A box with a mass of 2 kg is pushed by a 10 N force. The acceleration
is
_m/s^2?

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.

Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.

Elements: Nitrogen; Oxygen; Phosphorus; Selenium...

Answer:

4.79m/s

Explanation:

According to law of conservation of momentum;

The sum of momentum of the bodies before collision is equal to the momentum after collision.

m1u1 + m2u2 = (m1+m2)v

Given;

m1 = 0.3kg

u1 = 2.30m/s

m2 = 0.0225kg

u2 = 38m/s

Required

speed of the arrow after passing through the target v

Substituting the given data into the formula

0.3(2.3) + 0.0225(38) = (0.3 + 0.0225)v

0.69 + 0.855 = 0.3225v

1.545 = 0.3225v

v = 1.545/0.3225

v = 4.79m/s

Hence the speed of the arrow after passing through the target is 4.79m/s

Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,

Fs = m a = W a / g

(a = centripetal acceleration, m = mass, g = acceleration due to gravity)

We have

a = v ² / R

(v = tangential speed, R = radius of the curve)

so that

Fs = W v ² / (g R)

Solving for v gives

v = √(Fs g R / W)

Perpendicular to the road, the car is in equilibrium, so Newton's second law gives

N - W = 0

(N = normal force, W = weight)

so that

N = W

We're given a coefficient of static friction µ = 0.4, so

Fs = µ N = 0.4 W

Substitute this into the equation for v. The factors of W cancel, so we get

v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s

Answer:

  y =  - 0.050 sin (131.59t )

Explanation:

In this exercise we are told to approximate the movement of a piston to the simple harmonic movement

          y = A cos (wt + Ф)

in this case they indicate that the stroke (C) of the piston is twice the amplitude

          C = 2A

          A = C / 2

angular velocity is related to frequency

          w = 2π f

let's substitute

         y = [tex]\frac{C}{2}[/tex] cos (2π f t +Ф)

To find the phase (fi) we will use the initial conditions, the piston starts at the midpoint of the stroke, if we create a reference system where the origin is at this point

         y = 0 for  t = 0

we substitute in the equation

        0 = \frac{C}{2} cos (0 + Ф)

The we sew zero values ​​for the angles of Ф = π/2 rad

we substitute in the initial equation

      y = \frac{C}{2} cos (2π f t + π/2)

let's use the double angle relationship

     cos ( a +90) = cos a cos 90 - sin a sin 90

     cos (a+90) = - sin a

       y = -\frac{C}{2} sin (2πf t )

let's reduce the frequency to SI units

        f = 200 rpm (2π rad / 1rev) (1 min / 60s) = 20.94 rad / s

we substitute the given values

       y = - [tex]\frac{0.100}{2}[/tex]  sin (2π 20.94 t )

       y =  - 0.050 sin (131.59t )

Answer:

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

Explanation:

Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.

In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.

Therefore, to test the validity of the claim, the student should use the following measuring tools;

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.

b. A meterstick to measure the distance of the track that the car travels on.

Hence, with these two devices the student can effectively measure or determine the validity of the claim.

Answer:

40m/s

Explanation:

a=g

u=0

s=81

v²=u²+2as

v²=2(9.81)(81)

v=√1587.6=39.8446985181≈40m/s

The velocity of the roller coaster car each time it reaches the bottom is 40 ms⁻¹. The correct option is (A).

The rate at which the position of an object changes with respect to time is described by the physical quantity known as velocity. It has both magnitude and direction because it is a vector quantity.

Given:

Initial velocity, u = 0 m/s

Acceleration, a = -9.8 ms⁻²

Distance, d = 81 m

From the third equation of motion:

v² = u² - 2as

v² = 0 - 2×(-9.8)×81

v = 40 ms⁻¹

Hence, the velocity of the roller coaster car is 40 ms⁻¹. The correct option is (A).

To learn more about Velocity, here:

https://brainly.com/question/17127206

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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E

and, we know that, F = ma

So,

[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E

a = [tex]\frac{q_{p}.E }{m_{p} }[/tex]      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2[tex]at^{2}[/tex]

Here, u = 0.

