Which is a characteristic of diatoms?
A)whip-like tail
B)heterotrophic
C)multicellular
D)glass-ive cell wall

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:

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:

option C is correct

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

Answer:

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

Explanation:

Took the test

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:

Mass

Explanation:

Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.  

Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.  

Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2

Answer:

The frequency of the 4th harmonic of the string is 481.13 Hz.

Explanation:

When a stretch string fixed at both ends is set into vibration, it produces its lowest sound of possible note called the fundamental frequency.  Under certain conditions on the string, higher frequencies called harmonics or overtones can be produced.

The frequency of the forth harmonic is the third overtone of the string and can be determined by:

          f = [tex]\frac{2}{L}[/tex][tex]\sqrt{\frac{T}{m} }[/tex]

Given that; L = 48.0 cm = 0.48 m,

                 m = 0.3 g = 0.0003 Kg,

                 T = 4.0 N,

         f = [tex]\frac{2}{0.48}[/tex][tex]\sqrt{\frac{4}{0.0003} }[/tex]

         f = 4.1667 × 115.4701

           = 481.1252

        f = 481.13 Hz

The frequency of the 4th harmonic of the string is 481.13 Hz.

Answer:

So the induced current opposes the motion that induced it (from Lenz's Law). When we pull the magnet out, the left hand end of the coil becomes a south pole (to try and hold the magnet back). Therefore the induced current must be flowing clockwise.

hope this helps u...

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:

[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:

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:

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:

true. b, c and d

Explanation:

Let's review each statement separately  

a) False. The kinematics studies the position, speed and acceleration of the bodies, but not what causes these changes

b) True. Velocity is the displacement between time, displacement is a vector, and time is a scalar, so the division between them gives a vector

c) True. speed is the displacement that is a length between time, so its unit is length / time

d) true  desaceleration = - aceleration

Answer:Velocity

Explanation:

Velocity is directly proportional to the frequency of a wave.

Velocity=frequency x wavelength

Answer:

Check the explanation

Explanation:

Part A

F = CA

this drag force balances the weight = 6X 9.8

so

6X9.8 = 0.5 X A X0.5 X 1.2 X 532

A= 0.069 m2

Part B

here the sorce is moving and the observer is at rest

so f= f(- 1 - 1

f = 1.1X10 343 343 – 53

f' = 1.3 KHz

Part C:

given the intensity = 30 dB

we know that I dB = 10 log (I(W/m2))

so we get I (W/m2) = 1000

Part D : The catch

Given that U1 = 53 M1 = 6 kg

U2 =-10 M2=0.25

V1=V2

now conserving momentum

6 X 53 -0.25 X10 =(6+0.25)V

V= 50.48 m/sec

Answer:

The magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

Explanation:

Information from the question:

Number of turns of the coil = 100 turns

The diameter of the coil = 6 m

The radius of the coil = diameter / 2 = 3 m

The coil current = 2.5 A

Formula : The Magnetic field at the center of the coil =

                                  k * number of turns * current / 2 * radius

Therefore, The Magnetic field at the center of the coil=

                                 (4 * [tex]\pi[/tex] * 10 ^ -7 * 100 * 2.5 ) / (2 * 3)

The Magnetic field at the center of the coil = 5.23 * 10 ^ -5 T

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:depends on the masses of the objects and the distance between them

Explanation:

According to Newton's law of universal gravitation,the force of attraction between two objects depends on the masses of the objects and the distance between them

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:

Here's your answer :

hope it helps!

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]

Next, you use the formula for the magnetic force produced by the wires:

[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

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:

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:

Minority group - People who are singled  out for unequal treatment  and who regard  themselves as objects of  collective discrimination.

Ethnicity - A shared identity based on  cultural elements such as  heritage, language, and  religion.

Majority group - A group's subset that  consists of more than half  of the entire group's  members.

Race - A socially constructed  group of people who share  physical characteristics  that are considered  significant by a society  and that are used to  distinguish them from  other groups.

Explanation:

To better understand the above-mentioned definitions we would use some examples:

Elder people can be considered as a minority group in the United States because of their reduced status as a consequence of popular discrimination and prejudice against them. In contrast to them, young men consisting of more than half the population in the age category is an example of the majority group.

Groups of people like Hispanic Americans, Jews, Italian America, Irish are some examples of people belonging to different ethnic backgrounds residing in the U.S.

African Americans who are distinguished based on their skin color is an example of race.

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: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:

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:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

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]

Answer:

A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.

The hot air displacing the cold air is an example of  transfer by

Explanation:

also the answer is hit my dm on ig