A student throws a baseball upwards at an angle of 60 degrees to the horizontal. The initial
horizontal and vertical components are 12.5 m/s and 21.7 m/s, respectively.
60 degrees
Refer to the above information and diagram. What position will have the smallest magnitude
of vertical velocity?

There are 90 phones of volume, 10-7 W/m2 of sound intensity, and 0.0314 watts of source power.

A straightforward frequency analysis compares the values of the fields you provide and generates a report listing each value for those fields along with the frequency at which each value occurs.

The rate at which a sound power wave repeats itself, also known as frequency or pitch, is measured in cycles per second. Bullfrog calls and cricket chirps have lower frequencies than drum beats and whistles, respectively.

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If A box of mass 210 kg is pulled from rest with a string of tension 1300n inclined at 35° to the horizontal. if the box moved with a speed of 10m/s and the frictional force between the box and surface is 100 n, Then the distance covered by the box is 10.89 meters.

To calculate the distance covered by the box, we need to analyze the forces acting on it and apply the work-energy principle.

Given:

Mass of the box, m = 210 kg

Tension in the string, T = 1300 N

The angle of inclination, θ = 35°

Frictional force, f = 100 N

Initial speed, u = 0 m/s

Final speed, v = 10 m/s

First, let's resolve the tension force into components parallel and perpendicular to the incline. The parallel component of the tension force can be calculated as:

T_parallel = T * cos(θ)

Next, let's calculate the net force acting on the box along the incline. The net force is given by:

Net force = T_parallel - f

Now, using Newton's second law, we can calculate the acceleration (a) of the box:

Net force = m * a

From the given information, we have the final velocity (v), initial velocity (u), and acceleration (a). We can use the following kinematic equation to calculate the distance covered (s):

v^2 = u^2 + 2as

Rearranging the equation, we get:

s = (v^2 - u^2) / (2a)

Now, let's plug in the given values and calculate the distance covered:

T_parallel = 1300 N * cos(35°) ≈ 1067.35 N

Net force = 1067.35 N - 100 N = 967.35 N

a = (967.35 N) / (210 kg) ≈ 4.61 m/s^2

s = (10 m/s)^2 - (0 m/s)^2 / (2 * 4.61 m/s^2) ≈ 10.89 m

Therefore, the distance covered by the box is approximately 10.89 meters.

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Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= \(\frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}\)

\(= \frac{-1760}{563} + \frac{1760}{354}\)

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

Answer:

1.4km/h

Explanation:

distance/time = speed

69/45=1.4km/h

The ball that correctly relates the kinetic energy of a ball is;  Ball C

The kinetic energy of an object is the energy generated by an object in motion ( i.e. object whose speed is greater than zero )

and the formula for Kinetic energy = 1/2 mv^2 .

Ball B :

mass = 4kg , speed = 8 m/s . kinetic energy = 130 J

K.E = 1/2 ( 4 * 8^2 )

      = 128 J   ( 130 J ≠ 128 J ) ( does not correctly relate the kinetic energy

Ball C :

mass = 5 kg, speed = 8 m/s , kinetic energy = 160 J

K.E = \(1/2*\) \(( 5 * 8^2 )\)

      = 160 J  ( 160 J = 160 J )  ( correctly relates the kinetic energy )

Hence we can conclude that The ball that correctly relates the kinetic energy of a ball is;  Ball C.

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Answer:

b=c or b,c

Explanation:

The mass of Car B is -6000 kg.

