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1 year ago

Force Problem: Four concurrent forces are applied to a 5 Kg mass: 50 N North, 25 N East, 35 N South, and 35 N West.

Physics

1 answer:

Kaylis [27]1 year ago

6 0

A) Net horizontal force:
Fx= -35 + 25 + -10N

Net vertical force
Fy= 50 - 35 = 15N

Net force
F^2 = fX^2 +Fy^2
F= 18N

B.) Acceleration of system
a=18/5=3.6m/s^2

C). Equilibriant force must be equal to the net force above
F1= 18N

D.) Acceleration becomes half
a1= a/2 =1.8m/s^2

E.) Equilibrant force also doubles
F" = 2F' = 36N

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Shalnov [3]

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

$36=\frac{1}{2}m(12)^2$ and isolating the m:

$m=\frac{2(36)}{12^2}$ which gives us

m = .50 kg

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3 years ago

Which term describes the time it takes for half of a radioactive elements atoms to decay?

devlian [24]

I think that is half-life

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3 years ago

An object is 16.0cm to the left of a lens. The lens forms an image 36.0cm to the right of the lens.

CaHeK987 [17]

A) 11.1 cm

We can find the focal length of the lens by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p = 16.0 cm is the distance of the object from the lens

q = 36.0 cm is the distance of the image from the lens (taken with positive sign since it is on the opposide side to the image, so it is a real image)

Solving the equation for f:

$\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{36.0 cm}=0.09 cm^{-1}\\f=\frac{1}{0.09 cm^{-1}}=11.1 cm$

B) Converging

The focal length is:

- Positive for a converging lens

- Negative for a diverging lens

In this case, the focal length is positive, so it is a converging lens.

C) 18.0 mm

The magnification equation states that:

$\frac{h_i}{h_o}=-\frac{q}{p}$

where

$h_i$ is the heigth of the image

$h_o$ is the height of the object

$q=36.0 cm$

$p=16.0 cm$

Solving the formula for $h_i$ , we find

$h_i = -h_o \frac{q}{p}=-(8.00 mm)\frac{36.0 cm}{16.0 cm}=-18.0 mm$

So the image is 18 mm high.

D) Inverted

From the magnification equation we have that:

- When the sign of $h_i$ is positive, the image is erect

- When the sign of $h_i$ is negative, the image is inverted

In this case, $h_i$ is negative, so the image is inverted.

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3 years ago

Assume charge builds up on the cloud and on the ground until a uniform electric field of 4.00 ✕ 106 N/C throughout the space bet

kumpel [21]

Answer:

The value of the maximum charge the cloud can hold Q = 177.08 coulomb

Explanation:

Given data

Electrical Field E = 4 × $10^{6}$ $\frac{N}{C}$

the maximum charge the cloud can hold is given by

Q = E A x ---- (1)

where A = area = 5 × $10^{6}$

x = electrical Permittivity of air = 8.854 × $10^{-12}$

Q = (4 × $10^{6}$ ) (5 × $10^{6}$ ) ( 8.854 × $10^{-12}$ )

Q = 177.08 coulomb

This is the value of the maximum charge the cloud can hold.

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3 years ago

A 50.0 mg sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 19.7 days,

PilotLPTM [1.2K]

<h2>Answer: 4.928 days </h2><h2 />

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

<u></u>

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=3.13mg$ is the final amount of the material

$A_{o}=50mg$ is the initial amount of the material

$t=19.7days$ is the time elapsed

$h$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$3.13mg=(50mg)2^{\frac{-19.7days}{h}}$ (2)

$\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}})$ (4)

$-2.77=-\frac{19.7days}{h}ln(2)$ (5)

Clearing $h$ :

$h=\frac{-19.7days}{-2.77}(0.693)$ (6)

Finally:

$h=4.928days$

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3 years ago