S = 1/2[tex]at^{2}[/tex]

Put equation 1 into the above equation:

S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex]  )[tex]t^{2}[/tex]      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex]  )[tex]t^{2}[/tex]    Equation 3

We know that the charge of electron is equal to the charge of proton so,

[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q

By dividing the equation 2 by equation 3, we get:

[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]

Solve the above equation for S,

S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S

So,

S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]

Plugging in the values,

As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]

S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]

As we know the mass of chlorine is 35.5 and of sodium is 23

S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]

S = 2.56 cm

Answer:

the local atmospheric pressure is  93.63 kPa

the mass of the weights is 156.9 kg

Explanation:

Given that;

Initial pressure of gas = 100 kPa

mass of piston = 10 kg and diameter = 14 cm = 0.14 m

g = 9.81 m/s²

Now,

P_gas = P_atm + P_piston

100 = P_atm + P_piston --------- let this equation 1

P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²

P_piston = 98.1 / (π/4×( 0.14 )²)

P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa

now, from equation 1

100 = P_atm + P_piston

we substitute

100 = P_atm + 6.37

P_atm = 100 - 6.37

P_atm = 93.63 kPa

Therefore, the local atmospheric pressure is  93.63 kPa

Now for pressure of the gas in the cylinder ⇒ 2×initial pressure

Pgas_2 = 2 × 100 = 200 kPa

Pgas_2 = P_atm + P_piston + P_weight

Pgas_2 =  P_gas  + P_weight

we substitute

200 kPa =  100 kPa  + P_weight

P_weight =  200 kPa -  100 kPa

P_weight = 100 kPa =  100,000 Pa

Also;

P_weight = M×g / A

100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)

100,000 × 0.01539 = M × 9.81

1539 = M × 9.81

M = 1539 / 9.81

M = 156.9 kg

Therefore, the mass of the weights is 156.9 kg

Answer:

Explanation:

Fluid A :

Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³

volume strain = Δ V / V  = 196 x 10⁻⁶ / 3000 x 10⁻⁶

= .06533

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .

It is Acetone .

Fluid B :

Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³

volume strain = Δ V / V  = 138 x 10⁻⁶ / 3000 x 10⁻⁶

= .046

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43  x 10⁷ Pa = 1.3  GPa .

It is Gasoline  .

Fluid C :

Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³

volume strain = Δ V / V  = 84 x 10⁻⁶ / 3000 x 10⁻⁶

= .028

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14  GPa .

It is Water   .

Answer:

Explanation:

From the question we are told that

Work done [tex]W=2820[/tex]

a)Generally the heat flow for an adiabatic process is 0 (zero)

[tex]Q= U + W =>0[/tex]

[tex]Q=0[/tex]

b)Generally Change in internal energy of gas is mathematically given by

Since [tex]W=-2820J[/tex]

Therefore

[tex]U=2820[/tex]

Giving

[tex]Q= 2820 -2820[/tex]

[tex]Q=O[/tex]

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

Answer:

the way a mineral breaks apart

Explanation:

The way a mineral breaks apart involves fracture and cleavage.

These characteristics are very important in mineral identification especially during physical observations. Minerals have different cleavage properties and fracture planes.

Answer:

A:the way a mineral breaks apart

Explanation:

Answer:

a) vx = -12.7 m/s vy = -56.7m/s

b) v= 58.1 m/s

c) θ = 77.4º S of W

Explanation:

a)

       [tex]v_{ps} = -44m/s (1)[/tex]

       [tex]v_{we} = - 18m/s * cos (45) = -12.7 m (2)[/tex]

       [tex]v_{ws} = - 18m/s * cos (45) = -12.7 m (3)[/tex]

        [tex]v_{e} = v_{x} = -12.7 m/s (4)[/tex]

b)

        [tex]v = \sqrt{(-12.7m/s)^{2} + (-56.7m/s)^{2}} = 58.1 m/s (6)[/tex]

c)

