Write submultipels units of time​

Answer:

 ΔL = 3.82 10⁻⁴ m

Explanation:

This is a thermal expansion exercise

          ΔL = α L₀ ΔT

          ΔT = T_f - T₀

where ΔL is the change in length and ΔT is the change in temperature

Let's reduce the length to SI units

          L₀ = 90.5 mm (1m / 1000 mm) = 0.0905 m

let's calculate

          ΔL = 25.10⁻⁶ 0.0905 (154.6 - (14.4))

          ΔL = 3.8236 10⁻⁴ m

using the criterion of three significant figures

          ΔL = 3.82 10⁻⁴ m

By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:

W = ∆K

and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.

So we have

-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²

Solve for k to get k = 800 N/m.

Answer:

the pressure exerted by the object is 392,000 N/m²

Explanation:

Given;

mass of the object, m = 8 kg

area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²

acceleration due to gravity, g = 9.8 m/s²

The pressure exerted by the object is calculated as;

[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]

Therefore, the pressure exerted by the object is 392,000 N/m²

Explanation:

The area of a circle of radius r is given by

[tex]A = \pi r^2[/tex]

Taking the derivative of A with respect to time t, we get

[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]

We also know that

[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]

[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]

Answer "a" would be correct.

Answer:

d

Explanation:

There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D

Brainliest please~

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

Answer:

physical material from which something is made or which has discrete existence

Explanation:

Answer:

Explanation:

a)

The force on electron acts opposite to the velocity , and direction of force on electron is always opposite to direction of electric field .

Hence direction of electric field must be in the same  in which electrons travels.

Hence option iii is correct.

b )

deceleration a = force / mass

= qE / m

= 1.6 x 10⁻¹⁶ x 5.6 x 10⁵ / 9.1 x 10⁻³¹

= .98 x 10²⁰ m /s²

v² = u² - 2 a s

0 = (8.06 x 10⁶ )² - 2 x .98 x 10²⁰ s

s = 64.96 x 10¹² / 1.96 x 10²⁰

= 33.14 x 10⁻⁸ m

c ) time required

= 8.06 x 10⁶ / .98 x 10²⁰

= 8.22 x 10⁻¹² s .

d ) Its speed will be same as that in the beginning ie 8.06 x 10⁶ m/s .

Answer:

(a) Option (i)

(b) 6.6 x 10^-4 m  

(c) 8.2 x 10^-11 s

Explanation:

initial velocity, u = 8 .06 x 10^6 m/s

Electric field, E = 5.6 x 10^5 N/C

(a) The direction of field is opposite.

Option (i).

(b) Let the distance is s.  

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times \frac{qE}{m}\times s\\\\8.06\times 10^6\times 8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} s\\\\s = 6.6\times 10^{-4} m[/tex]

(c) Let the time is t.

Use first equation of motion.

[tex]v = u + a t \\\\0 = u - \times \frac{qE}{m}\times t\\\\8.06\times 10^6 = \frac {1.6\times 10^{-19}\times 5.6\times 10^5}{9.1\times 10^{-31}} t\\\\t = 8.2\times 10^{-11} s[/tex]

Answer:

The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.

Weight can be calculated using the equation:

weight = mass × gravitational field strength

This is when:

weight (W) is measured in newtons (N)

mass (m) is measured in kilograms (kg)

gravitational field strength (g) is measured in newtons per kilogram (N/kg)

Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.

Hope this helps!!!!

Answer:

electric current

Explanation:

The answer is electric current

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

Answer:

24 cm/s

Explanation:

Applying

Pythagoras theorem,

a² = b²+c²............. Equation 1

Where a = resultant, b = vertical component, c = horizontal component

From the question,

Given: a = 26 cm/s, c = 10 cm/s

Substitute these values into equation 1

26² = b²+10²

676 = b²+100

b² = 676-100

b² = 576

b = √576

b = 24 cm/s

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) g

• the upward normal force of the surface, magnitude n₁

• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction

• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction

• the applied force, mag. F, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) g, pointing downward

• normal force, mag. n₂, pointing upward

• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction

• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

n₂ - (3.0 kg) g = 0   ==>   n₂ = (3.0 kg) g   ==>   f₂ = 0.30 (3.0 kg) g

• net horizontal force:

c₂ - f₂ = 0   ==>   c₂ = 0.30 (3.0 kg) g ≈ 8.8 N

Answer:

Wheel and axle

Explanation:

Which simple machine is shown in the diagram?

a wheel and axle

From the given diagram, the machine shown is actually a wheel and axle

Description of wheel and axle

The wheel and axle is a machine consisting of a wheel attached to a smaller axle so that these two parts rotate together in which a force is transferred from one to the other.

Answer:

Wheel and axle

Explanation:

Answer:

C.  Rotational motion

Explanation:

The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).

Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.

"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."

Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.

Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.

The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.

Learn more about motion here:

https://brainly.com/question/22810476

#SPJ2

Answer:

Explanation:

For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.

Answer:

KInetic energy reduces.

Explanation:

Application of breaks reduces velocity. Reduction of velocity constitutes velocity reduction.

