A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?

The time spent by the campers when they turn around downstream is 15 minutes.

The concept of total distance traveled by Nick and Chloe can be used to determine the time they turn around downstream.

Let time for downstream = t1

Let time for upstream = t2

distance covered in upstream = distance covered in downstream = d

12(t1) = d

4(t2) = d

12t1 = 4t2

t1 + t2 = 1

t2 = 1 - t1

12t1 = 4(1 - t1)

12t1 = 4 - 4t1

16t1 = 4

t1 = 4/16

t1 = 0.25 hours

t1 = 0.25(60 min) = 15 mins

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Answer:

Explanation:

The first thing we are asked to find is the Force experienced by the box. That is found in the formula:

F - f = ma where F is the force exerted by the box, f is the friction opposing the box, m is the mass, and a is the acceleration (NOT the same as the pull of gravity). But F can be rewritten in terms of the angle of inclination also:

[tex]wsin\theta-f=ma[/tex] where w is the weight of the box. We will use this version of the formula because it will help us answer the second question, which is to solve for a. Filling in:

First we need the weight of the box. Having the mass, we find the weight:

w = mg so

w = 25.14(10) so

w = 251.4 N (I am not paying any attention at all to the sig fig's here, since I noticed no one on this site does!) Now we have the weight. Filling that in:

251.4sin(30) - f = ma Before we go on to fill in for f, let's answer the first question. F = 251.4sin(30) so

F = 125.7   And in order to answer what a is equal to, we find f:

f = μ[tex]F_n[/tex] where Fn is the weight of the object.

f = .25(251.4) so

f = 62.85. Filling everything in now altogether to solve for a, the only missing value:

125.7 - 62.85 = 25.14a and

62.85 = 25.14a so

a = 2.5 m/s/s

Now we have to move on to another set of equations to answer the last part. The last part involves the y-dimension. In this dimension, what we know is that

a = -10 m/s/s

v₀ = 0 (it starts from rest)

Δx = -13.30 m (negative because the box falls this fr below the point fro which it started). Putting all that together in the equation for displacement:

Δx = v₀t + [tex]\frac{1}{2}at^2[/tex] and we are solving for time:

[tex]-13.30=0t+\frac{1}{2}(-10)t^2[/tex] and

[tex]t=\sqrt{\frac{2(-13.30)}{-10} }[/tex] so

t = 1.6 seconds to reach the bottom of the slope from 13.30 m high.

Answer:

v₀ₓ = 24.24 m / s

Explanation:

This is a projectile launching exercise, where the windshield comes out with a horizontal initial velocity.

Y axis  

initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

when it reaches the ground its height is zero and the initial height is y₀=1.2m

          0 = y₀ + 0 - ½ g t²

          t = [tex]\sqrt{2 y_o/g}[/tex]

          t = [tex]\sqrt{2 \ 1.2 / 9.8}[/tex]

          t = 0.495 s

X axis

          x = v₀ₓ t

          v₀ₓ = x / t

          v₀ₓ = 12 / 0.495

          v₀ₓ = 24.24 m / s

Answer:

See annex

Explanation:

By convention potential at ∞    V(∞ ) = 0

As the distance from the sphere decreases the potential increases up to the point d = R  ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.

I am sorry I could not make a better graph

The graph that best  illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below

[tex]V = \frac{KQ}{R}[/tex]

for r <= R

[tex]V = \frac{KQ}{r}[/tex]

for  r > R  

Therefore the graph will be

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Answer:

The efficiency of Carnot's heat engine is 26.8 %.

Explanation:

Temperature of hot reservoir, TH = 100 degree C = 373 K

temperature of cold reservoir, Tc = 0 degree C = 273 K

The efficiency of Carnot's heat engine is

[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]

The efficiency of Carnot's heat engine is 26.8 %.

Answer:

the correct answer is C

Explanation:

This is a system with circular motion, there is a relationship between the linear and angular variables

         a = α r

with the cube going down the well, the tension of the leather is maintained therefore the acceleration of the cube is

          W = m a

          -mg = ma

          a = -g

this acceleration a is the same as that at the edge of the drum.

         α = a / r

where we can see that the angular acceleration is constant

consequently the correct answer is C

Explanation:

Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.

Answer:

we need the block

Explanation:

1×2 =4 lest 74 =345

Explanation:

ehb-pynw-ayo

joi n fast

Answer:

Explanation:

Step 1) Combine all resistors into an equivalent overall resistor. These are all in series so you just add them up. Req = 10Ω + 20Ω + 30Ω = 60Ω:

Step 2) Using Ohm's Law, I = V/R = 120/60 = 2 A

Now you know how much current is flowing, and that current flows through each resistor the same. So the current in the 30 Ω resistor is 2.00 amps.

Answer:

The electric field is given by 4.5 N/C.

