points all i have rn :/ thanks <3

Korolek [52]

Answer: c the last one

Explanation:

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2 years ago

Read 2 more answers

Write a linear equation that expresses the relationship between the temperature in degrees Celsius (C) and degrees Fahrenheit (F

UkoKoshka [18]

Answer:

$\displaystyle F=\frac{9}{5}C+32$

$\displaystyle C=\frac{5}{9}(F-32)$

Explanation:

Temperature Units Conversion

The conversion formula between Celsius and Fahrenheit temperature scales is well-known. But we'll use the provided data to derive the formula. Let's model the relationship between Fahrenheit (F) and Celsius (C) as a linear function like

$F=mC+b$

Where m and b must be computed according to the pair of conditions given. The values for each temperature scale are (C,F)=(0,32) and (100,212). Replacing the first value

$32=m\times 0+b$

It means that

$b=32$

By using the second point

$212=m\times 100+32$

Solving for m

$\displaystyle m=\frac{212-32}{100}=\frac{180}{100}$

Simplifying

$\displaystyle m=\frac{9}{5}$

So, the conversion formula is

$\displaystyle F=\frac{9}{5}C+32$

Which is the widely known formula for temperature conversion

Solving for C, we get the inverse relation

$\boxed{\displaystyle C=\frac{5}{9}(F-32)}$

3 0

3 years ago

As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe

spin [16.1K]

Answer:

1.19 m/s²

Explanation:

The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so

f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

g = 2π√(71.753/200 Hz)/3.16

g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

6 0

2 years ago

Hurry please I don’t have long for this for a test !!!

Anna35 [415]

I think the Answer is b

7 0

2 years ago