For more help contact +1 (304) 223-3136

Answer:

Explanation:

From the given information:

From the rotational axis, the distance of the force of gravity is:

d_g = 25+5.0 cm

d_g = 30.0 cm

d_g = 30.0 × 10⁻² m

However, the relative distance of FB  cos 75.9° from the axis is computed as:

d_B = 5.0 cm

d_B = 5.0 × 10⁻² m

The net torque rotational equilibrium = zero (0)

i.e.

[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]

= 772.4 N

Thus, the force exerted = 1772.4 N

Answer:

If the mass of one of the objects is tripled, then the force of gravity between them is tripled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

[tex]f = \mu R = \mu W[/tex]

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]

f = 347.08 N

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

[tex]K = 0.5 mv^2[/tex]

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

(a)  The kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)  The final speed of both the puck A and B are same.

Let the mass of puck A is m and the mass of puck B is 2 m.

Initial speed for both the pucks is same as u and the distance is same for both is s.

Let the tension is T for same.

Then, the kinetic energy is given as,

[tex]KE = \dfrac{1}{2}mv^{2}[/tex]

(a)

As the speed is same, so the kinetic energy depends on the mass.

Then,

[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]

So, kinetic energy of A : Kinetic energy of B = 1 : 2.

Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)

The final speed for the puck is given as,

v = s/t

here, s is the distance covered.

Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.

Thus, we can conclude that the final speed of both the puck A and B are same.

learn more about the kinetic energy here:

https://brainly.com/question/12669551

Explanation:

someone to check if the answer is correct

Answer: [tex]2.4705\ kg.m^2[/tex]

Explanation:

Given

length of the rod is L=0.9 m

Mass of the rod m=3.8 kg

Point masses has mass of m=2.3 kg

Moment of Inertia of the rod about the center is

[tex]\Rightarrow I_o=\dfrac{1}{12}ML^2[/tex]

Moment of inertia of combined system is the sum of rod and two point masses.

[tex]\Rightarrow I=I_o+2mr^2[/tex]

[tex]\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2[/tex]

Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).

At point A, the block has total energy

E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²

E (A) = 686 J + 1/2 (10.0 kg) v₀²

At point B, the block's potential energy is converted into kinetic energy, so that its total energy is

E (B) = 1/2 (10.0 kg) v₁²

The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,

E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J

Throughout this whole process, energy is conserved, so

E (A) = E (B) = E (C) = E (D)

(a) Solve for v₀ :

686 J + 1/2 (10.0 kg) v₀² = 2548 J

==>   v₀ ≈ 19.3 m/s

(b) Solve for v₁ :

1/2 (10.0 kg) v₁² = 2548 J

==>   v₁ ≈ 22.6 m/s

Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:

• net horizontal force:

∑ F = -f = ma

• net vertical force:

∑ F = n - mg = 0

where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :

n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N

f = µn = 0.500 (98.0 N) = 49.0 N

==>   - (49.0 N) = (10.0 kg) a

==>   a = - 4.90 m/s²

The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that

v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)

==>   v₂² = 490 m²/s²

and thus the block has total/kinetic energy

E (C) = 1/2 (10.0 kg) v₂² = 2450 J

(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so

2450 J = (10.0 kg) (9.80 m/s²) h

==>   h = 25.0 m

(d) At half the maximum height, the block has speed v₃ such that

2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²

==>   v₃ ≈ 15.7 m/s

The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by

v = v₁ + at = 22.6 m/s - (4.90 m/s²) t

The block comes to a rest when v = 0 :

0 = 22.6 m/s - (4.90 m/s²) t

==>   t ≈ 4.61 s

It covers a distance x after time t of

x = v₁t + 1/2 at ²

so when it comes to a complete stop, it will have moved a distance of

x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m

(e) The block crosses the rough region

(52.0 m) / (2.00 m) = 26 times

Answer:

[tex]M.E=41J[/tex]

Explanation:

From the question we are told that:

Amplitude [tex]a=4.20cm[/tex]

Spring Constant [tex]K=262N/m[/tex]

Mass [tex]m=560g[/tex]

Generally the equation for mechanical energy is mathematically given by

[tex]M.E=\frac{1}{2}km^2[/tex]

[tex]M.E=0.5*262*0.56^2[/tex]

[tex]M.E=41J[/tex]

Answer:

Plastic

Explanation:

Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.

This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.

Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).

Answer:

Explanation:

a)

v² = u² + 2 a s

v = 9 m/s

u = 4 m/s

s = 50 m

9² = 4² + 2 x a x 50

a = 0.65 m /s²

Acceleration is 0.65 m /s²

b )  

time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²

(v - u ) / t = a

(v - u ) / a = t

(9 - 4 ) / .65 = t

t = 7.7

time when passing the first sign will be 7.7 s earlier .

Reading of time indicator = 45 - 7.7

= 37.3 seconds.

Answer:

(a) 0.45 m/s^2

(b) 33.9 s

Explanation:

initial velocity, u = 4 m/s

final velocity, v = 9 m/s

distance, s = 50 m

(a) Let the acceleration is a.

Use third equation of motion

[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]

(b) Let the time is t.

Use first equation of motion

v = u + at

9 = 4 + 0.45 x t

t = 11.1 s

So, the initial time, t' = 45 - 11.1 = 33.9 s  

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T