A force of 1.5 N acts on a 0.2 kg cart. If the cart starts from rest, how fast is the cart going after accelerating over 30 cm?

Answer:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Explanation:

The force of gravity between two planets, is given by the following formula:

F = Gm₁m₂/r²   ----------- equation 1

where,

F = Force of gravity between two planets

G = Gravitational Constant

m₁ = Mass of one planet

m₂ = Mass of other plant

r = Distance between two planets

Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:

F' = Gm₁m₂/(4r)²

F' = (1/16) (Gm₁m₂/r²)

using equation 1:

F' = F/16

So, the force of gravity has become 16 times less than initial value.

Answer:

C is the correct answer

Explanation:

Answer:

578.2 N

Explanation:

The computation of reading on the scale is shown below:-

Data provided in the question

Weight of a girl = 59 kg

Constant rate = 8 m/s

Since the elevator is descended so the acceleration is zero

As we know that

Reading on the scale is

[tex]F = m\times g[/tex]

where,  m = 59 kg

g  [tex]= 59 \times 9.8 m/s^2[/tex]

So, the reading on the scale is

= 578.2 N

Therefore for computing the reading on the scale we simply applied the above formula.

Answer:

a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

Answer:

(a) ω = 12.57 rad/s

(b) a = 189.5 m/s²

(c) T = 9.47 N

Explanation:

(a)

The speed of rotation is given by the formula:

ω = θ/t

where,

ω = speed of rotation = ?

θ = angular displacement = (1 rotation)(2π rad/1 rotation) = 2π rad

t = time taken = 0.5 s

Therefore,

ω = 2π rad/0.5 s

ω = 12.57 rad/s

(b)

The centripetal acceleration of the object is given by the formula:

a = v²/r

where,

a = Centripetal Acceleration = ?

v = linear speed of object = rω

r = length of rope = 1.2 m

Therefore,

a = (rω)²/r

a = rω²

a = (1.2 m)(12.57 rad/s)²

a = 189.5 m/s²

(c)

The tension required to maintain the motion is equal to the centripetal force:

Tension = Centripetal Force

T = ma

where,

m = mass of object = 50 g = 0.05 kg

Therefore,

T = (0.05 kg)(189.5 m/s²)

T = 9.47 N

Answer:

O An increase in the temperature will increase the pressure.

Explanation:

Decrease in volume, more collisions, increase in temperature, more particles: all of these increase the pressure.

Answer: O An increase in the temperature will increase the pressure.

Answer:

[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]

Explanation:

Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:

[tex]a_{r} = \omega^{2}\cdot R[/tex]

Where:

[tex]\omega[/tex] - Angular speed, measured in radians per second.

[tex]R[/tex] - Radius of rotation, measured in meters.

The angular speed is first determined:

[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]

Where [tex]\dot n[/tex] is the angular speed, measured in revolutions per minute.

If [tex]\dot n = 3000\,rpm[/tex], the angular speed measured in radians per second is:

[tex]\omega = \frac{\pi}{30}\cdot (3000\,rpm)[/tex]

[tex]\omega \approx 314.159\,\frac{rad}{s}[/tex]

Now, if [tex]\omega = 314.159\,\frac{rad}{s}[/tex] and [tex]R = 0.1\,m[/tex], the resultant acceleration is then:

[tex]a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)[/tex]

[tex]a_{r} = 9869.588\,\frac{m}{s^{2}}[/tex]

If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:

[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]

Answer:

The object has an excess of [tex]10^{13}[/tex] electrons.

Explanation:

When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:

[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]

The object has an excess of [tex]10^{13}[/tex] electrons.

Answer:

T = 15,576 N

Explanation:

The speed of a wave on a string is given by

        v = √ T /ρ rho

also the speed of the wave is given by the relationship

       v = λ f

we substitute

     λ f = √ T /ρ

T = (lam f)² ρ

let's find the wavelength in a string, fixed at the ends, the relation that gives the wavelength is

       L= λ/2 n

       λ= 2L / n

we substitute

      T = (2L / n f)²ρ rho

let's calculate

      T = (2 1.20 / 2 590) 0.022

      T = 15,576 N

Complete Question

The complete question is gotten from OpenStax

A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s ? The player’s mass is 70.0 kg, and air resistance is negligible.

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball

Answer:

The  time it will take is  [tex]t = 1.4907 \ s[/tex]  

Explanation:

From the  question we are told that

    The force experienced by the player is  [tex]F = 126 \ N[/tex]

     The distance of the ball from the player is  [tex]d = 2.00 \ m[/tex]

      The initial velocity is  u =  0 m/s because the player stopped

From the Newton law the acceleration of the player is mathematically evaluated as

             [tex]a = \frac{F}{m }[/tex]    [i,e  F =  ma  ]

substituting values

             [tex]a = \frac{126}{70}[/tex]

             [tex]a = 1.8 \ m/s^2[/tex]

Now from the equation of motion  we have that

           [tex]s = ut + \frac{1}{2} at^2[/tex]

substituting values              

             [tex]2.0 = 0 + \frac{1}{2} * 1.8 * t^2[/tex]

             [tex]t = \sqrt{ \frac{2.0}{0.9} }[/tex]

            [tex]t = 1.4907 \ s[/tex]

Answer:

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Explanation:

Given;

Rate of inflation dV = 140pift^3/min

Radius r = 7 ft

Change in radius = dr

Volume of a spherical balloon is;

V = (4/3)πr^3

The change in volume can be derived by differentiating both sides;

dV = (4πr^2)dr

Making dr the subject of formula;

dr = dV/(4πr^2)

Substituting the given values;

dr = 140π/(4π×7^2)

dr = 0.714285714285 ft/min

dr = 0.71 ft/min

the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.

