A professional quarterback throws a 0.40 kg football. what is the force of weight?

Answer:

The resistance of A is 6 ohms and the resistance of B is 3 ohms

Explanation:

Step 1: For the first connection (parallel connection), the resistance of B will be calculated.

Note: in a parallel connection, the voltage through each resistor is the same.

[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]

Step 2: The resistance of A will be calculated from the second connection (series connection)

Note: in series connection, the current flowing in each resistor is the same

[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]

Explanation:

heat caoacity and heat is difference

The heat energy conducted per hour through the side walls of the cylindrical steel boiler  is 27708847 kJ.

The rate at which heat is transported by conduction through a material's unit cross-section area when a temperature gradient exits perpendicular to the area is known as thermal conductivity.

In the International System of Units (SI), thermal conductivity is measured by Wm⁻¹K⁻¹.

Diameter of the cylindrical steel boiler: d = 1.00m.

Length  of the cylindrical steel boiler: l = 3.00m.

thickness of the walls is = 6.0 mm = 0.006 m

Temperature gradient is = (140-40) °C/0.006 m = 1666.67 °C/m

Thermal conductivity of steel,  = 42 W/m-°C.

Hence, the heat energy conducted per hour through the side walls of the cylindrical steel boiler = 42×3600×1666.67 ×2π×0.5(0.5+3.0) Joule

= 27708847 kJ

Learn more about  thermal conductivity here:

https://brainly.com/question/23897839

#SPJ2

Answer:

Here are a few commonly used types of shooting in basketball.

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

hii

Explanation:

i hope this helps you

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it

Explanation:

hope it helps

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Answer:

The magnitude of the friction force is 8197.60 N

Explanation:

Using the definition of the centripetal force we have:

[tex]\Sigma F=ma_{c}=m\frac{v^{2}}{R}[/tex]

Where:

Now, the force acting in the motion is just the friction force, so we have:

[tex]F_{f}=m\frac{v^{2}}{R}[/tex]

[tex]F_{f}=2100\frac{18^{2}}{83}[/tex]

[tex]F_{f}=8197.60 \: N[/tex]

Therefore the magnitude of the friction force is 8197.60 N

I hope it helps you!

Answer:

0 N

Explanation:

Applying,

F = qvBsin∅................. Equation 1

Where F = Force on the charge, q = charge, v = Velocity, B = magnetic charge, ∅ = angle between the velocity and the magnetic field.

From the question,

Given: q = 4.88×10⁻⁶ C, v = 265 m/s, B = 0.0579 T, ∅ = 0°

Substitute these values into equation 1

F = ( 4.88×10⁻⁶)(265)(0.0579)(sin0)

Since sin0° = 0,

Therefore,

F = 0 N

Answer:

There was 7 continents in africa

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]

The maximum magnetic field is calculated using the following intensity formula;

[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

Answer:

NORMAL

Explanation:

In mathematics to define a plane you need a line perpendicular to the plane called NORMAL and the point of application of this line at a point on the plane.

Based on this definition you only need to specify the normal plane is perpendicular to this line, it should be noted that this does not define a single plane but the whole family of plane is contains the normal.

Consequently only the NORMAL is needed

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

Answer:

F₂ = -7.3 N

Explanation:

Given that,

The mass of an object, m₁ = 3.7 kg

First force, F₁ = 11 N

The net acceleration of the object is 1 m/s².

We know that,

F₁+F₂ = ma

11+F₂ = (3.7)(1)

F₂ = 3.7-11

F₂ = -7.3 N

so, the other force is 7.3 N and it is acting in west direction.

Answer:

[tex]\boxed {\boxed {\sf 29,400 \ Joules}}[/tex]

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

[tex]E_P= m \times g \times h[/tex]

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

Substitute the values into the formula.

[tex]E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m[/tex]

Multiply the three numbers and their units together.

[tex]E_p=1470 \ kg*m/s^2 \times 20 m[/tex]

[tex]E_p=29400 \ kg*m^2/s^2[/tex]

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

[tex]E_p= 29,400 \ J[/tex]

The crate has 29,400 Joules of potential energy.

Answer:

29,400 J

Explanation:

did the quiz <3

Answer:

[tex]t=2413s[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=1.5m/s[/tex]

Time [tex]t=30min=>30*60=>1800[/tex]

Distance [tex]d=24.3km[/tex]

Generally the Newton's equation for Speed going down the stream is mathematically given by

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]1.5+v=frac{24300}{1800}[/tex]

 [tex]v=12m/s[/tex]

Therefore

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]t=\frac{24300}{12-1.5}[/tex]

 [tex]t=2413s[/tex]

Answer:

[tex]\phi=4344.72Nm^2/c[/tex]

Explanation:

From the question we are told that:

Dimension of Wall:

 [tex](L*B)=(8.7 m * 3.2 m)[/tex]

Electric field [tex]B=210 N/C[/tex]

Angle [tex]\theta =42 \textdegree North[/tex]

Generally the equation for electric Flux is mathematically given by

 [tex]\phi=EAcos\theta[/tex]

 [tex]\phi=210*(8.7*3.2)*cos 42[/tex]

 [tex]\phi=4344.72Nm^2/c[/tex]

Answer:

the period of oscillation of the given object is 0.14 s

Explanation:

Given;

mass of the object, m = 3 kg

extension of the spring, x = 0.085 m

The spring constant is calculated as follows;

[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]

The angular speed of a 4 kg object is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]

Therefore, the period of oscillation of the given object is 0.14 s

Answer:

1. 0.625 g/mL

2. 2×10⁵ J

3. 5040 J

4. 75000 J

Explanation:

1. Determination of the density.

We'll begin by calculating the volume of the marble. This can be obtained as follow:

Volume of water = 60 mL

Volume of water + Marble = 68 mL

Volume of marble =?

