Help plz!:( it’s for physics but it has like math !!
(SHOW ALL WORK)

1) A 30-kg object is accelerated at a rate of 3m/s2. how much force does the object have?

2)A ball is thrown w/ a force of 50N and travels 4m. how much work is done to the ball?

3) Joey pushes on a door that is stuck with 400 N of force but is unable to move it. How much work has he done?

4) A toy car does 300j of work in 15 sec. how much power did the car use?

5)A 60w light bulb burns for 2 min. how much work was done?

6) how much power is used if it takes Frank (a 450 N boy) 3 sec to run 2 meters?

Complete Question:

A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will

A. be behind the package.

B. be over the package.

C. be in front of the package.

D. depend on the speed of the plane when the package was released.

Answer:

B.

Explanation:

As no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed, so when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.

The package hits the ground in such a way that the plane will be exactly over the package.

The given problem is based on the effect of air resistance on the motion of object. The resistive force exerted on any object on account of dust, and other air particles is known as air resistance or simply as air drag.

In the given problem, as no other horizontal forces are present, due to the horizontal movement and the vertical one are independent each other (as they are perpendicular), the plane and the package continue moving horizontally at the same speed.

So when the package hits the ground (due to the action of gravity in the vertical direction only) the plane will be exactly over the package.

Thus, we can conclude that the package hits the ground in such a way that the plane will be exactly over the package.

Learn more about the air drag here:

https://brainly.com/question/14755232

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

[tex]volume = \frac{mass}{density} \\[/tex]

From the question

mass = 10 g

density = 87 g/ml

We have

[tex]volume = \frac{10}{87} \\ = 0.114942528...[/tex]

We have the final answer as

Hope this helps you

Answer:

6.7 seconds

Explanation:

d=(1/2)at^2

equation

1000=(1/2)45t^2.

substitute

2000=45t^2.

multiply by 2 for both sides

44.44=t^2.

divide both sides by 45

6.7=t

take the square root of both sides

Answer:

apex consumers

Explanation:

they are top

Answer:

they don't strech so they tear a muscle when they perform

Explanation:

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

Answer:

adding bulbs:bulbs will glow in parallel circuit.

but the brightness of each bulb differs in series circuit.

removing bulbs:if u remove a bulb from parallel circuit,the other bulbs continue glowing in the same brightness

but if u remove a bulb in series circuit,the brightness of each bulb increases.

Answer:

The person above is correct.

Explanation:

In a parallel sequence, the light bulbs have electrons that flow equally. Though, in a series circuit, it's mainly connected to one wire, and the electrons get weaker.

Answer: It will be about 44.1m

Explanation:

Answer:

Dividing the silicon density by 1000 and then multiply it by 1000000.

Explanation:

A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:

[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]

[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]

In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.

Answer:

(a) [tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

(b) [tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

(c) [tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Explanation:

Vectors

Given a vector

[tex]\vec A=6\hat i-2\hat j+3\hat k[/tex]

We must determine the following:

a) A vector in the same direction as A with double magnitude 2A.

If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:

[tex]2\vec A=12\hat i-4\hat j+6\hat k[/tex]

b) A unit vector in the same direction of A.

The unit vector needs to compute the magnitude of the vector:

[tex]\mid A\mid=\sqrt{6^2+2^2+3^2}[/tex]

[tex]\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7[/tex]

[tex]\mid A\mid=7[/tex]

The unit vector is:

[tex]\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}[/tex]

[tex]\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}[/tex]

[tex]\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k[/tex]

c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.

The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:

[tex]-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)[/tex]

[tex]-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k[/tex]

Answer:

Δt = 1.1 s

Explanation:

Given information:

H= 50.0 m

g= 9.8 m/s²

   [tex]H = \frac{1}{2} * g* t^{2}[/tex]

Solving for t, we get:

t₁= 3.2 s

We can find the final velocity for the object when it hits the ground, using the following expression:

[tex]v_{f}^{2} - v_{o}^{2} = 2*g*H[/tex]

Solving for vf, we get:

vf = 33.9 m/s

Applying the definition of acceleration, being this acceleration the one due to gravity (g), we can write the following equation:

[tex]v_{f} = v_{o} + g*t[/tex]

(Assuming the downward direction to be positive).