To solve this problem, we can apply the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Therefore, we can write the equation for the conservation of momentum as:

(mass of Car A * velocity of Car A) + (mass of Car B * velocity of Car B) = (mass of Car A + mass of Car B) * velocity after collision

Let's substitute the given values into the equation:

(2000 kg * 10 m/s) + (mass of Car B * 0 m/s) = (2000 kg + mass of Car B) * (-5 m/s)

Simplifying the equation:

20000 kg*m/s = -5 m/s * (2000 kg + mass of Car B)

Dividing both sides by -5 m/s:

-4000 kg = 2000 kg + mass of Car B

Subtracting 2000 kg from both sides:

mass of Car B = -4000 kg - 2000 kg

mass of Car B = -6000 kg

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Answer:

The ratio of the model size is 1 : 2000

Explanation:

Given

Real Diameter =  0.012 um

Scale Diameter =  24 um

Required

Determine the scale ratio

The scale ratio is calculated as follows;

\(Scale = \frac{Real\ Measurement}{Scale\ Measurement}\)

Substitute values for real and scale measurements

\(Scale = \frac{0.012\ um}{24\ um}\)

Divide the numerator and the denominator by 0012um

\(Scale = \frac{1}{2000}\)

Represent as ratio

\(Scale = 1 : 2000\)

Hence, the ratio of the model size is 1 : 2000

Answer:

Explanation: The Answer Is D Two Thousand & The Number



Answer:

17.6 m/s², 26.2° above x axis

Explanation:

Apply Newton's second law.

Sum of forces in the x direction:

∑Fₓ = ma

40 N + 55.0 N cos 45° = (5.00 kg) aₓ

aₓ = 15.8 m/s²

Sum of forces in the y direction:

∑Fᵧ = ma

55.0 N sin 45° = (5.00 kg) aᵧ

aᵧ = 7.78 m/s²

Use Pythagorean theorem to find the magnitude of the resultant.

a² = aₓ² + aᵧ²

a² = (15.8 m/s²)² + (7.78 m/s²)²

a = 17.6 m/s²

Use trigonometry to find the angle.

tan θ = aᵧ / aₓ

tan θ = (7.78 m/s²) / (15.8 m/s²)

θ = 26.2°

Answer:

A ,C

Explanation:

Shows

A scale which reads 0 in the vacuum of space is placed on the surface of planet Physica. On the planet's surface, the scale indicates a force of 10,000 Newtons. The surface area of the scale is 0.125 square meters.

To calculate the surface area of the scale, we can use the formula:

P = F/A

where P is the pressure, F is the force, and A is the surface area.

Given that the atmospheric pressure on the surface of Physica is 80,000 Pascals and the scale indicates a force of 10,000 Newtons, we can plug in these values into the equation:

80,000 Pa = 10,000 N / A

To solve for A, we can rearrange the equation:

A = 10,000 N / 80,000 Pa

A = 0.125 m²

In summary, based on the given information, the surface area of the scale on the surface of planet Physica is calculated to be 0.125 square meters.

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Answer:

a. 13.96 m/s in the +x direction

b. -1.76 m/s that is 1.76 m/s in the -x direction

Explanation:

(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.

Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = + 37.0 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck

So, making v₂ subject of the formula, we have

v₂ = v₁ + Ft/m

substituting the values of the variables into the equation, we have

v₂ = v₁ + Ft/m

v₂ = 2.40 m/s + 37.0 N × 0.050 s/0.160 kg

v₂ = 2.40 m/s + 1.85 Ns/0.160 kg

v₂ = 2.40 m/s + 11.56 m/s

v₂ = 13.96 m/s

(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?

Since Impulse, I = Ft = m(v₂ - v₁) where F = force applied = - 13.3 N(positive since it is applied to the right), t = duration of applied force = 0.050 s, m = mass of hockey puck = 0.160 kg, v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right) and v₂ = final velocity of hockey puck

So, making v₂ subject of the formula, we have

v₂ = v₁ + Ft/m

substituting the values of the variables into the equation, we have

v₂ = v₁ + Ft/m

v₂ = 2.40 m/s + (-13.3 N) × 0.050 s/0.160 kg

v₂ = 2.40 m/s - 0.665 Ns/0.160 kg

v₂ = 2.40 m/s - 4.156 m/s

v₂ = -1.756 m/s

v₂ ≅ -1.76 m/s

(a). The velocity of the puck is 13.96 m/s in the +x direction

(b). The final velocity of the puck is -1.76 m/s which is 1.76 m/s in the -x direction

The velocity of the object is defined as the movement of the object in a particular direction with respect to time.

