A 50 kg bicyclist starts his ride down the road with an acceleration of 1m/s2 in air with a density of 1.2 kg/m3. If his velocity at a given moment is 2m/s, how much force is he exerting? Assume the area of his body is 0.5m^2.
a. The bicyclist is exerting 1.1 N of force.
b. The bicyclist is exerting 49 N of force.
c. The bicyclist is exerting 50 N of force.
d. The bicyclist is exerting 51 N of force.

Answers

Answer 1

Answer:

b. The bicyclist is exerting 49 N of force

Explanation:

Given;

mass of bicyclist start, m = 50 kg

acceleration, a = 1 m/s²

density of air, ρ = 1.2 kg/m³

velocity, v = 2 m/s

Area of the bicyclist body, A = 0.5 m²

The drag force on the bicyclist is given by ;

Fd = 0.5CρAv²

where;

C is drag coefficient = 0.9 for bicycle

Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²

Fd = 1.1 N

The force of the bicyclist is given by;

F = ma

F = 50 x 1

F = 50 N

The effective force exerted by the bicyclist is given by;

Fe = F - Fd = 50 N - 1.1 N

Fe = 49 N

Therefore, the force exerted by the bicyclist is 49 N


Related Questions

On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)

Answers

Answer:

The height of the cliff is 121.276 m

Explanation:

Given;

initial velocity of the projectile, v₁ = 75 m/s

final velocity of the projectile, v₂ = 90 m/s

spring compression = 5 m

Apply the law of conservation of energy;

mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²

gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²

gh₁  - gh₂ = ¹/₂v₂² - ¹/₂v₁²

g(h₀  - h₂) = ¹/₂ (v₂² - v₁²)

h₀  - h₂ = ¹/₂g (v₂² - v₁²)

h₀ = h(cliff) + 5m

when the projectile hits the ground, Final height, h₂ = 0

[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]

Therefore, the height of the cliff is 121.276 m

An electric bulb rated 100 W, 100 V has to be

operated aross 141.4 V, 50 Hz A.C. supply. The

capacitance of the capacitor which has to be

connected in series with bulb so that bulb will

glow with full intensity is [NCERT Pg. 251]

Answers

Answer:

The capacitance of the capacitor is 31.84 μF.

Explanation:

Given;

power rating of the bulb, P = 100 W

voltage rating of the bulb, Vr = 100 V

operating voltage of the bulb, V= 141.4 V

frequency of the AC = 50 Hz

P = IV = 100 W

V = 100 V

I =

Ic = 1 A

The voltage across the capacitor is given by;

[tex]V_c = \sqrt{V^2 - V_R^2} \\\\V_c = \sqrt{141.4^2 - 100^2} \\\\V_c =99.97 \ V[/tex]

[tex]V_c = I_cX_c\\\\V_c = I_C* \frac{1}{2\pi fC}\\\\ 99.97 = 1 * \frac{1}{2\pi *50 *C}\\\\ C=\frac{1}{2\pi *50*99.97}\\\\ C = 31.84*10^{-6} \ F\\\\C = 31.84 \ \mu F[/tex]

Therefore, the capacitance of the capacitor is 31.84 μF.

A motorboat is a lot heavier than a pebble. Why does the boat float?

Answers

Answer:

The boat has more buoyancy

Explanation:

It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Answers

Answer:

102 m

Explanation:

Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Let the stopping distance be equal to S.

According to the definition of speed,

Speed = distance / time.

make time the subject of the formula

Time = distance / speed

then, the equivalent time is:

48.96 / 12 = S / 25

Cross multiply

12S = 48.96 x 25

12S = 1224

S = 1224 / 12

S = 102 m

Therefore, the stopping distance is 102 m

Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years

Answers

Answer:

T = 1.4696 10⁴ years

Explanation:

For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law

               F = ma

               G m M / r² = m a_c = m v² / r

                G M / r = v²

the speed of the circular orbit is

               v = 2π r / T

           

we substitute

               G M / r = 4π² r² / T²

               T² = (4π² / G M)  r³

Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse

               T² = (4π² /GM)  a³

the constant is worth

                (4π² / GM) = 2.97 10⁻¹⁹    s² / m³

let's reduce the distance to SI units

AU is the distance from the Earth to the Sun

               a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)

               a = 8.976 10¹³ m

               T² = 2.97 10⁻¹⁹ (8.976 10¹³)³

               T² = 21.4786 10²²

               T = 4.63 10¹¹ s

Let's reduce to years

               T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)

               T = 1.4696 10⁴ years

Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?

Answers

Answer:

Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?

A battery is used to charge a parallel-plate capacitor, after which it is disconnected. Then the plates are pulled apart to twice their original separation. This process will double the: __________A. capacitance
B. surface charge density on each plate
C. stored energy
D. electricfield between the two places
E. charge on each plate"

Answers

Answer: C.