Answer:

Frequency= 0.25m

Period= 4.0 secs

Explanation:

Clay is surfing on a wave with a speed of 5.0m/s

The wave crests are 20m apart

Therefore, the frequency of the wave can be calculated as follows

Frequency= wave speed/distance

= 5.0/20

= 0.25m

The period (T) can be calculated as follows

T= 1/frequency

T = 1/0.25

T= 4.0secs

Hence the frequency is 0.25m and the period is 4.0 secs

Answer: The frequency heard is 562.7 Hz.

Explanation: Doppler Effect happens when there is shift in frequency during a realtive motion between a source and the observer of that source.

It can be calculated as:

[tex]f_{o} = f_{s}(\frac{c+v_{o}}{c-v_{s}} )[/tex]

where:

c is the speed of light (c = 332m/s)

all the subscripted s is related to the Source (frequency, velocity);

all the subscripted o is related to the Observer (frequency, velocity);

As the source is moving towards the observer and the observer is moving towards the source, the velocities of each are opposite related to direction.

So, the frequency perceived by the observer:

[tex]f_{o} = 439(\frac{332+32}{332-48} )[/tex]

[tex]f_{o} = 439(\frac{364}{284} )[/tex]

[tex]f_{o} = 439(1.282 )[/tex]

[tex]f_{o}[/tex] = 562.7 Hz

At this condition, the observer hears the train's horn in a perceived frequency of 562.7 Hz

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

Answer:

final velocity = 39.3m/s

distance travelled = 589.5m

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of object = [tex]6.0g/ml[/tex]

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

[tex]6.0g/ml=\frac{3.0g}{\text{Volume of substance}}[/tex]

[tex]{\text{Volume of substance}}=0.50ml[/tex]

Thus the volume of an irregularly shaped object is 0.50 ml

Answer:

axb

Explanation:

because the singular order

Answer:

V = 0.45 Volts

Explanation:

First we need to find the total current passing through the wire. That can be given by:

Total Current = I = (Current Density)(Surface Area of Wire)

I = (Current Density)(2πrL)

where,

r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m

L = Length of Wire = 6.5 m

Therefore,

I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]

I = 1.25 x 10⁻⁴ A

Now, we need to find resistance of wire:

R = ρL/A

where,

ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm

A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²

Therefore,

R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)

R = 0.36 Ω

From Ohm's Law:

Voltage = V = IR

V = (1.25 x 10⁻⁴ A)(0.36 Ω)

V = 0.45 Volts

Answer:

The force experienced  is 0.6 N

Explanation:

Given data

length of wire L= 2 m

current in wire I= 0.6 A

magnetic field B= 0.5

The force experienced can be represented as

[tex]F= BIL[/tex]

[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]

Answer:

3g/(8π²) ≈ 0.372 m

Explanation:

Draw a free body diagram.  There is a weight force at the center of the pendulum.

Sum of the torques about the pivot:

∑τ = Iα

mg (½ L sin θ) = (⅓ mL²) α

3g sin θ = 2L α

α = 3g/(2L) sin θ

For small θ, sin θ ≈ θ.

α = 3g/(2L) θ

θ" = 3g/(2L) θ

The solution of this differential equation is:

θ = θ₀ cos(√(3g/(2L)) t)

So the period is:

T = 2π / √(3g/(2L))

If the period is 1 second:

1 = 2π / √(3g/(2L))

√(3g/(2L)) = 2π

3g/(2L) = 4π²

L = 3g/(8π²)

L ≈ 0.372 m

The length of the pendulum rod is 0.37 m.

The time period of a pendulum is defined as the time taken by the pendulum to complete one oscillation.

Here,

The mass of the pendulum, m = 2 kg

Time period of the pendulum, T = 1 s

Since, the pendulum is suspended from the pivot and is oscillating, at the position of the pendulum when it makes an angle θ with the pivot, there is a force of weight acting at the center of the pendulum.

The length of pendulum at that point = L/2

The perpendicular distance at that point, r = (L/2) sinθ

Therefore, the torque acting on the pendulum at that point,

τ = Iα

where I is the moment of inertia of the pendulum and α is the angular acceleration.

mg (L/2 sinθ) = (mL²/3)α

1/2 gsinθ = 1/3 Lα

Therefore,

α = (3g/2L) sinθ

For smaller values of θ, we can take sinθ = θ

So, α = (3g/2L) θ

We know that α = θ''

where θ is the angular displacement.