(a) Calculate the magnitude of the velocity of the puck after a force of 37.0 N directed to the right has been applied for 0.050 s.

It is given that

Impulse, I = Ft = m(v₂ - v₁)

where F = force applied = + 37.0 N(positive since it is applied to the right),

t = duration of applied force = 0.050 s,

m = mass of hockey puck = 0.160 kg,

v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right)  

v₂ = final velocity of a hockey puck

So, making v₂ subject of the formula, we have

\(V_2=v_1+\dfrac{Ft}{m}\)

substituting the values of the variables into the equation, we have

\(v_2=2.40+\dfrac{37\times 0.050}{0.160}\)

\(v_2=2.40+11.56\)

\(v_2=13.96\ \dfrac{m}{s}\)

(b) If instead, a force of 13.3 N directed to the left is applied from t = 0 to t =0.050 s, what is the final velocity of the puck?

Since Impulse, I = Ft = m(v₂ - v₁)

where F = force applied = - 13.3 N(positive since it is applied to the right),

t = duration of applied force = 0.050 s,

m = mass of hockey puck = 0.160 kg,

v₁ = initial velocity of hockey puck =  + 2,40 m/s (positive since it is moving to the right)  

v₂ = final velocity of a hockey puck

So, making v₂ subject of the formula, we have

\(v_2=v_1+\dfrac{Ft}{m}\)

substituting the values of the variables into the equation, we have

\(v_2=2.40+\dfrac{-13.3\times 0.050}{0.160}\)

\(v_2=2.40-4.156\)

\(v+2=-1.76\ \frac{m}{s}\)

Thus

(a). The velocity of the puck is 13.96 m/s in the +x direction

(b). The final velocity of the puck is -1.76 m/s which is 1.76 m/s in the -x direction

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The examples of balanced forces include;

Balanced forces are those that are opposite in direction and equal in size. Balanced forces are considered to be in a state of equilibrium.

A tow truck tries and fails to pull a car out of a ditch. Neither the car nor the tow truck move. (forces are balanced that's why they don't move).

Jada and Mya run into each other in bumper cars. They bounce off each other and go in new directions. (forces are  balanced that's why they bounce off in opposite direction).

Zeke rests at the top of Stone Mountain and enjoys the view. (forces are balanced that's why he is at rest).

Haleigh throws a baseball from first base to second. (forces are not balanced that's why the baseball is in motion).

Ryan is playing ice hockey and glides across the ice in a straight line at a constant speed. (forces are not balanced that's why he is in motion).

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The tangential speed of the keys is approximately 4.99 m/s, calculated using the formula v = rω where ω is found using the formula ω = √(a/r).

To determine the tangential speed of your keys, we can use the formula:

v = rω

Where:

v = tangential speed

r = radius (0.34 m in this case)

ω = angular velocity

To find the angular velocity, we can use the formula:

a = rω²

Where:

a = centripetal acceleration (145 m/s² in this case)

r = radius (0.34 m in this case)

Rearranging the formula, we get:

ω = √(a/r)

Substituting the values, we get:

ω = √(145 m/s² / 0.34 m) ≈ 14.697 rad/s

Now we can find the tangential speed by substituting the values into the formula:

v = rω

v = 0.34 m × 14.697 rad/s ≈ 4.99 m/s

Therefore, the tangential speed of your keys is approximately 4.99 m/s.