       [tex]tg_{\theta} =\frac{v_{y} }{v_{x} } = \frac{(-56.7m/s)}{(-12.7m/s} = 4.46 (7)[/tex]

        tg⁻¹ θ = tg⁻¹ (4.46) = 77.4º S of W. (8)

This question is incomplete, the complete question is;

According to information found in an old hydraulics book, the energy loss per unit weight of fluid flowing through a nozzle connected to a hose can be estimated by the formula; h= (0.04 to 0.09)(D/d)⁴V²/2g

where h is the energy loss per unit weight, D the hose diameter, d the nozzle tip diameter, V the fluid velocity in the hose, and g the acceleration of gravity.

Do you think this equation is valid in any system of units

Answer:

YES, the equation is a general equation that is valid in any system of units

Explanation:

Given the data in the question;

h = (0.04 to 0.09)(D/d)⁴ × [tex]\frac{V^{2} }{2g}[/tex]

so

[ N.m/N ] = (0.04 to 0.09) ( m/m)² × (m²/s²)1/2 × (s²/m)

[ N.L/N ] = (0.04 to 0.09) ( L⁴/L⁴) × (L²/T²)1/2 × (T²/L)

∴ [ L ] = (0.04 to 0.09) [L]

So as each term in the equation must have the same dimensions, the constant term (0.04 to 0.09) must be without dimension.

Therefore, YES, the equation is a general equation that is valid in any system of units

Answer:

e.

Explanation:

       [tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]

        [tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]

Answer:

5766.7 K

Explanation:

We are given that

Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]

Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]

Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]

We have to find the temperature at the surface of the Sun.

We know that

Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]

Where [tex]K_{sc}=1350 W/m^2[/tex]

[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]

Using the formula

[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]

T=5766.7 K

Hence, the temperature at the surface of the sun=5766.7 K

Answer: take less time to go the same distance

Explanation:

Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.

Answer: he can only make it over 5 buses

Explanation:

Given the data in the question;

we know that range is expressed as;

R = (V₀²sin2∅₀)/g

V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),

so we substitute

R = ((25.4)²sin2(64.8))/9.8

R = 50.7 m

now, them number of buses will be;

n = R / bus length

given that bus length is 10.0 m

we substitute

n = 50.7 m / 10.0

n = 5.07 ≈ 5

Therefore, he can only make it over 5 buses

Answer:

The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.

Explanation:

Let suppose that car accelerates uniformly in a rectilinear motion. Given that initial and final speeds and travelled distances are known, then the acceleration needed by the vehicle ([tex]a[/tex]), measured in feet per square second, is determined by the following kinematic formula:

[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot \Delta x }[/tex] (1)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds, measured in feet per second.

[tex]\Delta x[/tex] - Travelled distance, measured in feet.

If we know that [tex]v_{o} = 34.907\,\frac{ft}{s}[/tex], [tex]v_{f} = 67.027\,\frac{ft}{s}[/tex] and [tex]\Delta x = 852\,ft[/tex], then acceleration needed to accomplish the task is:

[tex]a = 1.921\,\frac{ft}{s^{2}}[/tex]

The constant rate of acceleration required in order to accomplish this is 1.921 feet per square second.

Answer:

Zero

Explanation:

Appearing to throw the ball but still holding on to it means the recoil velocity will be zero because the recoil velocity is defined as the backward velocity as a result of throwing an object or shooting a bullet. In this case the object was not thrown and thus there is no recoil velocity.

Answer:

static friction

Explanation:

static friction is the friction force that acts on objects at rest

Answer:

1. The force of the car engine.

Explanation:

We shall see the effect and role played by each force, one by one, as follows:

1. The force of car engine:

The engine produces a force through combustion that is converted to mechanical work through the shaft. This work is then transmitted to the wheels of the car that cause the motion in the car and increase its kinetic energy.

2. Air Resistance:

It is the opposing force of air that tries to reduce the motion of the car and as a result, reduce its kinetic energy.