Solving for t, we get:

t₂ = 2.1 s

So the difference in time when both objects hit the ground, it's simply

Δt = t₂ - t₁ = 3.2 s - 2.1 s = 1.1 s

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

Answer:

The maximum torque on the loop is 395.80 N.m.

Explanation:

Given;

number of turns of the wire, N = 150 turns

length of the square loop, L = 18.0 cm = 0.18 m

current in the wire, I = 50.9 A

Magnetic field, B = 1.6 T

Maximum torque on the loop is given by;

τ = NIAB

τ = (150)(50.9)(0.18²)(1.6)

τ = 395.80 N.m

Therefore, the maximum torque on the loop is 395.80 N.m.

Answer:

1. [tex] E = 1.14 \cdot 10^{13} N/C [/tex]

2. [tex]E_{k} = 9.1 \cdot 10^{-17} J[/tex]      

Explanation:

1. The strength of the nucleus' electric field (E):

[tex]E = \frac{kq}{r^{2}}[/tex]

Where:

k: is the Coulomb constant = 9x10⁹ Nm²/C²

q: is the proton charge = 1.6x10⁻¹⁹ C

r: is the radius = 10⁻¹⁰ m

[tex]E = \frac{kq}{r^{2}} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*79*1.6 \cdot 10^{-19} C}{(10^{-10} m)^{2}} = 1.14 \cdot 10^{13} N/C[/tex]

2. The kinetic energy (Ek) of an electron is the following:

[tex] E_{k} = \frac{1}{2}mv^{2} [/tex]    

Where:

m is the electron's mass = 9.1x10⁻³¹ kg

v: is the speed of the electron

We can find the speed of the electron by equaling the centripetal force (Fc) and the electrostatic force (Fe):

[tex] F_{c} = F_{e} [/tex]  

[tex] \frac{mv^{2}}{r} = \frac{kq^{2}}{r^{2}} = qE [/tex]

[tex] v^{2} = \frac{qEr}{m} = \frac{1.6 \cdot 10^{-19} C*1.14 \cdot 10^{13} N/C*10^{-10} m}{9.1 \cdot 10^{-31} kg} = 2.00 \cdot 10^{14} m^{2}/s^{2} [/tex]                  

Now, we can find the kinetic energy:

[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}9.1 \cdot 10^{-31} kg*2.00 \cdot 10^{14} m^{2}/s^{2} = 9.1 \cdot 10^{-17} J [/tex]    

I hope it helps you!

Answer:

The correct option is  B

Explanation:

From the question we are told that

   The mass of the pile is  M

   The  height is  H  =  4 y

    The vertical distance achieve during the first lift is  [tex]h_1  =  3 y[/tex]

     The time taken is  [tex]t_1 =  4T [/tex]

    The vertical distance achieve during the second lift is  [tex]h_2  =  y[/tex]

     The time taken is [tex] t_2 =  T [/tex]

Generally the velocity of the crane during the first lift is  

      [tex]v _1 =  \frac{h_1}{t_1 }[/tex]

=>    [tex]v _1 =  \frac{3 y}{4T }[/tex]    

Generally the velocity of the crane during the second  lift is    

      [tex]v _2 =  \frac{h_2}{t_2 }[/tex]

=>    [tex]v _2 =  \frac{ y}{T}[/tex]  

Generally the power generated by  the crane during the first lift is    

    [tex]P_1 =  F_1 *  v_1[/tex]

Here [tex]F_1[/tex] is the weight of the brick which is mathematically represented as

      [tex]F_1 =  M  * g [/tex] , g is the acceleration due to gravity

 So

       [tex]P_1 =  Mg  *  \frac{3y}{4T}[/tex]

Generally the power generated by  the crane during the first lift is    

    [tex]P_1 =  F_2 *  v_2[/tex]

Here [tex]F_2[/tex] is the weight of the brick which is mathematically represented as

      [tex]F_2 =  M  * g [/tex] , g is the acceleration due to gravity

 So

       [tex]P_1 =  Mg  *  \frac{y}{T}[/tex]