Explanation:

For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.

Now, the stored energy can be written as:

E = (1/2)*Q^2/C

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.

If the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.

The given problem is based on the concept of parallel plat capacitor. For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

here.

e is the permittivity of free space.

Since, the distance is inversely proportional then if we double the distance, the capacitance halves.  Now, the stored energy can be given as,

E = (1/2)*Q^2/C

here,

Q is the charge stored in the capacitor.

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

So, the energy is proportional to the distance between the plates.

Thus, we can conclude that if the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.

Learn more about the energy stored in a capacitor here:

https://brainly.com/question/3611251

Jumping on a trampoline cause you to fly up in the air. What type of newton’s law is it ?

Answers

Answer:

The Third law

Explanation:

For every action there is an equal and opposite reaction.

Answer:

First Law

Explanation:

An object at rest (not moving) will stay at rest unless an unbalanced force acts on it.

An object in motion will stay in motion (in a straight line and at a constant speed) unless an unbalanced force acts on it.

You jump down on a trampoline and fly up in the air as a result.

the diagram shows a contour map. letter a through k are reference points on the map. which points are located at the same elevation above sea level?

Answers

Answer:

K and I

Explanation:

Contour maps use lines that represent spaces in a map that have the same elevation, this means that all the lines should be continuous and closed, in this case, we are not able to see the full extent of most of the lines, but since the points are located in different lines we can assume that they are at different heights, so since only point K and point I are on the same line, we know that these two points are at the same height.

A particle moves along a path described by y=Ax^3 ​​ and x = Bt, where tt is time. What are the units of A and B?

Answers

Answer:

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Explanation:

From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:

[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)

[tex][l] = B\cdot [t][/tex] (Eq. 2)

Now we finally clear each constant:

[tex]A = \frac{[l]}{[l]^{3}}[/tex]

[tex]A = \frac{1}{[l]^{2}}[/tex]

[tex]B = \frac{[l]}{[t]}[/tex]

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Calculate the force a 75 kg high jumper must exert in order to produce an acceleration that is 3.2 times the acceleration due to gravity.

Answers

Answer:

Explanation

According to Newton's second law of motion,

F = ma

m is the mass

a is the acceleration

If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g

The formula becomes;

F = m(3.2g)

F = 3.2mg

m= 75kg

g = 9.81m/s²

F = 3.2(75)(9.81)

F = 2,354.4N

Hence the force exerted by the jumper is 2,354.4N

cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?

Answers

Answer:

Displacement is 3 km North

Explanation:

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground

Answers

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

Part D

Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three

possible combinations. In two of the combinations, the two ends are the same. In one combination, the

two ends are different. Describe the force you feel in each combination

Answers

Answer:

i. The magnetic force of repulsion.

ii. The magnetic force of attraction.

Explanation:

A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.

i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.

ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.

The scientific method is the only way of learning about Nature used by scientist today *

A. true
B. false

Answers

Answer:

false

Explanation:

In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by:
A) applying a known force and measuring it's acceleration.
B) measuring the volume.
C) weighing it.

Answers

Answer:

A) By applying a known force, and measuring it's acceleration.

Explanation:

This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.

Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.

The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value

Answers

Complete Question

The complete  question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The original voltage is  [tex]V_o[/tex]

     The new voltage is [tex]V  =\frac{V_o}{2}[/tex]

     The capacitance is  [tex]C = 150\ nF = 150 *10^{-9} \  F[/tex]

     The first resistance is  [tex]R_i =  26 \Omega[/tex]

      The second resistance is [tex]R_E =  200 \Omega[/tex]

Generally the equivalent resistance is  

        [tex]R_e =  R_1 + R_E[/tex]

=>     [tex]R_e =  26 +200 [/tex]

=>     [tex]R_e = 226 \ \Omega [/tex]

Generally the time constant is mathematically represented as

     [tex]\tau  =  RC[/tex]

=>  [tex]\tau  =  226 * 150 *10^{-9}[/tex]

=>  [tex]\tau  =  3.39 *10^{-5} \  s [/tex]

Generally the voltage is mathematically represented as

    [tex]V =  V_o  e^{-\frac{t}{\tau} }[/tex]

=>   [tex]\frac{V_o}{2} =  V_o  e^{-\frac{t}{\tau} }[/tex]

=>   [tex]0.5 =    e^{-\frac{t}{\tau} }[/tex]

=>   [tex]ln(0.5) =    {-\frac{t}{ 3.39 *10^{-5} } }[/tex]

=>  [tex]ln(0.5)   * 3.39 *10^{-5}  =   -t [/tex]

=>  [tex]t = 2.35*10^{-5} \  s  [/tex]

A negative charge -Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth. Is any excess charge induced on the inner surface of the metal? Is there any excess charge on the outside surface of the metal? Why or why not? Would someone outside the solid measure an electric field due to the charge -Q? Is it reasonable to say that the grounded conductor has shielded the region outside the conductor from the effects of the charge -Q? In principle, could the same thing be done for gravity? Why or why not?

