A professional golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00410 s. After the collision, the ball leaves the club at a speed of 32.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

Answer:

electromagnetic radiation moves at the speed of light

Answer: b or d

Explanation: b or d

Answer:

a

Explanation:

F= ma

interestingly

when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases

(man, PHYSICS IS JUST THE BESY)

A: a

object with the smallest mass has largest acceleration

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = 4

c. The signal is even, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

[tex] \frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda[/tex]

Which gives;

E = 0.5 × 1.571² × 1² × 4 = 4.935 J/kg

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = 1.234 W/kg

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Answer:

I don't understand your question ❓,the object.....of what experiment

For A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed of 4.20 m/s  is mathematically given as

F = 193.2N

Generally, the equation for the  magnitude of the average force mathematically given as

F = m(v1+v2)/t

F = 8.40(4.2+O.4)/t

F = 193.2N

In conclusion magnitude of the average force is

F = 193.2N

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Answer:

If an object is electrically neutral it has no net charge becuase it has the same number of protons as it does electrons, which are opposite charges that offset each other. No, that just means that the sum of all its positive and negative amounts of charge equals zero.

Explanation:

Explanation:

neutron star, any of a class of extremely dense, compact stars thought to be composed primarily of neutrons. Neutron stars are typically about 20 km (12 miles) in diameter. Their masses range between 1.18 and 1.97 times that of the Sun, but most are 1.35 times that of the Sun.

The final temperature of the mixture is 33.123 °C.

Temperature can be defined as the hotness or coldness of a thing or place.

To calculate the final temperature of the mixture, we use the formula below.

Formula:

Where:

make t₃ the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, The final temperature of the mixture is 33.123 °C.

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Answer:

Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.

Explanation:

Answer:

A bar of soap when weighed in air has a weight of 5,2N . When completely immersed in water , however it has a weight of 3,7N . What is the volume of the bar of soap

As a conservation biologist for the Chesapeake Bay, you and your

colleagues have been conducting a research study that tracks the decrease

in the bald eagle population over the past few years.

What evidence can you find for the decrease in the bald eagle population?

(a) The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.

(b) The force diagram of a car slowing down as it approaches a stop sign includes force of the car and frictional force opposing the motion.

(c) The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.

(d) The force diagram of baseball before it leaves the bat incudes only the weight of the baseball acting downwards.

The force diagram of a coaster sitting under a cup of coffee includes the weight of the coater plus the weight of coffee acting downwards.

                                        ↑ Fn

                                        Ф                  Fn = W

                                        ↓ W

Where;

The force diagram inlcudes the applied force and frictional force opposing the motion.

                                      Ff ←  Ф  → F

where;

The force diagram includes the force of the test and action of the fridge which are eqaul and opposite.

                               Fb ←  Ф → Fa

where;

The force diagram includes only the weight of the baseball acting downwards.

                                 Ф

                                  ↓

                                 W = mg

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B. kilogram

I hope this helps you

:)

Answer:

B. kilogram

Explanation:

When stated in the unit J s, which is equal to kg m2 s1, the kilogram (kg) is defined by considering the fixed numerical value of the Planck constant h to be 6.62607015 1034 when expressed in the unit J s, which is equivalent to kg m2 s1. The United States Prototype Kilogram 20, a platinum-iridium cylinder held at NIST, is the country's principal mass standard. The kilogram was initially known as the Kilogram of the Archives, and it was defined as the mass of one cubic decimeter of water at its greatest density temperature. It was superseded by the International Prototype Kilogram following the International Metric Convention in 1875, which became the unit of mass without reference to the mass of a cubic decimeter of water or the Archives Kilogram. National Prototype Meters and Kilograms were allocated to each country that signed the International Metric Convention. Learn more about the kilogram's history and current definition. The kilogram (kg) is the only SI basic unit whose name and symbol incorporate a prefix for historical reasons. The SI prefix for 1000 or 103 is "kilo." Prefix names and symbols are attached to the unit name "gram," and prefix symbols are attached to the unit symbol "g," to create names and symbols for decimal multiples and submultiples of the unit of mass. Find out more about this historical oddity.

Units of Mass

10 milligrams (mg) = 1 centigrams (cg)

10 centigrams = 1 decigrams (dg) = 100 milligrams

10 decigrams = 1 gram (g)

10 decigrams = 1000 milligrams

10 grams = 1 dekagrams (dag)

10 dekagrams = 1 hectogram (hg)

10 dekagrams = 100 grams

10 hectograms = 1 kilogram (kg)

10 hectograms = 1000 grams

1000 kilograms = 1 megagram (Mg) or 1 metric ton (t)

A body's mass is a measurement of its inertial property, or the amount of stuff it contains. The force imposed on a body by gravity or the force required to maintain it is measured by its weight. On Earth, gravity accelerates a body downward at around 9.8 m/s2. In the context of weights and measurements, weight is frequently used as a synonym for mass. The verb "to weigh," for example, meaning "to ascertain the mass of" or "to have a mass of." Weight should be phased out in favor of mass, and the term mass should be used when mass is indicated. The kilogram is the SI unit of mass (kg). The weight of a body in a given reference frame is defined in science and technology as the force that causes the body to accelerate at the same rate as the local acceleration of free fall in that reference frame. As a result, the newton is the SI unit for the amount weight defined in this way (force) (N).

