Ali kicks a soccer ball upward at 45° angle
with an initial speed of 32 m/s. What
expression should Ahmed use to calculate
the magnitude of the ball's initial velocity
the horizontal direction?

Select one:
a. 32 sin(45°) m/s

b. 32 cos(45°) m/s

c. 32 tan(45°) m/s

d. 45 m/s

Answer: 25 m/s

Explanation:

1. rearrange equation p = mv to find v

2. v = p/m - then convert 800 g into kg (= 0.8 kg)

3. plug in p and m - v = 20 / 0.8 = 25 m/s

Answer:

V= 25m/s

Explanation:

Mass=800g

Momentum=20kgm/s

V= ?

Convert 800g to kg

800/1000 = 0.8kg

Momentum=mass× Velocity

Momentum/Mass = Velocity

20/0.8 =25m/s

Answer:

A) I_laser = 70.74 W/m²

B) I_bulb = 1.989 W/m²

C) it is not advisable to look at the laser beam directly.

Explanation:

We are given;

Power; P = 0.50 mW = 0.5 × 10^(-3) W

Diameter; d = 3 mm = 0.003 m

Radius; r = d/2 = 0.003/2 = 0.0015 m

A) Area of beam; A = πr²

A = 0.0015²π

Now, formula for average intensity is;

I = P/A

I = (0.5 × 10^(-3))/0.0015²π

I = 70.74 W/m²

B) We are told to find the intensity of a lightbulb producing 100-W.

Thus, P = 100 W

A light bulb is spherical in shape. Thus;

Area; A = 4πr²

We are told it's 2 m away.

Thus; r = 2 m

A = 4π(2)²

A = 16π

Thus, I = P/A = 100/16π

I = 1.989 W/m²

C) The intensity of the laser beam is far greater than that of the light bulb. Thus, it is not advisable to look at the laser beam directly.

Answer:

most evaporation and precipitation in the water cycle occus over the ocean

The option Most evaporation and precipitation in the water cycle occur over the ocean. is the correct answer

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

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

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

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Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

Answer:

hope fully it help s

Answer:

4.097 Joules

Explanation:

Applying,

P.E = mgh............... Equation 1

Where P.E = Potential energy, m = lemming mass, h = height, g = acceleration due to gravity

From the question,

Given: m = 0.0780 kg, h = 5.36 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

P.E = 0.0780(9.8)(5.36)

P.E = 4.097 Joules.

Hence the potential energy is 4.097 Joules

Answer:

4.097 J

Explanation:

Answer:

option b is correct..................

Answer:

Explanation:

F = ma, so filling in:

15000 = 30000a and

a = .50 m/s/s

The acceleration of a train is 0.50 [tex]m/s^{2}[/tex].

Acceleration is the rate of change of the velocity of an object with respect to time.

F = ma

a = F/m

a = 15000/30000

a = 0.50 [tex]m/s^{2}[/tex]

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Answer:

The Reference standard with which we carry out the measurement of any physical quantity of the same kind is know as unit .

I HOPE I HELP YOU.

Answer:

The standard quantity which is used for the comparison with an unknown quantity is called unit.

Unit is a known or standard quantity in terms of which other physical quantities are measured.

for examples;kilogram,metre,second,etc.

Explanation:

hope it helps you

Explanation:

B. Doubling the heat flow into the engine while halving the work done

hope this helps you

have a nice day

Answer:

Maximum height, H = 35 meters

Explanation:

Given the following data;

Time = 7 seconds

Acceleration due to gravity, g = 10 m/s²

To find the maximum height;

Mathematically, the maximum height is given by the formula;

H = ½gt²

H = ½ * 10 * 7

H = 5 * 7

Maximum height, H = 35 meters

Answer:

F1=26N and F2=09N ..this is from the two simultaneously equations

Answer:

11760 J

Explanation:

cuz potential engery is PE = MHG

SO 80×15×9.8= 11760 J

Answer:

11769j

Explanation:

here,

mass(m)=80 kg

height(h)=15m

acceleration due to gravity(g)=9.8m/s^2

now,

potential energy = m×g×h

= 80×9.8×15

= 11760j

Answer:

I. Vs = 8.0 Volts.

II. Is = 4.5 Amperes.

Explanation:

Given the following data;

Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex]  ..... equation 1

Ip = 3.0 A

Vp = 12 V

To find the voltage and current on the secondary coil;

I. For the voltage in the secondary coil (Vs), we would use the following formula;

[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex]  ...... equation 2.

Substituting eqn 1 into eqn 2, we have;

[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]

[tex] \frac {12}{V_{S}} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] V_{S} * 1.5 = 12 [/tex]

[tex] V_{S} = \frac {12}{1.5} [/tex]

Vs = 8.0 V

II. For the current in the secondary coil (Is), we would use the following formula;

[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex]  .... equation 3

Substituting eqn 1 into eqn 3, we have;

[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]

[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] I_{S} = 1.5 * 3.0 [/tex]

Is = 4.5 A

The frequency of the other car is 669HZ

Let start off by writing out the parameters given in the question

frequency= 595 HZ

Speed of sound(v)= 343m/s

Speed of the car(vs)= 20m/s

Speed of the second car(vo) = 20m/s

The frequency of the other car can be calculated as follows

(v+vo/v-vs) f

= (343+20/343-20)595

= (363/323)595

= 1.1238(595)

= 669 HZ

Hence the frequency of the other car is 669 HZ

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Answer:

Obeying to weights & measurement regulations in both national and international metrology legislation, standards and test procedures is a requirement to participate in any market because it's aimed to safeguard the consumers and promote fair competition, which provides efficiency and saves unnecessary costs to U.S. businesses and stakeholders.