3. Frictional Force between road and tire:

It is also the opposing force of air that tries to reduce the motion of the car and as a result, reduce its kinetic energy.

4. Gravity:

Gravity pulls everything towards the center of Earth so it does not have much significant role in horizontal motion like this.

Hence, the force of the car engine did the work that increased the car's kinetic energy.

Answer: you divide total distance by time. To get the time, divide total distance by speed. To get distance,  multiply speed times the amount of time.

Explanation:

I hope this helps

Answer:

Explanation:

Magnetic field due to current at a distance of 4.4 cm

B = 10⁻⁷ x 2 x 5.2 / 4.4 x 10⁻²           [ B = 10⁻⁷ x 2i / r = ]

= 2.36 x 10⁻⁵ T.

Force on moving electron = Bqv , B is magnetic field , q is charge and v is velocity of charge .

Force = 2.36 x 10⁻⁵  x 1.6 x 10⁻¹⁹ x 6.2 x 10⁴

= 23.41 x 10⁻²⁰ N .

This force will be perpendicular to the direction of current .

Answer:

The roof and walls of the auditorium or cinema hall are generally covered with sound absorbent materials like draperies or compressed fibreboard to reduce reverberation. These materials reduce the formation of echoes by absorbing sound waves.

Explanation:

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Answer:

a) Average speed is 24.04 m/s

b) the rms speed is 25.84 m/s

c) the most probable speed is 16 m/s

Explanation:

Given the data in the question;

a) Determine the average speed.

To determine the average speed, we simply divide total some of speed by number of particles;

Average speed =  [(2×11 m/s)+(7×16 m/s)+(4×19 m/s)+(3×26 m/s)+(6×31 m/s)+(1×37 m/s)+(2×45 m/s)] / 25    

= 601 / 25

= 24.04 m/s

Therefore, Average speed is 24.04 m/s

b) Determine the rms speed

we know that  (rms speed)² = sum of square speed / total number of particles

so

(rms speed)² =  [(2×11²)+(7×16²)+(4×19²)+(3×26²)+(6×31²)+(1×37²)+(2×45²)] / 25

(rms speed)² =  16691 / 25

(rms speed)² =  667.64

(rms speed) = √ 667.64

(rms speed) = 25.84 m/s

Therefore, the rms speed is 25.84 m/s

c) Determine the most probable speed.

Most particles (7) have velocity 16 m/s

i.e 7 is the maximum number of particle for a particular speed ,

Therefore, the most probable speed is 16 m/s

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm=[tex]1900\times 10^{-9}m[/tex]

Where [tex]1nm=10^{-9} m[/tex]

[tex]\lambda_2=400nm=400\times10^{-9}m[/tex]

[tex]\lambda_1=700nm=700\times 10^{-9}m[/tex]

We have to find  the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

[tex]\theta=sin^{-1}(\frac{m\lambda}{d})[/tex]

Using the formula

m=1

[tex]\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})[/tex]

[tex]\theta=21.62^{\circ}[/tex]

Now, m=2

[tex]\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})[/tex]

[tex]\theta_2=24.90^{\circ}[/tex]

[tex]\Delta \theta=\theta_2-\theta_1[/tex]

[tex]\Delta \theta=24.90-21.62[/tex]

[tex]\Delta \theta=3.28^{\circ}[/tex]

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

Answer:

25.3J

Explanation:

Given parameters:

Mass of aluminum  = 3.05g

Initial temperature  = 10.8 °C

Final temperature  = 20 °C

Specific heat  = 0.9J/g °C

Unknown:

Amount of heat needed for the temperature to change  = ?

Solution:

To solve this problem, we use the expression:

       H  = m C Ф  

H is the amount of heat

m is the mass

C is the specific heat capacity

Ф is the change in temperature

     H  = 3.05 x 0.901 x (20 - 10.8) = 25.3J

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

            Work done  = Weight x height  = 700 x 828

       Work done  = 579600J