Can someone please answer how to convert mass into weight?

Answer:

Approximately [tex]35.2\; \rm m[/tex].

Explanation:

Given:

Initial velocity: [tex]u = 13\; \rm m \cdot s^{-1}[/tex].

Acceleration: [tex]a = -2.40\; \rm m \cdot s^{-2}[/tex] (negative because the car is slowing down.)

Implied:

Final velocity: [tex]v = 0\; \rm m \cdot s^{-1}[/tex] (because the car would come to a stop.)

Required:

Displacement, [tex]x[/tex].

Not required:

Time taken, [tex]t[/tex].

Because the time taken for this car to come to a full stop is not required, apply the SUVAT equation that does not involve time:

[tex]\begin{aligned} x &= \frac{v^2 - u^2}{2\, a} \\ &= \frac{{\left(0\; \rm m \cdot s^{-1}\right)}^2 - {\left(13\; \rm m \cdot s^{-1}\right)}^2}{2\times \left(-2.40\; \rm m\cdot s^{-2}\right)} \approx 35.2\; \rm m \end{aligned}[/tex].

In other words, this car would travel approximately [tex]35.2\; \rm m[/tex] before coming to a stop.

Answer:

[tex]P=2.57\times 10^{-7}\ N/m^2[/tex]

Explanation:

Given that,

A radio wave transmits 38.5 W/m² of power per unit area.

A flat surface of area A is perpendicular to the direction of propagation of the wave.

We need to find the radiation pressure on it. It is given by the formula as follows :

[tex]P=\dfrac{2I}{c}[/tex]

Where

c is speed of light

Putting all the values, we get :

[tex]P=\dfrac{2\times 38.5}{3\times 10^8}\\\\=2.57\times 10^{-7}\ N/m^2[/tex]

So, the radiation pressure is [tex]2.57\times 10^{-7}\ N/m^2[/tex].

Answer:

Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.

Explanation:

Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

Answer:

Yes, geothermal energy is a renewable energy resource because the water can be heated by pumping it through the rocks.

Answer:

2

Explanation:

there are 2 significant figures in there

Answer:

The magnitude of the bugs displacement is 3.87 m

Explanation:

An illustrative diagram for the scenario is given in the attachment below.

In the diagram, the bug's displacement is given by x. The diagram shows a right angle triangle with x as the hypotenuse. We can determine x from the Pythagorean theorem which states that " the square of the hypotenuse equals sum of squares of the other two sides". That is

x² = 2.25² + 3.15²

x² = 5.0625 + 9.9225

x² = 14.985

x = √14.985

x = 3.87 m

Hence, the magnitude of the bugs displacement is 3.87 m.

Answer:82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles  (n)  = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

    = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

    = 140 K

B) Calculate the change in its internal energy

ΔU = Q  this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

                        = ( 3 * 8.3144 * 140 ) / 1.20

                        = 2910 Pa

A) The change in the temperature of the gas is; ΔT = 139.5 K

B) The change in internal energy of the gas is; ΔU = 5.22 × 10³ J

C) The change in pressure of the gas is; ΔP = 2899.5 Pa

We are given;

Volume of Monatomic ideal gas; V = 1.2 m³

Amount of heat added; Q = 5.22 × 10³ J

number of moles; n = 3

A) To calculate the change in temperature of the monatomic idea gas, we will use formula;

Q = ³/₂nRΔT

Where R is a constant = 8.314 J/mol.K

ΔT is the change in temperature

Making ΔT the subject of the formula;

ΔT = ²/₃(Q/(nR))

ΔT = ²/₃(5.22 × 10³)/(3 × 8.314)

ΔT = 139.5 K

B) Due to the fact that no work was done, then from first law of thermodynamics, we can say that;  

ΔU = Q

Thus;

change in internal energy; ΔU = 5.22 × 10³ J

C) The change in pressure will be calculated from the formula;

ΔP = (n*R*ΔT)/V

ΔP = (3 * 8.314 * 139.5)/1.2

ΔP = 2899.5 Pa

Read more at; https://brainly.com/question/14122248

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

[tex]mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s[/tex] ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.

Answer:

wavelength =  0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 *  0.220 s/2

x= 162.8 m

Answer:

option 3

Explanation:

can i get brainliest

Answer:

The maximum speed of the car is 35 m/s

The total distance traveled by the car is 658.33 m

Explanation:

Given;

initial velocity of the car, u = 15 m/s

acceleration of the car, a = 2 m/s²

time of car motion, t = 10 s

(i)

Initial distance traveled by the car is given by;

d₁ = ut + ¹/₂at²

d₁ = (15 x 10) + ¹/₂(2)(10)²

d₁ = 150 + 100

d₁ = 250 m

The maximum speed of the car during this is given by;

v² = u² + 2ad₁

v² = (15)² + (2 x 2 x 250)

v² = 1225

v = √1225

v = 35 m/s

(ii)

The final distance cover by the car during the deceleration of 1.5 m/s².