Answers

Answer:

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

Explanation:

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.

A daring stunt woman sitting on a tree limb
wishes to drop vertically onto a horse gallop-
ing under the tree. The constant speed of the
horse is 6.8 m/s, and the woman is initially
1.91 m above the level of the saddle.
How long is she in the air? The acceleration
of gravity is 9.8 m/s.
Answer in units of s.

Answers

Answer:

she is in the air for approximately 0.62 seconds

Explanation:

We want to find the time for a free fall under the acceleration of gravity, covering a distance of 1.91 m, and considering that the woman doesn't impart initial velocity in the vertical direction. So we use the kinematic equation:

[tex]d=v_i\,t+ \frac{g}{2} \,t^21.91 = 0 +4.9\, t^2\\t^2=1.91/4.9\\t=\sqrt{1.91/4.9} \\t\approx 0.624\,\,sec[/tex]

Then she is in the air for approximately 0.62 seconds

ionic bonds form when electrons?

Answers

Answer:

when the electron transferred permanently to another atom

A 0.75 m3 rigid tank initially contains air whose density is 1.18 kg/m3. The tank is connected to a high-pressure supply link through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 4.95 kg/m3. Determine, in kg, the mass of air that has entered the tank..

Answers

Answer:

2.83kg

Explanation:

Answer:

2.83kg

Explanation:

Given

initial density = 1.18 kg/m3

Final density in the tank = 4.95 kg/m3.

Let us write the mass balance first.

Change in the mass of the system=mass of the air entering the system - Mass of air out the system

Mass that entered= M2 - M1

But DENSITY= MASS/ VOLUME

Mass= volume × Density

We can expressed the mass in terms of density since density is given in the question.

Mass that entered= (volume × density)2 - ( volume × density)1

= (V ρ)2 - (V ρ)1

But V1= V2 the volume remains the same

= ( ρ2 - ρ1)v

= (4.95 kg/m3 - 1.18 kg/m3) 0.75 m3

= 3.77× 0.75

= 2.8275kg

Mass that entered= 2.83kg

therefore, mass of air that has entered the tank= 2.83kg

a car is moving eastward and speeding up. the momentum of the car is

Answers

Accelerating.. I hope that’s the answer your looking for


Weight of a body becomes greater at the pole than that at the equator . why ?

Answers

Answer:

Because the Earth rotates it is wider around at the equator than around the pole.
The distance from the Pole to the centre is smaller than the distance from the equator to the centre of the Earth. The weight decreases the further away from the centre of the Earth.

m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.

Answers

Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor:  T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

T₂ = 49.3°C

A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?

Answers

Answer:

Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.

 

Explanation:

To find the distance at which the first package will land we need to calculate the time:

[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]

Where:

Y(f) is the final position = 0

Y(0) is the initial position = 160 m

V(0y) is initial speed in "y" direction = 0

g is the gravity = 9.81 m/s²

t is the time=?                                          

[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]

[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]

Now we can find the distance of the first package:

[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]

Then, after 2 seconds the distance traveled by plane is (from the initial position):

[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]

Now, the distance of the second package is:

[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]

The distance between the packages is:

[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]

Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.

I hope it helps you!

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:

29 m/s2

31 m/s2

33 m/s2

30 m/s2

32 m/s2

Answers

Explanation:

The question is incomplete. Here is the complete question.

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of  30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...

Given

initial velocity of rocket u = 0m/s

final velocity of rocket = 30m/s

Height reached by the rocket = 98m

Required

upward acceleration of the rocket

Using the equation of motion below to get the acceleration a:

[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]

Hence  upward acceleration of the rocket during the burn phase is closest to 5m/s²

Note that the velocity used in calculation was assumed.

What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

Answers

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)

Answers

Answer:

III) [tex]d_{1}+ d_{2}=d_{t}[/tex]

Explanation:

I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5  with the the y being any other value.

II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)

III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]

(i) using coordinate grid notation to represent d₁, d₁ = 5y

(ii) using coordinate grid notation  to represent d₂, d₂ = 5x + 5y

(ii) The sum of d₁ and d₂ is written as 5x + 10y

In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.

Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.

(i) Using coordinate grid to represent d = (0, 5)

[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]

(ii) Using coordinate grid to represent d₂ = (5, 5)

[tex]d_2 = 5x + 5y[/tex]

(iii) The total vector is written as;

[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]

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Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons

Answers

Answer:

A group of protons and neutrons that are surrounded by electrons  I think that's the answer...

Explanation:

^ what they said
explanation: i don’t remember

What are the standard international (si) units of distance

Answers

Answer:

meter

Explanation:

Answer: The International System of Units is a system of measurement based on 7 base units

Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.

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