The ratio of the first power generated to the second power is  

       [tex]\frac{P_1}{P_2} =  \frac{Mg  *  \frac{3y}{4T} }{ Mg  *  \frac{y}{T} }[/tex]

=>    [tex]\frac{P_1}{P_2}  =  \frac{3}{4}[/tex]

=>   [tex]P_2 = \frac{4}{3} P_1[/tex]

Answer:

We are given:

V1 = 5.5L          T1 = -38 C   or   235 k

V2 = 1.3L           T2 = T

From the gas equation:

PV = nRT

Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:

V / T = k  (where k is a constant)

so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables

Changing the variables for the first case:

V1 / T1 = k   (where k is the same constant) ----------------(1)

Similarly,

V2 / T2 = k  (again, k has the same value)----------------(2)

From (1) and (2):

k is the common value

V1 / T1 = V2 / T2

Replacing the variables

5.5 / 235 = 1.3 / T

T = 1.3 * 235 / 5.5

T = 55.54 k

Therefore, at 55.54 K the gas will have a volume of 1.3L

Answer:

I believe the electromagnetic field should be reversed.

Explanation:

When a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.

An electromagnetic wave may be defined as a type of wave that is significantly created as a result of vibrations between an electric field and a magnetic field. These waves are composed of oscillating magnetic and electric fields.

According to the context of this question, when an individual is constructing an electromagnetic wave and then reverses the direction of the current, it will eventually affect the direction of the magnetic field in the same direction with respect to the current. So, if the direction of the current is reversed, the direction of the magnetic field would also be reversed.

Therefore, when a student creates an electromagnetic wave and then reverses the direction of the current, the direction of the magnetic field will be reversed.

To learn more about Magnetic fields, refer to the link:

https://brainly.com/question/14411049

#SPJ6

Your question seems incomplete. The most probable complete question is as follows:

A student creates an electromagnetic wave and then reverses the direction of the current. Which of the following will happen to the magnetic field?

Answer:

B) 200 [J]

Explanation:

In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]

F = m*g

where:

m = mass = 5 [kg]

g = gravity acceleration = 10 [m/s^2]

F = 5*10 = 50 [N]

w = F*d

where:

F = force = 50 [N]

d = 4 [m]

w = 50*4 = 200 [J]

Answer:

1. p = 562.3 kg*m/s

2. F = 142.7 N

Explanation:

1. The linear impulse (p) is given by:

[tex] p = mv [/tex]

Where:

m: is the passenger's mass = 73.7 kg

v: is the speed = 7.63 m/s

[tex] p = mv = 73.7 kg*7.63 m/s = 562.3 kg*m/s [/tex]

Hence, the magnitude of the linear impulse experienced by a passenger is 562.3 kg*m/s.

2. The average force can be calculated using the following equation:

[tex] F = \frac{m(v_{f} - v_{0})}{t} = \frac{73.7 kg(7.63 m/s - 0)}{3.94 s} = 142.7 N [/tex]  

Therefore, the average force experienced by the passenger is 142.7 N.

I hope it helps you!

Answer:

A professional basketball game depends on the association presiding over the game. An NBA game lasts for 48 minutes whereas FIBA games take 40 minutes. The total time taken to play for any specialized game is over 2 hours 15 minutes. The time includes the time disruptions like fouls, timeouts, and breaks.

I hope it helps you...

Answer:

The period of that same pendulum on the moon is 12.0 seconds.

Explanation:

To determine the period of that same pendulum on the moon,

First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].

From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²

∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]

[tex]g_{M}[/tex] = 1.63 m/s²

From the question, T=2π√L/g

[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]

We can write that,

[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)

Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity

and

[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)

Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.

Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.

Dividing equation (1) by (2), we get

[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]

From the question,

[tex]T_{E} = 4.9secs[/tex]

[tex]g_{E}[/tex] = 9.8 m/s²

[tex]g_{M}[/tex] = 1.63 m/s²

[tex]T_{M}[/tex] = ??

