An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s?

Answer:

i) Mr. Dunn arrives to home first.

ii) 3 min

Explanation:

i. To find who arrives first to home you calculate the time, by using the following formula:

[tex]t=\frac{x}{v}[/tex]

x: distance

v: velocity

Mr. Dunn:

[tex]t=\frac{64.8km}{48km/h}=1.35h[/tex]

Mrs. Dunn:

[tex]t=\frac{81.2km}{58km/h}=1.4h[/tex]

Hence, Mr. Dunn arrives to home first.

ii. To calculate the difference in minutes, you convert hours to minutes:

[tex]1.35h*\frac{60min}{1h}=81min\\\\1.40h*\frac{60min}{1h}=84min\\\\\Delta\ t=(84-81)min=3min[/tex]

the difference between the times is 3min

(i) Mr. Dunn takes less time so he arrives at home first.

(ii) The second person arrives 3 min late.

(i) We have to calculate the time taken to reach home by Mr. Dunn and Mrs. Dunn.

t = x/v

where x is the distance

and v is the velocity

Time taken by Mr. Dunn:

distance x = 64.8 km

speed v  = 48 km/h

t = 64.8 / 48

t = 1.35 h

Time taken by Mrs. Dunn:

distance x = 81.2 km

speed v  = 58 km/h

t' = 81.2 / 58

t' = 1.4 h

Hence, Mr. Dunn arrives at home first.

(ii) To calculate the difference in minutes, you convert hours to minutes:

The time taken by Mr. Dunn in minutes is:

t = 1.35×60 = 81 minutes

The time taken by Mrs. Dunn in minutes is:

t' = 1.4×60 = 84 minutes

the difference between the times is 3min

Learn more about distance and time :

https://brainly.com/question/4199102?referrer=searchResults

Answer:

Magnetic field can be used to produce current, infact a changing magnetic field can produce current.

A changing magnetic field in a loop causes the flux linked with the loop to change in turn generating a emf in the loop and therefore a current.

For a loop of area A and resistance R.

I =dPhi/dt/R

В. А

I = AcosФ/R .dB /dt

But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).

We were unable to transcribe this image

Ф= В. А

I = Acos dB Rd

Answer:

The time period for this pendulum is 1.68 seconds

Explanation:

Solution

Given that:

The length of the pendulum is measured from the axis of rotation to the center of mass of the bob of the pendulum

Now,

In this case, the length becomes:

L= 80 - 15+5

L = 70 cm

The time period = T = 2π √L/g

T = 2* 3.14 *√0.7/9.8

= 1.68 seconds

Note: Kindly find an attached work to the part of the solution of the given question

Answer:

option C is correct

................

Answer:

C- in an inelastic collision, both momentum & kinetic energy are conserved

Explanation:

Took the test

Answer:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.Explanation:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.

Each atom is made up of a nucleus and one or more electrons that are linked to it. One or more protons and a significant number of neutrons make up the nucleus. Only the most prevalent type of hydrogen is neutron-free.

Atoms that are neutral or ionized make up every solid, liquid, gas, and form of plasma. Atoms are incredibly tiny, measuring typically 100 picometers across. The nucleus of an atom contains more than 99.94% of its mass.

Therefore, An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

To learn more about atom, refer to the link:

https://brainly.com/question/1566330

#SPJ2

Answer:

A) 0.188

B) 53.1 ⁰

Explanation:

taking moment about 0

∑ Mo = Lo∝

mg 1/2 sin∅ = 1/3 m L^2∝

note ∝ = w[tex]\frac{dw}{d}[/tex]∅

forces acting along t-direction ( ASSUMED t direction)

∑ Ft = Ma(t) = mr∝

mg sin ∅ - F = m* 1/2 * 3g/2l sin∅

therefore F = mg/4 sin∅

forces acting along n - direction ( ASSUMED n direction)

∑ Fn = ma(n) = mr([tex]w^{2}[/tex])

= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )

hence N = mg/2 ( 5cos∅ -3 )