Therefore,

θ'' = (3g/2L) θ

So, ω = √3g/2L

Therefore, the equation of motion of the pendulum can be written as,

θ = θ₀ cos(ωt)

θ = θ₀ cos [(√3g/2L) t]

So, time period of the pendulum,

T = 2[tex]\pi[/tex]/ω

T = 2[tex]\pi[/tex]/(√3g/2L)

2[tex]\pi[/tex]/(√3g/2L) = 1

(√3g/2L) = 2[tex]\pi[/tex]

Therefore, length of the rod,

L = 3g/8[tex]\pi[/tex]²

L = 0.37 m

Hence,

The length of the pendulum rod is 0.37 m.

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#SPJ2

Attaching the image file here.

Answer:

F2 = 282.28N

Explanation:

F = force

E = extension

Answer:

a. ω = 14.1 radians per day

b. ω = 2.24 rotations per day

Explanation:

a.

The relationship between linear and angular velocity is given as:

v = rω

where,

v = linear speed of point = 527787 miles/day

r = radius of Saturn = diameter/2 = 74900 miles/2 = 37450 miles

ω = angular velocity of point = ?

Therefore,

527787 miles/day = (37450 miles)ω

ω = (527787 miles/day)/(37450 miles)

ω = 14.1 radians per day

b.

The number of rotations per day can be found by converting the units of angular acceleration:

ω = (14.1 radians per day)(1 rotation/2π radians)

ω = 2.24 rotations per day

Answer:

Electric intensity will not necessarily be zero

Explanation:

E=−dV/dl

​where E= Electric field intensity

V= Electric Potential ;

l= distance traveled in direction of electric field

Electric intensity at a point is the negative of rate of change of the electric potential at a point. So if a function is zero at a given point, the slope will not necessarily be zero at that point.

Answer:

The work required to move this charge is 0.657 J

Explanation:

Given;

magnitude of charge, q = 4.4 x 10⁻⁶ C

Electric field strength, E =  3.9 x 10⁵ N/C

distance moved by the charge, d = 50 cm = 0.5m

angle of the path, θ = 40°

Work done is given as;

W = Fd

W = FdCosθ

where;

F is the force on the charge;

According the coulomb's law;

F = Eq

F = 3.9 x 10⁵ x 4.4 x 10⁻⁶  = 1.716 N

W = FdCosθ

W = 1.716 x 0.5 x Cos40

W = 0.657 J

Therefore, the work required to move this charge is 0.657 J

Answer:

L = 0.60 m

The length in metres should be 0.60 m

Explanation:

A pipe open at both ends can have a standing wave pattern with resonant frequency;

f = nv/2L ........1

Where;

v = velocity of sound

L = length of pipe

n = 1 for the fundamental frequency f1

Given;

Fundamental frequency f1 = 294 Hz

Velocity v = 350 m/s

n = 1

From equation 1;

Making L the subject of formula;

L = nv/2f1

Substituting the given values;

L = 1×350/(2×294)

L = 0.595238095238 m

L = 0.60 m

The length in metres should be 0.60 m

Answer:

(a) 62.3 N

(b) 1.89 N

(c) 0.430 kg m²

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B.  There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A.  There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

(c) Draw a free body diagram of the pulley.  There are two forces:

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].

Given :

a) First, determine the acceleration of the B block.

[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]

[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]

[tex]\rm a = 0.9\; m/sec^2[/tex]

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

[tex]\rm \sum F=ma[/tex]

[tex]\rm mg-T_b=ma[/tex]

[tex]\rm T_b = m(g-a)[/tex]

[tex]\rm T_b = 7\times (9.8-0.9)[/tex]

[tex]\rm T_b = 62.3\;N[/tex]

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

[tex]\rm \sum F=ma[/tex]

[tex]\rm T_a=ma[/tex]

[tex]\rm T_a = 2.1\times 0.9[/tex]

[tex]\rm T_a = 1.89\;N[/tex]

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

[tex]\rm \sum \tau = I\alpha[/tex]

[tex]\rm T_br-T_ar = I\alpha[/tex]

[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]

Now, substitute the values of the known terms in the above expression.

[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]

[tex]\rm I = 0.430 \; kg\;m^2[/tex]

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Answer:

q = 2.997*10^-4C

Explanation:

In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.

You use the second Newton law to sum both electrical and gravitational forces:

[tex]F_e-W=ma\\\\qE-mg=ma[/tex]             (1)

Fe: electric force

W: weight of the penny

q: required charge = ?

m: mass of the penny = 3g = 0.003kg

E: magnitude of the electric field = 100N/C

g: gravitational acceleration = 9.8m/s^2

a: acceleration of the penny = 0.19m/s^2

You solve the equation (1) for q, and replace the values of the other parameters:

[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]

It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2

Explanation:

[tex]1.2 \mathrm{N} ; 2 \mathrm{N}[/tex]

2.[tex]200 \mathrm{N} ; 200 \mathrm{N}[/tex]

4.[tex]2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}[/tex]

[tex]5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}[/tex]

[tex]6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}[/tex]

[tex]8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}[/tex]

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of [tex]256 \mathrm{m} / \mathrm{s}[/tex]

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.