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Answer:

The angle is  \(\phi = 0.45 0 ^o\)

Explanation:

From the question we are told that  

     The objective  focal length   \(f = 1.33 \ m\)

     The  eyepiece focal length is  \(f_o = 2.82 \ cm = 0.0282 \ m\)

      The diameter of the sunlight is  \(d = 25000km = 2.5 *10^{7} \ m\)

       The distance of the sun from from the earth is  \(D = 1.5 *10^8 km = 1.5 *10^{11} \ m\)

  Generally the magnification of the object is mathematically evaluated as

        \(m = -\frac{f_o }{f_e }\)

The negative  sign is because the lens of the telescope is  diverging light

 substituting values  

      \(m = -\frac{1.33 }{0.0282 }\)

      \(m = - 47.16 3\)

Now we can obtain the angle made by the object (sunlight ) with respect to the telescope  as follows  

        \(tan \theta = \frac{d}{D}\)

substituting values

      \(tan \theta = \frac{2.5 *10^{7}}{1.5*10^{11}}\)

      \(tan \theta = 0.0001667\)

      \(\theta= tan^{-1}[0.0001667]\)

     \(\theta= 0.00955^o\)

The  magnification can  also be mathematically represented as

      \(m = \frac{\phi }{\theta }\)

Where \(\phi\) is the angle the image made with telescope

Since the negative sign indicate direction of light movement we will remove it from the calculation below

      =>   \(47.163 = \frac{\phi}{0.00955}\)

     =>    \(\phi = 0.45 0 ^o\)

1.00 kg of ice at -24.0°C is placed in contact with a 1.00 kg block of a metal at 5.00°C. They come to equilibrium at -8.88°C.

We can use the principle of conservation of heat to solve this problem. The heat lost by the metal must equal the heat gained by the ice.

The heat lost by the metal is given by

Q1 = m1c1ΔT1

Where m1 is the mass of the metal, c1 is its specific heat, and ΔT1 is the change in temperature.

The heat gained by the ice is given by

Q2 = m2c2ΔT2

Where m2 is the mass of the ice, c2 is its specific heat, and ΔT2 is the change in temperature.

Since the two objects come to thermal equilibrium, we can set Q1 equal to Q2

m1c1ΔT1 = m2c2ΔT2

Solving for c1, we get

c1 = m2c2ΔT2 / (m1ΔT1)

By putting these values we get

c1 = (1.00 kg)(2.06 kJ/kg·K)(-24.0°C - (-8.88°C)) / [(1.00 kg)(5.00°C - (-8.88°C))]

c1 = 0.902 kJ/kg·K

Hence, the specific heat of the metal is 0.902 kJ/kg·K.

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Answer:

The mass is the resistance to the acceleration of a moving object.

Explanation:

hope it helps

Answer:

to changr in motion is called

Answer:

5.92 *10^-6 C

Explanation:

For the two charges q3 the force between them is given by

Now we know that

F = 48.1 N, d = 0.0809 m, and k = 8.99 *10^9 kg⋅m^3⋅s^−2⋅C^-2; therefore, the above gives

Now we solve for q_3.

Dividing both sides by 8.99 * 10^9 gives

multiplying both sides by (0.0809)^2 gives

finally, taking the square root of both sides gives

Evaluating the right-hand side gives

Hence, the charge q_3 is 5.92 x 10^-6 C.

Answer:

t = 2.94 x 10⁶ years

Explanation:

The equation used in the case of constant speed is:

s = vt

t = s/v

where,

s = distance = 12 light years

s = (12 light years)(9.461 x 10¹² km/light year) =  113.532 x 10¹² km

v = speed = 440 km/hr

t = time passed = ?

Therefore,

t = (113.532 x 10¹² km)/(440 km/hr)

t = 2.58 x 10¹¹ hr

Now, converting it to years:

t = (2.58 x 10¹¹ hr)(1 year/8766 hr)

t = 2.94 x 10⁶ years

Answer:

Endurance, or aerobic

Strength, or resistance training

Balance

Flexibility

Explanation:

Answer:

Walking. Walking is simple, yet powerful. It can help you stay trim, improve cholesterol levels, strengthen bones, keep blood pressure in check, lift your mood, and lower your risk for a number of diseases (diabetes and heart disease, for example).

