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netineya [11]
3 years ago
7

An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 5 N, F2 = 8 N, and F3 = 5

N. When answering the questions below, assume the x-direction is to the right, and the y-direction is straight upwards.
a) What is the net force in component form, in terms of F1, F2, F3, and the unit vectors i and j
b) What is the magnitude of the net force, in newtons?
c) What is the angle θ, in degrees, of the net force, measured from the +x-axis? Enter an angle between -180° and 180°.
d) What is the magnitude, |a| of the acceleration, in meters per square second, if the block has a mass of 6.9 kg?

Physics
1 answer:
DIA [1.3K]3 years ago
4 0

The figure is missing: find it in attachment.

a) (3i - 5j) N

First of all, we have to write each force as a vector with a direction:

- Force F1 points downward, so along the negative y-direction, so we can write it as

F_1 = -5 j

- Force F2 points to the right, so along the positive x-direction, so we can write it as

F_2 = +8 i

- Force F3 points to the left, so along the negative x-direction, so we can write it as

F_3 = -5 i

Now we can write the net force, by adding the three vectors and separating the x-component from the y-component:

F=F_1+F_2+F_3 = -5j+8i-5i = (8-5)i+(-5)j=(3i-5j)N

b) 5.8 N

The magnitude of the net force can be calculated by applying Pythagorean's theorem on the components of the net force:

F=\sqrt{F_x^2+F_y^2}

where

F_x = 3 N is the x-component

F_y = -5 N is the y-component

Substituting into the equation,

F=\sqrt{(3)^2+(-5)^2}=5.8 N

c) -59.0^{\circ}

The angle of the net force, measured with respect to the positive x-axis (counterclockwise), can be calculated by using the formula

\theta=tan^{-1} (\frac{F_y}{F_x})

where

F_x = 3 N is the x-component

F_y = -5 N is the y-component

Substituting into the equation, we find

\theta = tan^{-1} (\frac{-5}{3})=-59.0^{\circ}

d) 0.84 m/s^2

The acceleration can be found by using Newton's second law:

F=ma

where

F is the net force on an object

m is its mass

a is the acceleration

For the object in the problem, we know

F = 5.8 N

m = 6.9 kg

Solving the equation for a, we find the magnitude of the acceleration:

a=\frac{F}{m}=\frac{5.8}{6.9}=0.84 m/s^2

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Explanation:

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3 years ago
2. A jack exerts a vertical force of 4.5 X 103
skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

\\  \\

Given :-

\star  \sf  \small force = 4.5 \times  {10}^{3}  \: newton

\star  \sf  \small distance = 0.25 \: meter

\\  \\

To find:-

\sf \star \: work = \: ?

\\  \\

Solution:-

we know :-

\bf \dag \boxed{ \rm work = force \times distance}

\\  \\

So:-

\dashrightarrow \sf work = force \times distance

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{0 \cancel.5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \cancel \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work =  \dfrac{4\cancel.5}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{3 - 1} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{1}  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times  \cancel{10}}{ \cancel2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times 5}{ 1} \\

\\  \\

\dashrightarrow \sf work =225 \times 10

\\  \\

\dashrightarrow \bf work =\red{2250\: joule}

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2 years ago
Does the atmosphere contain 21% nitrogen in 78% oxygen
Alex777 [14]

Answer:

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7 0
3 years ago
In a liquid with a density of 1400 kg/m3 , longitudinal waves with a frequency of 370 Hz are found to have a wavelength of 8.40
VARVARA [1.3K]

Answer:

Bulk modulus = 1.35 × 10^{10} Pa

Explanation:

given data

density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

we get here bulk modulus of the liquid that is

we know Bulk Modulus = v^2*\rho   ...............

here \rho is density i.e 1400 kg/m³

and v is =  frequency × wavelength

v = 370 × 8.40 = 3108 m/s

so here bulk modulus will be as

Bulk modulus = 3108² × 1400

Bulk modulus = 1.35 × 10^{10} Pa

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3 years ago
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