Explanation:

Charge density = 80 nC/m3

inner radius, r' = 1 mm

outer radius, r'' = 3 mm

distance,  r = 4 mm

The linear charge density is given by

[tex]\lambda =\rho \times\pi\times (r''^2 - r'^2)\\\\\lambda = 80\times 10^{-9}\times 3.14\times 10^{-6}\times(9-1)\\\\\lambda = 2\times 10^{-12}\\[/tex]

The electric field is given by

[tex]E = \frac{\lambda }{4\pi\varepsilon_or}\\E=\frac{9\times 10^9\times 2 \times 10^{-12}}{0.004}\\\\E=4.5 N/C[/tex]

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

Therefore, the electric field is  3.91 x 10⁶ N/C

F = 153 N

[tex]\theta = 11.3°[/tex]

Explanation:

Let us define first our directional convention. Anything pointing up or to the right is considered positive and anything pointing down or to the left is considered negative. Now let's look at the components [tex]F_{x}[/tex] and [tex]F_{y}[/tex]:

[tex]F_{x}[/tex] = 350 N - 200 N = 150 N

[tex]F_{y}[/tex] = 180 N - 150 N = 30 N

The magnitude of the resultant force F is given by

[tex]F = \sqrt{F_{x}^{2}+F_{y}^{2}}[/tex]

[tex]\:\:\:\:\:\:= \sqrt{(150\:N)^{2}+(30\:N)^{2}}[/tex]

[tex]\:\:\:\:\:\:=153\:N[/tex]

To find the direction [tex]\theta[/tex], we use

[tex]\tan \theta = \dfrac{F_{y}}{F_{x}}=\dfrac{30\:N}{150\:N}=0.2[/tex]

or

[tex]\theta = \tan^{-1}(0.2) = 11.3°[/tex]

Answer:

  T = 9056 K

Explanation:

In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation

                 λ T = 2,898 10⁻³

where lam is the wavelength of the maximum emission

                T = 2,898 10⁻³ /λ

let's calculate

                 T = 2,898 10⁻³ / 320 10⁻⁹

                  T = 9.056 10³ K

                  T = 9056 K

Answer:

171.5J

Explanation:

K=1/2 *m *U²

K=1/2 *7 *7²

K=171.5 J

Answer:

12+2=24+30+2=66

Explanation:

Answer:

A. the motion of electrically charged particles

Explanation:

Given that,

35 m/s at 57° from the x-axis.

Speed, v = 35 m/s

Angle, θ = 57°

Horizontal component,

[tex]v_x=v\cos\theta\\\\=35\times \cos(57)\\\\=19.06 m/s[/tex]

Vertical component,

[tex]v_y=v\sin\theta\\\\v_y=35\times \sin(57)\\\\=29.35\ m/s[/tex]

Hence, this is the required solution.

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =[tex]F^>[/tex] × [tex]x^>[/tex]

W = [tex]Fxcos\theta[/tex]    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = [tex]Fxcos([/tex] 0° [tex])[/tex]

W = [tex]Fx[/tex] ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

[tex]W_{net[/tex] = ΔKE

= [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu²

we make F, the subject of formula

F = [tex]\frac{m}{2x}[/tex]( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

Answer:

In linear motion, the directions of all the vectors describing the system are equal and constant which means the objects move along the same axis and do not change direction. Correct Answer: In a straight line

Explanation:

Answer:

In a straight line. we can also have translational motion which is also a kind of linear motion .

Answer:

M L1 = m L2       torques must be zero around the fulcrum

M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg

Answer:

the horizontal component of the force is 50 N

Explanation:

Given;

force applied by the man, F = 100 N

angle of inclination of the force, θ = 60⁰

mass of the dog, m = 20 kg

The horizontal component of the force is calculated as;

[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]

Therefore, the horizontal component of the force is 50 N

Answer:

the correct answer is D

Explanation:

In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.

       R = δ + ε

where

       δ = 2.0 m

       ε = 7.0 m

the positive sign indicates that it is heading east

        R = 2.0 + 7.0

        R = 9.0 m

the direction is east

the correct answer is D

Answer:

Force = 1.632 Newton

Explanation:

Given the following data;

Pressure = 102 kPa

Area = 0.016 m²

To find what force the atmosphere exert on the palm of your hand;

Mathematically, pressure is given by the formula;

[tex] Pressure = \frac {Force}{area} [/tex]

Force = 102 * 0.016

Force = 1.632 Newton

Answer:

0.444atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (

P2 = final pressure (

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question,

P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm

P2 = ?

V1 = 80L

V2 = 160L (double of V1)

T1 = 34°C = 34 + 273 = 307K

T2 = 0°C = 0 + 273 = 273K

Using P1V1/T1 = P2V2/T2

0.999 × 80/307 = P2 × 160/273

79.92/307 = 160P2/273

Cross multiply

307 × 160P2 = 79.92 × 273

49120P2 = 21818.16

P2 = 21818.16 ÷ 49120

P2 = 0.444

P2 = 0.444atm

Answer:

sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.

hope it is helpful to you

Answer: Hazmat products are allowed in your FC are:

Explanation:

Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.  

Hazardous material allowed in FC are as follows.

So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).

Thus, we can conclude that hazmat products are allowed in your FC are:

Answer:

- the coating’s index of refraction is 1.25

- the required thickness is 104.1667 nm

Explanation:

Given the data in the question;

Thickness of coating t = 100 nm

wavelength λ = 500nm

we know that refractive index is;

t = λ/4n

make n, the subject of formula

t4n = λ

n = λ / 4t

we substitute

n = 500 / ( 4 × 100 )

n = 500 / 400

n = 1.25

Therefore, the coating’s index of refraction is 1.25

2)

given that;

Index of refraction of the coating; n = 1.20

λ = 500 nm

thickness of coating t = ?

t = λ / 4n

we substitute

t = 500 / ( 4 × 1.2 )

t = 500 / 4.8

t = 104.1667 nm

Therefore, the required thickness is 104.1667 nm