The elements that are metals are as follows:

Metals are any of a number of chemical elements in the periodic table that form a metallic bond with other metal atoms.

Metals are known to be generally shiny, somewhat malleable, hard and often a conductor of heat and electricity.

Some of the elements on the periodic table are as follows:

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In general, option c - warming it up on the stove - is often an effective method to increase the reaction rate.

Increasing the temperature of a reaction generally leads to faster reaction rates. This is because higher temperatures provide more thermal energy to the reactant particles, causing them to move faster and collide more frequently. The increased collision frequency and energy lead to more successful collisions and a higher likelihood of effective molecular interactions, which speeds up the reaction. On the other hand, options a and b - putting it in the fridge where it is cold or covering it with a blanket to make it dark - are unlikely to have a significant effect on the reaction rate. While temperature can influence reaction rates, cooling the reaction or making it dark typically reduces the kinetic energy of the particles, resulting in slower reaction rates. Option d - there is nothing you can do to change the speed of the reaction - is not accurate. The reaction rate can be influenced by various factors such as temperature, concentration, catalysts, and surface area, among others. By manipulating these factors, it is often possible to control and change the speed of a reaction. Hence option c, is correct

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Answer:

Explanation:

The orbital radius of the satellite orbiting planet Earth at the given linear speed is 26,000 km.

The orbital radius of the planet is the distance of the planet from the center of the Earth. The orbital radius of the planet is calculated as follows;

\(v = \sqrt{\frac{GM}{r} }\)

where;

r = GM/v²

r = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴)/(3,914²)

r = 2.6 x 10⁷ m

r = 26,000 km

Thus, the orbital radius of the planet is 26,000 km.

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Answer:

theta = 90°

Explanation:

hope it's helpful

Explanation:

A • B = (2i - 3j + k) • (i + j + k)

= (2)(1) + (- 3)(1) + (1)(1)

= 0 = AB cos θ

∴ cos θ = 0 (as A ≠ 0,B ≠ 0)

or θ = 90°

or the vectors A and B are mutually perpendicular.

-TheUnknownScientist

Answer:

Answer is B. Non renewable resources are easy to store and transport.

Answer:

It's A

Explanation:

sound waves are longitudinal they need a medium to travel through

Answer:

24 mph

Explanation:

Speed = Distance / Time

Speed = 20 ÷ 5/6

Speed = 24 mph

NOTE: 5/6 is a fraction in this case and is equivalent to 50/60 and 60 minutes = 1 hour of course.

Hope this helped!

Answer:

\(\textsf {speed = 24 mph}\)

Explanation:

\(\textsf {Formula used :}\\\implies \textsf {speed = distance/time}\)

\(\mathsf {Given :}\)

\(\implies \textsf {distance = 20 miles}\)

\(\implies \textsf {time = 50 minutes = 5/6 hour}\)

\(\textsf {Solving :}\)

\(\implies \mathsf {speed = 20 \div 5/6 }\)

\(\implies \mathsf {speed = 20 \times 6/5}\)

\(\implies \mathsf {speed = 4 \times 6}\)

\(\implies \textsf {speed = 24 mph}\)

A photon is a tiny energy packet of electromagnetic radiation, also referred to as a light quantum.

1) Ultraviolet radiation is commonly used in sanitizing hospitals because, viruses, methicillin-resistant Staphylococcus aureus, and vancomycin-resistant enterococci are all susceptible to the germicidal effects of UV-C.

2) According to the photon theory, electrons that have been stimulated return to the ground state to create atomic emission spectra. Light or the photon is the energy that is released when electrons drop to a lower energy level.

3) To characterise the atomic characteristics of light, Albert Einstein used Planck's quantum theory.

Planck's hypothesis is supported by Einstein's demonstration that electromagnetic radiation, such as light, has both wave-like and particle-like properties.

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