If a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr. Details about average speed can be found below.

Average speed can be calculated by dividing the distance moved by a body by the time taken. That is;

Average speed = Distance/Time

According to this question, a jogger runs a 10 kilometer race in 60 minutes. The average speed is calculated as follows:

Average speed = 10km/1hr

Average speed = 10km/hr.

Therefore, if a jogger runs a 10 kilometer race in 60 minutes, her average speed is 10km/hr.

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The magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

The magnitude of the force each plate experiences due to the other plate is determined as follows;

F = U/d

where;

[tex]F = \frac{1}{2} \frac{Q^2}{C} \times \frac{1}{d} \\\\[/tex]

Q = CV

[tex]F = \frac{1}{2} \frac{C^2V^2}{C d} = \frac{CV^2}{2d}[/tex]

where;

The capacitance is given as;

[tex]C = \frac{\varepsilon _o A }{d}[/tex]

[tex]F = \frac{\varepsilon _o A }{d} \times \frac{V^2}{2d} \\\\F = \frac{V^2 A \varepsilon _o }{2d^2}[/tex]

Thus, the magnitude of the force each plate experiences due to the other plate as a function of V, the potential drop across the capacitor is determined as [tex]\frac{V^2 A \varepsilon _o }{2d^2}[/tex].

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is one of these the question

Answer:

-------1

Explanation:

beacuse that is what i know

Answer:

Explanation:

You can use the cross product. Let the vector that perpendicular to a and c is [tex]\vec{d}[/tex], so:

[tex]\vec{d}=\vec{a}\times\vec{c}=\left|\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\5&4&-6\\4&3&2\end{array}\right] \right|=(8+18)\hat{i}-\hat{j}(10+24)+\hat{k}(15-16)=26\hat{i}-34\hat{j}-\hat{k}[/tex]

To check that c is perpendicular with a and b, do the dot product between c and a and also c and b and if the result is zero, you're true.

[tex]\vec{d}.\vec{a}=(26*5)-(34*4)+(6)=0[/tex]  (c perpendicular to a)

[tex]\vec{d}.\vec{c}=(4*26)-(34*3)-(2*1)=0[/tex] (d perpendicular to c)

Answer:

-259322.03N

Explanation:

[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]

Voltage = 5 volts

[tex]\sf \dfrac{number \ of \ turns \ in \ primary \ coil}{voltage\ in \ primary \ coil} = \dfrac{number \ of \ turns \ in \ secondary\ coil}{voltage\ in \ secondary\ coil}[/tex]

[tex]\hookrightarrow \sf \dfrac{400}{100} = \dfrac{20}{V_2}[/tex]

[tex]\hookrightarrow \sf 400V_2}{} = 20*100[/tex]

[tex]\hookrightarrow \sf V_2 = 5 V[/tex]

Answer:

did you get it done if not lmk I will help you out tomorrow when I get up

Explanation:

formula is ˠ=vf

f=1/T

1/7

f=0.14

wavelength=27Ⅹ0.14

=3.78m

OR

7Ⅹ27

=189m

The charge on each particles which are 1 m apart and feeling a repulsive force of 1 N is 1.05×10¯⁵ C

Let the charge on each particles be q

The charge on each particle can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = Kq² / r²

1 = (9×10⁹ × q²) / 1²

1 = 9×10⁹ × q²

Divide both side by 9×10⁹

q² = 1 / 9×10⁹

Take the square root of both side

q =  √(1 / 9×10⁹)

q = 1.05×10¯⁵ C

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Answer:

50 N

Explanation:

Let the natural length of the spring = L

so

100 = k(40 - L)       (1)

200 = k(60 - L)       (2)

(2)/(1):   2 = (60 - L)/(40 - L)

60 - L = 2(40 - L)

60 - L = 80 - 2L

2L - L = 80 - 60

L = 20

Sub it into (1):

100 = k(40 - 20) = 20k

k = 100/20 = 5 N/in

Now

X = k(30 - L) = 5(30 - 20) = 50 N

The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.

Explanation:

simple machine can multiplayer of speed and force

The final speed of the block, after being lifted 2.00 m is 3.39 m/s

Speed can be defined as the rate of change in the distance of a body.

To calculate the speed of the block after being lifted 2.00 m,  first, we need to calculate the acceleration of the block using the formula below

Formula:

Where:

Restructuring the formula above,

From the question,

Given:

Substitute these values into equation 2

Finally, to calculate the speed of the block, we use the formula below.

Where:

From the question,

Given:

Substitute these values into equation 3

Hence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.

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(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

[tex]\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F[/tex]

[tex]C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F[/tex]

[tex]\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F[/tex]

[tex]\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F[/tex]

Ct + C(78) = 2 μF + 3 μF = 5 μF

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

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