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3 years ago

HELP PLEASE!!!! What is the net force on this object?

Physics

2 answers:

Ede4ka [16]3 years ago

4 0

The 20N straight up exactly cancels the 20N straight down. So we can see that there's no net vertical force on the object, and if there's ANY net force on it at all, it's going to be horizontal.

The 14N force toward the left can't cancel the whole 22N toward the right. It can only cancel 14N of it, and then there's still another 8N more toward the right.

So the net force on the object is 8N toward the right.

rusak2 [61]3 years ago

3 0

the answer to this question is 8 newtons

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4. Compare the disturbance before and after the waves pass through the slit.

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How does the sun's energy most directly influence precipitation in an area?

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3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago

In a lab, four balls have the same velocities but different masses.

olya-2409 [2.1K]

Answer:

New Momentum of Ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Explanation:

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Momentum of Ball A=2.2 $\frac{k g m}{s}$

Momentum of Ball B=4.4 $\frac{k g m}{s}$

Momentum of Ball C=11 $\frac{k g m}{s}$

Momentum of Ball D=15 $\frac{k g m}{s}$

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2 $\frac{m}{s}$

Initial Momentum of Ball B=4.4 $\frac{k g m}{s}$

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

=3×2=6kg

Thus we get Mass of Ball B=6kg

According to the formula,

Change in momentum of Ball B $\Delta p=m \times \Delta v$

Where $\Delta p$ =change in momentum

m=mass of the ball B

$\Delta v$ =change in velocity ball B

And $\Delta v=v,$ since all balls, have same velocity

Thus the above equation, changes to

$\Delta p=m \times v$

Substitute all the values in the above equation we get

$\Delta p=6 \times 2.2$

$=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

Result:

Thus the New Momentum of ball B $=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}$

3 0

3 years ago

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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon

telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

$L = \frac{\lambda}{2} \\\\\lambda = 2L$

λ = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀ is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N + N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0

3 years ago