Note: the final or maximum speed of the car becomes the initial velocity during deceleration.

v² = u² + 2ad₂

where;

v is the final speed of the car when it stops = 0

0 = u² + 2ad₂

0 = (35²) + (2 x - 1.5 x d₂)

0 = 1225 - 3d₂

3d₂ = 1225

d₂ = 1225 / 3

d₂ = 408.33 m

The total distance traveled by the car is given by;

d = d₁ + d₂

d = 250 m + 408.33 m

d = 658.33 m

Answer:

36 m/s

Explanation:

t = 3.6s

u = 0m/s

a = +g = 10m/s²

v = ?

using,

v = u + at

v = 0 + 10(3.6)

v = 36 m/s

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  [tex]X = 38.3 \ m[/tex]

Explanation:

From the question we are told that

The acceleration along the x axis is  [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  [tex]v_{0}x = 7.10 m/s[/tex]

Generally from the equation for acceleration along x axis we have that

     [tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]

=>   [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]

=>   [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

At  t =0  s   and  [tex]v_{0}x = 7.10 m/s[/tex]

=>   [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]

=>   [tex]K_1 = 7.10[/tex]      

So  

      [tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

=>  [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]

=>  [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]

At  t =0  s   and   x = -14.0 m

  [tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]

=>   [tex]K_2 = -14[/tex]

So

     [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]

At  t = 10.0 s

      [tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]

=>   [tex]X = 38.3 \ m[/tex]

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]

Answer:

b. matter

Explanation:

Waves disturb matter but do not transmit it.

Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.

Answer:

See explanation

Explanation:

Given that the density of the unknown substance is 0.79 g/cm3 and is soluble in water, the possible substances it could be are;

i) t-butanol

ii) ethanol

iii) 2-propanol

iv) acetone

However, the actual identity of the unknown substance can be obtained by carrying out a boiling point test. The four substances listed above have different boiling points. Hence the boiling point of the unknown substance ultimately discloses its identity.

Answer:

Explanation:

m1v1=m2v2

m1=70 kg

m2=10 g=0.01 kg

v2=500 m/s

m1v1=m2v2

v1=m2v2/m1

v1=0.01*500/70

v1=0.07

Answer:

A plane mirror is placed to the right of an object. The image formed by the mirror will be a virtual image that appears to be on the left of the mirror.

Explanation:

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

Learn more : https://brainly.com/question/21444366

Answer:

Acceleration = 18g

Explanation:

Given the following data;

Initial velocity, u = 26m/s

Final velocity, v = 0

Time = 0.15 secs

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation, we have;

[tex]a = \frac{0 - 26}{0.15}[/tex]

[tex]a = \frac{26}{0.15}[/tex]

Acceleration = 173.33m/s2

To express it in magnitude of g;

Acceleration = 173.33/9.8

Acceleration = 17.7 ≈ 18g

Acceleration = 18g

Answer:

0.25 m.

Explanation:

mass of the block = 7.40 kg, height = 0.83 m, force constant of the spring = 9.50 x [tex]10^{2}[/tex] N/m.

The maximum compression on the spring can be determined by;

Potential energy stored in the spring = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]

But, potential energy = mgh

So that,

mgh = [tex]\frac{1}{2}[/tex] K[tex]x^{2}[/tex]

7.4 x 9.8 x 0.83 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]

60.1916 = 9.50 x [tex]10^{2}[/tex] x [tex]x^{2}[/tex]

[tex]x^{2}[/tex]= [tex]\frac{60.1916}{9.50*10^{2} }[/tex]

  = 0.06336

x = 0.2517

x = 0.25 m

The maximum compression of the spring is 0.25 m.

Answer:

sorry I wish I could it help you

Answer:

The final velocity of the car is 16.893 m/s

The final velocity of the truck is 4.393 m/s

Explanation:

Given;

mass of the car, m₁ = 715 kg

mass of the truck, m₂ = 1490 kg

initial velocity of the car, u₁ = 0

initial velocity of the truck, u₂ = 12.5 m/s

let the final velocity of the car, = v₁

let the final velocity of the truck, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(715 x 0) + (1490 x 12.5) = 715v₁ + 1490v₂

18625 = 715v₁ + 1490v₂ -----equation (1)

Apply one-directional velocity formula;

u₁ + v₁ = u₂ + v₂

0 + v₁ = 12.5 + v₂

v₁ = 12.5 + v₂

Substitute v₁ into equation (1)

18625 = 715(12.5 + v₂) + 1490v₂

18625 =8937.5 + 715v₂ + 1490v₂

18625 - 8937.5 = 715v₂ + 1490v₂

9687.5 = 2205v₂

v₂ = 9687.5 / 2205

v₂ = 4.393 m/s

solve for v₁

v₁ = 12.5 + v₂

v₁ =  12.5 + 4.393

v₁ = 16.893 m/s

Answer:

metals like iron or nickel

Explanation:

Answer:

option c is correct

Explanation:

we know that

2as=vf^2-vi^2

vf=24 m/s

vi= 0 m/s

a=g= 9.8 m/s^2

s=vf^2-vi^2/2a

s=(24)²-(0)²/2*9.8

s=576/19.6

s=29.4 m

therefore option c is correct

Answer:

Option 1

Explanation:

I always see animals do that