Volume of marble = (Volume of water + Marble) – (Volume of water)

Volume of marble = 68 – 60

Volume of marble = 8 mL

Finally, we shall determine the density of the marble. This can be obtained as follow:

Mass of marble = 5 g

Volume of marble = 8 mL

Density of marble =?

Density = mass / volume

Density of marble = 5 / 8

Density of marble = 0.625 g/mL

2. Determination of the work done.

Mass (m) = 1000 Kg

Height (h) = 20 m

Acceleration due to gravity (g) = 10 m/s²

Workdone (Wd) =?

The work done can be obtained as follow:

Wd = mgh

Wd = 1000 × 10 × 20

Wd= 2×10⁵ J

3. Determination of the potential energy.

Mass (m) = 12 Kg

Height (h) = 42 m

Acceleration due to gravity (g) = 10 m/s²

Potential energy (PE) =?

The potential energy can be obtained as follow:

PE = mgh

PE = 12 × 10 × 42

PE = 5040 J

4. Determination of the kinetic energy.

Mass (m) = 1500 Kg

Velocity (v) = 10 m/s

Kinetic energy (KE) =?

The kinetic energy can be obtained as follow:

KE = ½mv²

KE = ½ × 1500 × 10²

KE = 750 × 100

KE = 75000 J

Answer:

Tx not but mybe

Explanation:

for that reason its just trying to help

I suppose you meant to say the radius of the curve is 260 m, not mm?

There are 3 forces acting on the car as it makes the turn,

• its weight mg pulling it downward;

• the normal force exerted by the road pointing upward, also with magnitude mg since the car is in equilibrium in the vertical direction; and

• static friction keeping the car from skidding with magnitude µmg (since it's proportional to the normal force), pointing horizontally toward the center of the curve.

By Newton's second law, the net force on the car acting in the horizontal direction is

F = ma   =>   µmg = ma   =>   a = µg

where a is the car's radial acceleration given by

a = v ^2 / R

with v = the car's tangential speed and R = radius of the curve. At the start, the car's radial acceleration is

a = (32 m/s)^2 / (260 m) ≈ 3.94 m/s^2

(a) If µ were reduced by a factor of 2, then the radial acceleration would also be halved:

1/2 a = 1/2 µg

Then the car can have a maximum speed v of

1/2 a = v ^2 / R   =>   v = √(aR/2) = √((3.94 m/s^2) (260 m) / 2) ≈ 22.6 m/s

(b) If µ were increased by a factor of 2, then the acceleration would also get doubled. Then the maximum speed v would be

2a = v ^2 / R   =>   v = √(2aR) = √(2 (3.94 m/s^2) (260 m)) ≈ 45.3 m/s

Answer:

[tex]d =3.7*10^{-3} m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.2kg[/tex]

Spring constant [tex]\mu=22Nm^{-1}[/tex]

Amplitude [tex]A=5cm=0.05m[/tex]

Generally the equation for displacement d is mathematically given by

 [tex]d = Asin(\omega t)[/tex]

Where

 [tex]\omega=angular\ velocity[/tex]

 [tex]\omega=\sqrt{k/m}[/tex]

 [tex]\omega=\sqrt{22/1.2}[/tex]

 [tex]\omega=4.2817rads^{-1}[/tex]

Therefore  

 [tex]d = 0.05*sin(4.2817*1)[/tex]  

 [tex]d =3.7*10^{-3} m[/tex]

Answer:

1 x 10^-9 kilometers

Hope this helps

Have a good day :)

Explanation:

Answer:

a.  F = 14,733.33 N

b. F = 221,000 N

Explanation:

Given;

mass of the cars, m = 1300 kg

velocity of the cars, v = 17 m/s

time taken for the first car to stop after hitting a barrier, t = 1.5 s

time taken for the second car to stop after hitting a barrier, t = 0.1 s

The average forces exerted on each car is calculated as follows;

a. the car that hits a line of water barrels and takes 1.5 s to stop

[tex]F = ma = m\times \frac{v}{t} = 1300 \times \frac{17}{1.5} = 14,733.33 \ N\\\\F = 14,733.33 \ N[/tex]

b. the car that hits a concrete barrier and takes 0.10 s to stop

[tex]F = ma = m\times \frac{v}{t}= 1300 \times \frac{17}{0.1} = 221,000 \ N\\\\F = 221,000 \ N[/tex]

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]

E = 1461.95 N/C

c) The electric field E is calculated as:

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]

E = 239.76 N/C

Answer:

D. digital

Explanation:

Digital signals are transmitted through electromagnetic waves.

I'm not sure

Answer:

false

Explanation:

Answer:

A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...

Explanation:

One example of a physical entity that is quantized is:

The amount of money in your pocket.

The amount can't have any fraction of 1 cent.  

Its value must be an integer-multiple of cents, or 0.01 dollar.    

When it increases or decreases, it jumps from one integer number of cents to the next integer number.  It doesn't "slide" from one to the next.  It can never have a value between two integer numbers of cents.

From the water whole water cycle starts again.

Answer:

25.5×10_3 1928 82i93874 89_/ 9299