A ) Angle given = 30⁰c find coefficient of static friction

∪ = F/N

  = [tex]\frac{\frac{mg}{4}sin30 }{\frac{mg}{2}(5cos30 -3) }[/tex]  = 0.188

B) when there is no slip

N = O

   = 5 cos ∅ -3 =0

   therefore cos ∅ = 3/5  hence ∅ = 53.1⁰

Answer:30

Explanation:

Current=60 milliamps

Current=(voltage)/(resistance)

60=(voltage)/(resistance)

Doubling the resistance means multiplying both sides by 1/2

60x1/2=(voltage)/(resistance) x 1/2

30=(voltage)/2(resistance)

Therefore the resistance would be 30 milliamp if we double the resistance

Answer:

C. is always perpendicular to the surface of the conductor

Explanation:

On a charged conductor , electric charge is uniformly distributed on its surface . The lines of forces are also uniformly  distributed on all directions . They repel each other so they emerge perpendicular to the surface so that they do nor cut each other and at the same time they remain at maximum distance from each other.

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

Answer:

.7934[tex]m/s^{2}[/tex]

Explanation:

Acceleration = change in velocity / change in time

A = 10.98[tex]m/s[/tex] / 13.84[tex]s[/tex]

A = .7934[tex]m/s^{2}[/tex]

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s    (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

t = 319.47 s

Answer:

Gold, silver, platinum. Gold is the most malleable and ductile.

Explanation:

The elements which are malleable and ductile include the following:

Malleability is the ability of a substance to be hammered into thin sheets

while ductility involves the deformation of a substance without any

breakage occurring in it.

Transition metals are the group of elements which have both

characteristics and examples are listed above.

Read more about Transition metals here https://brainly.com/question/7102290

Answer:

1.47*10^{-8}s

Explanation:

You first calculate the distance traveled by the electron:

[tex]x=vt\\\\x=(0.941(3*10^8m/s))(15.7*10^{-9}s)=4.43m[/tex]

Next, you calculate the relative speed as measure by an observer in the positron, of the electron:

[tex]u'=\frac{u+v}{1+\frac{uv}{c^2}}\\\\u'=\frac{0.941c+0.941c}{1+\frac{(0.941)^2c^2}{c^2}}\\\\u'=0.99c[/tex]

with this relative velocity you calculate the time:

[tex]t=\frac{x}{u'}\\\\t=\frac{4.43m}{0.99c}=1.47*10^{-8}s[/tex]

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

Answer:

Explanation:

Initial volume of gas V₁ = 9.30 x 10⁻⁴ m³

final volume V₂ = 1 / 6 x  9.30 x 10⁻⁴

= 1.55 x 10⁻⁴ m³

Atmospheric pressure P = 1.013 x 10⁵ Pa .

temperature T .

PV = n RT

nRT = 1.013 x 10⁵ x 9.3 x 10⁻⁴

= 94.21

work done in isothermal process

= 2.303 nRT log V₁ / V₂

= 2.303 x 94.21 log 6

= 168.83 J .

Answer:

[tex]W=5.16 J[/tex]  

Explanation:

Using the Hooke's law we can find the elasticity constant:

[tex]F=-k\Delta x[/tex]

[tex]30.8=-k*0.177[/tex]

[tex]k=|-\frac{30.8}{0.177}|[/tex]

[tex]k=174 N/m[/tex]

Now, we know that the work done is equal to the elastic energy, so we will have:

[tex]W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})[/tex]

x2 is the final distance (x2 = 0.177+0.124 = 0.301 m)

x1 is the initial distance (x1 = 0.177 m)

[tex]W=\frac{1}{2}*174(0.301^{2}-0.177^{2})[/tex]

[tex]W=5.16 J[/tex]    

I hope it helps you!