From,

[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]

[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]

[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]

[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]

[tex]T_{M} = 12.01 secs[/tex]

∴ [tex]T_{M} = 12.0secs[/tex]

Hence, the period of that same pendulum on the moon is 12.0 seconds.

Answer:

The period of that same pendulum on the moon is 12.0 s

Explanation:

Given;

period of a pendulum’s swinging, T=2π√L/g

the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)

period of pendulum on Earth, T₁ = 4.9 s

period of pendulum on moon, T₂ = ?

The length of the pendulum is constant, make it the subject of the formula;

[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]

Therefore, the period of that same pendulum on the moon is 12.0 s

This question is incomplete, the complete question is;

A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.

If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation

Answer: the time constant of the damped oscillation is 47.44s

Explanation:

Given that;

t = 5.0s

Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao

EXPRESSION for amplitude is  A(t) = Ao e^-t / T

t is time while T is time constant

so

0.9Ao = Ao e^-t / T  

0.9 = e^ -t/T

So we take the natural log of both the sides

ln (0.9) = -t/T

-0.1054 =  -t/T

0.1054 =  t/T

WE now substitute our value of t

0.1054 =  t/T

0.1054T =  5.0

T = 5 / 0.1054

T = 47.44s

therefore the time constant of the damped oscillation is 47.44s

Answer:

(a) 22 kN

(b) 36 kN, 29 kN

(c) left will decrease, right will increase

(d) 43 kN

Explanation:

(a) When the truck is off the bridge, there are 3 forces on the bridge.

Reaction force F₁ pushing up at the first support,

reaction force F₂ pushing up at the second support,

and weight force Mg pulling down at the middle of the bridge.

Sum the torques about the second support.  (Remember that the magnitude of torque is force times the perpendicular distance.  Take counterclockwise to be positive.)

∑τ = Iα

(Mg) (0.3 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L)

F₁ = ½ Mg

F₁ = ½ (44.0 kN)

F₁ = 22.0 kN

(b) This time, we have the added force of the truck's weight.

Using the same logic as part (a), we sum the torques about the second support:

∑τ = Iα

(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)

F₁ = ½ Mg + ⅔ mg

F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)

F₁ = 36.0 kN

Now sum the torques about the first support:

∑τ = Iα

-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)

F₂ = ½ Mg + ⅓ mg

F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)

F₂ = 29.0 kN

Alternatively, sum the forces in the y direction.

∑F = ma

F₁ + F₂ − Mg − mg = 0

F₂ = Mg + mg − F₁

F₂ = 44.0 kN + 21.0 kN − 36.0 kN

F₂ = 29.0 kN

(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):

F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)

As x increases, F₁ decreases and F₂ increases.

(d) Using our equation from part (c), when x = 0.6 L, F₂ is:

F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)

F₂ = ½ Mg + mg

F₂ = ½ (44.0 kN) + 21.0 kN

F₂ = 43.0 kN

Answer:

Explanation:

given:

weight of bridge = 44 kN

weight of truck = 21 kN

a) truck is off the bridge

since the bridge is symmetrical, left support is equal to right support.

Left support = Right support = 44/2

Left support = Right support = 22 kN

b) truck is positioned  as shown.

to get the reaction at left support, take moment from right support = 0

∑M at Right support = 0

Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0

Left support = 44 (0.3) + 21 (0.4)  

                                  0.6

Left support = 36 kN

Right support = weight of bridge + weight of truck - Left support

Right support = 44 + 21 - 36

Right support = 29 kN

c)

as the truck continues to drive to the right, Left support will decrease

as the truck get closer to the right support,  Right support will increase.

d) truck is directly under the right support, find reaction at Right support?

∑M at Left support = 0

Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0

Right support = 44 (0.3) + 21 (0.6)  

                                  0.6

Right support = 43 kN

Answer:

1.3 m/s

Explanation:

It is given that,

Mass of bird A, [tex]m_A=2.2\ kg[/tex]

Mass of bird B, [tex]m_B=1.7\ kg[/tex]

Initial speed of bird A is 0 as it was at rest

Initial speed of bird B is 3 m/s

We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,

[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]

So, the center of mass for this system is 1.3 m/s.

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

Answer:

jenny

Explanation: