Marisa’s car accelerates at an average rate of 2.6m/s^2. Calculate how long it takes her car to accelerate from 24.6m/s to 26.8m/s? Show your work.

Answer:

B 5.0 A .

Explanation:

Hello.

In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:

[tex]I=\frac{Q}{t}[/tex]

Then, for the given data, we obtain:

[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]

Therefore, answer is B 5.0 A .

Best regards!

Answer:

Maybe

Explanation:

I say maybe because it will help them still but not quite

Answer:

Please find the answer in the explanation

Explanation:

Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.

What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.

What happens above the coil?

the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines

Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.

Below the coil?

The needle will move in an opposite direction.

Answer:

Kindly check the explanation section.

Explanation:

For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.

From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.

Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.

Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).

If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.

The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500

The fracture strength = .75 × Ah × Fhb = 309 kips.

The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.

Answer:

Explanation:

Given the density of silicon as 2.33g/cm³

We are to convert this to kg/cm³

We will be using the following conversion factors

1000g = 1kg

2.33g = x

Cross multiply

1000x = 2.33

x = 2.33/1000

x = 0.00233kg

Also we need to convert 1cm³ to 1m³

1cm = 0.01m

1cm³ = 0.01×0.01×0.01

1cm³ = 0.000001m³

Substituting into the density value of silicon

2.33g/cm³ = 0.00233kg/0.000001m³

= 2330kg/m³

Answer:

v = 66 m/s

Explanation:

Given that,

The initial velocity of a car, u = 0

Acceleration of the car, a = 11 m/s²

We need to find the final velocity of the toy after 6 seconds.

Let v is the final velocity. It can be calculated using first equation of motion. It is given by :

v = u +at

v = 0 + 11 m/s² × 6 s

v = 66 m/s

So, the final velocity of the car is 66 m/s.

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

Answer:

168

Explanation:

Answer: a 15 kg object would weigh the most on Neptune

168 N

Answer:

We are given:

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

Solving for Displacement:

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

v² - u² = 2 a ∆x

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

∆x = (27 m/s)² / (16 m/s²)

∆x ≈ 45.6 m

The stopping distance of car achieved during the braking is of 45.56 m.

Given data:

The initial speed of car is, u = 27 m/s.

The final speed of car is, v = 0 m/s. (Because car comes to stop finally)

The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].

In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.

Therefore,

[tex]v^{2}=u^{2}+2(-a)s[/tex]

Here, s is the stopping distance.

Solving as,

[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]

Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.

Learn more about the kinematic equation of motion here:

https://brainly.com/question/11298125

Answer:

The average force exerted on the superball by the sidewalk is 0.00122 N.

Explanation:

Given;

mass of the super ball, m = 50 g = 0.05 kg

initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)

final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)

time of motion, t = 1800 s

The average force exerted on the superball by the sidewalk is given by;

[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]

Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.

Answer:

The acceleration of the car is 10.16m/s²

Explanation:

Given parameters:

  Initial velocity = 8.77m/s

   Final velocity = 47.8m/s

   Time duration  = 3.84s

Unknown:

Acceleration of the car = ?

Solution:

To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;

     Acceleration  = [tex]\frac{V - U}{T}[/tex]

V is the final velocity

U is the initial velocity

T is the time taken

  Input the parameters and solve for acceleration;

      Acceleration  = [tex]\frac{47.8 - 8.77}{3.84}[/tex]   = 10.16m/s²

The acceleration of the car is 10.16m/s²

Answer:

I think it would be 50 I am not really sure

Explanation:

I think you would have to divid 1000 by 20 Again I'm not sure

Answer:

Explanation:

Magnetic flux is expressed as the product of magnetic field and cross sectional area.

Φ = BAsintheta

Given

B = 0.72T

A = 8.2cm×14cm

A = 0.082m × 0.14m

Area = 0.01148m²

Theta = 90°

Substitute into the formula

Φ = BAsintheta

Φ = 0.72(0.01148)sin90°

Φ = 0.72(0.01148)(1)

Φ = 0.0082656

Hence the magnetic flux through the loop is 8.2656 × 10^-3 Weber

Answer:

there is only a small amount of gravity present.

Explanation:

this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.

Answer:

6.7 m/s

Explanation:

Given:

Δx = 5 m

v₀ = 5 m/s

a = 2 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (5 m/s)² + 2 (2 m/s²) (5 m)

v = 6.7 m/s

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]

Explanation:

From the question we are told that

The weight of the mountain climber is m = 555 N

Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as  

       [tex]T_a*  cos 65 -555 + T_b * cos(85) =  0[/tex]

Here  [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side

 So

    [tex]0.423T_a   + 0.0871T_b  =  555[/tex]

=>   [tex] 0.0871T_b  =  555 - 0.423T_a[/tex]

=>   [tex] T_b  =  \frac{555 - 0.423T_a}{0.0871}[/tex]

Generally from the diagram , the total amount of force acting on the rope along the horizontal  axis at equilibrium is mathematically represented as

      [tex]T_a*  sin 65 - T_b * sin(85) =  0[/tex]

=>     [tex] 0.9063T_a - 0.9962T_b =  0[/tex]

=>     [tex] 0.9063T_a =   0.9962T_b [/tex]

=>     [tex] 0.9063T_a =   0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]

=>     [tex] 0.9063T_a =   [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]

=>    [tex] 0.0789T_a =   [552.891 - 0.421T_a[/tex]

=>    [tex] 0.4999T_a =   552.891 [/tex]

=>      [tex] T_a = 1106 \ N [/tex]

Answer:

options C is correct

Explanation:

asking questions is super in this education life

Answer:

option c should be the answer

Answer:

Explanation:

graphing is helpful

helps visualize the line

of your equation

The complete question is;

A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?

A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?

B) What is the power dissipated in his body?

C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?

Answer:

A) I = 1.379 A

B) P = 19016.41 W

C) r = 15990000 Ω

Explanation:

A) We are given;

Internal resistance of the power supply; r = 1600 Ω

Body resistance between hands; R = 10kΩ = 10000 Ω

Power supply voltage; E =16 kV = 16000 V

Formula for the current through the person's body with internal resistance is given by;

I = E/(R + r)

Thus;

I = 16000/(10000 + 1600)

I = 1.379 A

B) Formula for power dissipated is;

P = I²R

P = 1.379² × 10000

P = 19016.41 W

C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A

Thus, from I = E/(R + r) and making r the subject, we have;

r = (E/I) - R

r = (16000/0.001) - 10000

r = 15990000 Ω

Answer: f= M×A

1.75kg×24= 42N

Explanation:

Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!

Answer:

[tex](1630.13\pm 300.10)\ kg/m^3[/tex]

Explanation:

Given that,

The radius of a sphere is (6.45 ± 0.30) cm

Mass of the sphere is (1.79 ± 0.08) kg

Density = mass/volume

For sphere,

[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]

We can find the uncertainty in volume as follows :

[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]

Uncertainty in mass,

[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]

Now, the uncertainty in density of sphere is given by :

[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]

Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is  [tex]W_c =  P_o V_o ln (R_v)[/tex]  

Explanation:

Generally smallest workdone done by  a gas is mathematically represented as

          [tex]dW  =  PdV[/tex]

Generally for an isothermal process

    [tex]PV  =  nRT = constant [/tex]

=>   [tex]P = \frac{nRT}{V}[/tex]

Generally the total workdone is mathematically represented as

   [tex]W_c =  \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]

=> [tex]W_c = nRT  \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]

=>  [tex]nRT [lnV]   | \left \ {V_f}} \atop {V_o}} \right.[/tex]

=>  [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]

=>  [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]

From the question [tex]\frac{V_f}{V_o }  =  R_v[/tex]

=> [tex]W_c =  P Vln (R_v)[/tex]

at initial  state

[tex]W_c =  P_o V_o ln (R_v)[/tex]  

Answer:

The value  is [tex]  \mu  = 0.72  [/tex]

Explanation:

From the question we are told that

   The mass of the first car is  [tex]m_1  =  1870\ kg[/tex]

    the initial  speed of the car  is  [tex]u  =  13.5 \  m/s[/tex]

    The  mass of the second car is  [tex]m_2 =  2970\  kg[/tex]

    The distance move by both cars is  s =  1.93  m

Generally from the law of momentum conservation

    [tex]m_1 * u_1 + m_2 *  u_2  =  (m_1 + m_2 ) *  v_f[/tex]

Here [tex]u_2  =  0[/tex] because the second car is at rest

and  [tex]v_f[/tex] is the final  velocity of the the two car

So

    [tex]1870*  13.5+ 0=  ( 1870 + 2970 ) *  v_f[/tex]      

=> [tex]v_f  =  5.22\  m/s[/tex]

Generally from kinematic equation

    [tex]v_f^2 = u_2^2  +  2as[/tex]

here a is the deceleration

So

    [tex]5.22^2 = 0  +  2 *a  *  1.93[/tex]

=> [tex]a =  7.06 \  m/s^2 [/tex]

Generally the frictional  force is equal to the force propelling the car , this can be mathematically represented as

   [tex]F_f  =  F[/tex]

Here  F is mathematically represented as

[tex]F =  (m_1 + m_2) *  a[/tex]

[tex]F =  (1870 + 2970) *  7.06 [/tex]    

[tex]F =34170.4 \ N[/tex]  

and

[tex]F_f  =  \mu *  (m_1 + m_2 ) *  g[/tex]

[tex]F_f  =  47432 * \mu [/tex]

So

[tex]  47432 * \mu   = 34170.4  [/tex]

=> [tex]  47432 * \mu   = 34170.4  [/tex]

=> [tex]  \mu  = 0.72  [/tex]

Answer:

9 ft*lb

Explanation:

super simple but you just have to understand that the integral is going with the curve

work = integral a to b of f(x)dx = integral 0 to 9 of 10/(1+x)^2dx = 9ft*lb

Answer: D 160kg object moving 126 m/s

Explanation:

An object having a mass of 162 kg and moving with a velocity of 124 m/sec will require the greatest amount of force to change its motion. The correct option is D.

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

If the object has to stop, the final velocity must be zero. If the time is constant, the amount of force only depends on the mass and the velocity at which the body is moving.

The amount of force on the object depends on the momentum of the body.

The momentum of the body is;

P = mv

Object D will require the greatest amount of force to change its motion. Because the momentum of the body for option D is the greatest.

Hence, the correct option is D.

Learn more about the Force, here;

https://brainly.com/question/26115859

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Answer:

The magnitude of the net force acting on an object is equal to the mass. and the direction is in 20N

Explanation:

Answer:

(1,)

Explanation:

Answer:

magnitude: 21.6; direction: 33.7°

Explanation: