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2 years ago

If I keep F constant in F=ma, what is the relationship between m and a?

Physics

1 answer:

Jobisdone [24]2 years ago

6 0

Answer:

If F is a constant, we can take f = 1

f = m×a

ma = 1

therefore we can say that force is hence proportinal to the product of mass and acceleration.

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Which refers to the number of wavelengths that pass a fixed point in a second?

pav-90 [236]

Answer:

The word that refers to the number of wavelengths that pass a fixed point in a second is:
C.) Frequency

Frequency is also known as hertz.

Explanation:

Have a great rest of your day
#TheWizzer

7 0

2 years ago

A Force of 20N displaces a body about 45m how much work is done?

Orlov [11]

Answer:

159 ofc

Explanation:

7 0

3 years ago

Read 2 more answers

Mỗi giây có 2,1.1018 ion+

Alenkasestr [34]

A I seen it on the quiz

4 0

2 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

9.The force of gravity between two asteroids is 10,000 newtons (N).

MakcuM [25]

Answer:

F1 = G m1 m2 / R^2 force of attraction

F2 = G m1 m2 / (R/2)^2

F2 / F1 = 4 the force of gravity will be quadrupled

6 0

2 years ago