Explanation:

In the U.S., the National Institute of Standards and Technology (NIST) has an Office of Weights and Measures (OWM) that represents the country in the International Organization of Legal Metrology (OIML)

Answer:

Explanation:

As,

In Option,

A:

There are two same positive ions so they will move away from each other.

B:

There is one negative and positive ion so they will move towards each other.

C:

Again there is one negative and positive ion so they will move towards each other.

D:

Here, there is neutral ions so they will not move and its the correct option.

Explanation:

light travel slower in daimond

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = [tex]\frac{Q}{\Delta V}[/tex]

        C = ε₀ [tex]\frac{A}{d}[/tex]

we solve for the charge (Q)

        [tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = [tex]\epsilon_o \ \frac{A \ \Delta V_1 }{d_1}[/tex]

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          [tex]\frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}[/tex]

          ΔV₂ = [tex]\frac{d_2}{d_1 } \ \Delta V_1[/tex]

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

we calculate

         K = 1 10⁻⁹  12  [tex]\frac{0.005}{0.003}[/tex]  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.

Work = force x distance

Work = 5500 x 1.8

Work = 9900 N

therefore, work does 9900 N

Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).

Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.

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Answer:

Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. An industrial application of magnetic force is an electromagnetic crane that is used for lifting metal objects.

Answer:

Examples of magnetic force is a compass, a motor, the magnets that hold stuff on the refrigerator, train tracks, and new roller coasters. All moving charges give rise to a magnetic field and the charges that move through its regions, experience a force.

I Hope this will help you if not then sorry :)

Answer:

3 to 7 seconds.

Explanation:

Answer:

Torque; τ = 4.712 × 10^(-3) J

Magnetic moment; M = 0.0248 J/T

Explanation:

Torque is gotten from the formula;

τ = BIA

Where;

B is magnetic field

I is current

A is area

We are given;

B = 0.19T

I = 6.2A

Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m

Thus;

Area; A = 0.05m × 0.08m = 0.004 m²

Thus;

τ = 0.19 × 6.2 × 0.004

τ = 4.712 × 10^(-3) J

Formula for the magnetic moment is given by;

M = IA

M = 6.2 × 0.004

M = 0.0248 J/T

Answer:

La fuerza resultante sobre q₃ es  -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N

Explanation:

q₁ = -45 μC = -45 × 10⁻⁶ C

r₁₂ = 30 mm = 30 × 10⁻³ m

q₂ = 25 μC = 25 × 10⁻⁶ C

r₂₃ = 50 mm = 50 × 10⁻³ m

q₃ = 20 μC = 20 × 10⁻⁶ C

k = 9×10⁻⁹ N·m²/C²

Por lo tanto;

r₁₃ = √(50² + 30²) = 10·√(34)

F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴

F₁₂ = -1.125 × 10⁻¹⁴ N

F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j

F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵

Los componentes de F₁₃ son;

-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j

-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i

La fuerza resultante sobre la carga q₃, [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = [tex]\underset{F_{13}}{\rightarrow}[/tex] + [tex]\underset{F_{23}}{\rightarrow}[/tex]

∴ [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j

La fuerza resultante sobre q₃ es [tex]\left | \underset {F_3} \rightarrow \right |[/tex]  = -1.2245 × 10⁻¹⁵ i  + -0.24 × 10⁻¹⁵ j

La magnitud de la fuerza resultante sobre q₃,

[tex]\left | F_3 \right |[/tex] = √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵

La magnitud de la fuerza resultante sobre q₃, [tex]\left | F_3 \right |[/tex] ≈ 1.25 × 10⁻¹⁵ N.

Answer:

The total heat required to melt the ice is approximately 3.473 MJ

Explanation:

The given parameters for the layer of ice are;

The thickness of the layer of ice, t = 0.80 cm = 0.008 m

The area of the wind shield, A = 1.4 m²

The initial temperature of the ice, T₁ = -2.0 °C = 271.15 K

The density of the ice, ρ = 917 kg/m

The temperature at which ice melts, T₂ = 0 °C = 273.15 K

We have;

The mass of the ice, m = ρ × t × A

∴ m = 917 kg/m³ × 0.008 m × 1.4 m² = 10.2704 kg

The specific heat capacity of ice, c = 2,090 J/(kg·K)

∴ The equation for the heat capacity of the ice to melt, is given as follows;

ΔQ = m·c·ΔT

Where;

ΔT = T₂ - T₁

∴ ΔT = T₂ - T₁ = 273.15 K - 271.15 K = 2 K

ΔQ = 10.2704 kg × 2,090 J/(kg·K) × 2 K = 42.930272 kJ

The latent heat to melt the ice, Q = The latent heat of fusion of ice, L × Mass of ice, m

The latent heat of fusion of ice, L = 334 kJ/kg

∴ Q = 334 kJ/kg × 10.2704 kg = 3,430.3136 kJ

The total heat required to melt the ice, H = ΔQ + Q

∴ H = 42.930272 kJ + 3,430.3136 kJ = 3,473.243872 kJ ≈ 3.473 MJ.

Answer:

Amplitude.

Explanation:

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In Science, there are two (2) types of wave and these include;

I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.

II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.

An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.

Hence, an amplitude is a word that describes the maximum displacement a point moves from its rest position when a wave passes.

On a graph, the vertical axis (y-axis) is the amplitude of a waveform and this simply means that, it's measured vertically.

Mathematically, the amplitude of a wave is given by the formula;

x = Asin(ωt + ϕ)

Where;

x is displacement of the wave measured in meters.

A is the amplitude.

ω is the angular frequency measured in rad/s.

t is the time period measured in seconds.

ϕ is the phase angle.