Answer:

Explanation:

The two pictures attached here shows the solution to the two questions from the problem. thank you and I hope it helps you

Answer:

Explanation:

The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds . During .75 second duration . package undergoes free fall due to which additional vertical velocity is added

velocity added = a x t

= 9.8 x .75

= 7.35 m /s

Total vertical velocity

= 3 sin27 + 7.35

= 8.71 m /s

Horizontal component = 3 cos 27

= 2.67 m /s

If v be the resultant velocity of these components

v² = 2.67² + 8.71²

v² = 7.13 + 75.86

v = 9.11 m /s .

Answer:

Explanation:

Let the bullets speed be V .

Kinetic energy = 1/2 mV² where m is mass of bullet

This energy is converted into heat Q which raises the temperature of target by Δ T .

Q = mc Δ T  , m is mass , c is specific heat and Δ T is rise in temperature .

heat absobed by bullet

= .0075 x 130 x .040

= .039 J

heat absorbed by block of wood

= 17.5 x 1700 x .04

= 1190 J

Total heat absorbed

= 1190.039 J

So kinetic energy = heat absobed

= 1/2 x .0075 x V² = 1190.039

V² = 317343.73

V  = 563.33 m /s

Answer:B

Explanation:

Power=1000 watts

Time=5 seconds

Distance=5 meters

Force=(power x time) ➗ distance

Force=(1000 x 5) ➗ 5

Force=5000 ➗ 5

Force=1000

Force=1000N

Answer:1,000

Explanation:

ape.x

Answer:Velocity

Explanation:

Velocity is directly proportional to the frequency of a wave.

Velocity=frequency x wavelength

Answer:

Explanation:

given equation

aA + bB + C = 0

Putting the given values

a(6i-8j) +b (-8i-3j) +(26i-19j) = 0

i ( 6a - 8b ) - j ( 8a + 3 b ) = - 26 i + 19 j

comparing the coefficients of i and j

6a - 8b = -26

8a + 3b = -19.

multiplying first equation by 4 and second equation by 3  

24a - 32 b = - 104

24a + 9b = -57

9b + 32b = -57 + 104

41 b = 47

b = 1.41

6 a - 8 x 1.41 = -26

6a = -14.72

a = - 2.45  

Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

     Wₓ - fr = m a

the friction force has the expression

     fr = μ N

y Axis

     N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components the weight

     sin 27 = Wₓ / W

     Wₓ = W sin 27

     cos 27 = W_{y} / W

     W_{y} = W cos 27

     N = W cos 27

     W sin 27 - μ W cos 27 = m a

     mg sin 27 - μ mg cos 27 = m a

      μ = (g sin 27 - a) / (g cos 27)

      very = tan 27 - a / g sec 27

      μ = 0.510 - 0.0344

      μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

         μ = tan 25 - 0.3 / 9.8 sec 25

         μ = 0.466 -0.03378

         μ = 0.433

Answer:Convex lens spectacles is required for reading purpose..

Explanation:

I don't say you have to mark my ans as brainliest but if it has really helped you please don't forget to thank me...

Answer:

The horizontal distance between two adjacent crests or troughs is known as the wavelength.

Answer: Wavelength

Explanation:

From crest to crest, it is one full wavelength

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

[tex]\theta=0.260t^2[/tex]

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]

Then, you evaluate for t=2.40 s:

[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]

(b) The linear velocity is calculated by using the following formula:

[tex]v=\omega r[/tex]

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]

(c) The tangential acceleration is:

[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]

(d) The radial acceleration is:

[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]

Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm

k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 35969.4 x 10⁻³ N .

force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

= 118.7 N.

Answer:

Here's your answer :

hope it helps!

Answer:

a) 315 m

b) 3025.6 m

Explanation:

The picture attached shows the full explanation for the problem.

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

[tex]\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N[/tex]

Thus, the net force over the body is:

[tex]F=(913.14N)\hat{i}+(274.87N)\hat{j}[/tex]

Next, you calculate the magnitude of the force:

[tex]F=\sqrt{(913.14N)+(274.87N)^2}=953.61N[/tex]

and the direction is:

[tex]\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°[/tex]