Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.

Answer: Emotional

Explanation:

Answer:

Physical

Explanation:

I love going going for a jog or a run so does my dog

Answer:

Earth.

Explanation:

Center of gravity can be defined as the specific point where all of the weight of an object is concentrated.

Generally, all the objects found around the world all have a center of gravity.

When an object is balanced so that a displacement lowers its center of gravity, the object is said to be in stable equilibrium.

Hence, the force of gravity acting on an object is directed through this center of gravity and toward the center of the earth.

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, weight is given by the formula;

[tex] Weight, W = mg [/tex]

Where;

m is the mass of an object.

g is acceleration due to gravity.

Answer:

C. A heat engine must deposit some energy in a cold reservoir.

Explanation:

The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."

This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.

Then we have the equation:

Q = W + q

From this we can conclude that the correct option is:

C. A heat engine must deposit some energy in a cold reservoir.

There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.

C. A heat engine must deposit some energy in a cold reservoir.

Therefore, option C is correct.

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Explanation:

(10) Mass of a soccer player, m = 0.42 kg

Initial speed, u = 0

Final speed, v = 32.5 m/s

Time, t = 0.21 s

We need to find the force that sends soccer ball towards the goal.

Force, F = ma

[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.42 \times (32.5-0)}{0.21}\\\\F=65\ N[/tex]

So, 65 N of force soccer ball sends towards the goal.

(11) Mass of the satellite, m = 72,000 kg

Initial speed, u = 0 m/s

Final speed, v = 0.63 m/s

Time, t = 1296 s

We need to find the force is exerted by the rocket on the satellite.

Force, F = ma

[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{72,000\times (0.63-0)}{1296}\\\\F=35\ N[/tex]

So, 35 N of the force is exerted by the rocket on the satellite.

Hence, this is the required solution.

Answer:

The answer should be A) 0m/s

Explanation:

It is stopped on the train tracks therefore it is not moving.

Please tell me if I am wrong because I'm not 100% sure on this. Hope it's right and that it helped you.

Answer:

a. Q = 1881.73 x [tex]10^{-13}[/tex] C

b. As battery is not removed so, potential difference will remain same.

c. E = 21.42 x [tex]10^{3}[/tex] V/m

d. Q = 895.5 x [tex]10^{-13}[/tex] C

e. Again the potential difference will not change it will remain same as 9 V

f. E = 45 x [tex]10^{3}[/tex] V/m

Explanation:

Solution:

Here, Teflon is used so, the dielectric constant of the Teflon K = 2.1

Diameter = 2.1 cm

Radius = 2.1/2 cm

Radius = 1.05 cm

Radius = 0.015 m

Now, we need to find the area of each plate:

A = [tex]\pi r^{2}[/tex]

A = (3.14) ([tex]0.015^{2}[/tex])

A = 0.000225 [tex]m^{2}[/tex]

A = 2.25 x [tex]10^{-4}[/tex] [tex]m^{2}[/tex]

We are given the thickness of the plate which equal to the distance between the two plates.

d = 0.20 mm = 0.2 x [tex]10^{-3}[/tex] m

d = 0.2 x [tex]10^{-3}[/tex]  m = distance between two plates.

Hence, the capacitance of the dielectric without the dielectric

C = [tex]\frac{E.A}{d}[/tex]

Putting up the values we get,

E = 8.85 x [tex]10^{-12}[/tex]

C = [tex]\frac{8.85 . 10^{-12} x 2.25 . 10^{-4} }{0.002}[/tex]    

C = 99.5 [tex]10^{-13}[/tex]

If dielectric is included then,

[tex]C^{'}[/tex] = K C

[tex]C^{'}[/tex] = (2.1) ( 99.5 x [tex]10^{-13}[/tex])

[tex]C^{'}[/tex] = 209.08 x [tex]10^{-13}[/tex] F

As we know the voltage of the battery V = 9V So,

a) Charge before the Teflon is removed:

Q = CV

Q = [tex]C^{'}[/tex]V

Q = (209.08 x [tex]10^{-13}[/tex] F) (9V)

Q = 1881.73 x [tex]10^{-13}[/tex] C

b) Potential Difference before the Teflon is removed = ?

As battery is not removed so, potential difference will remain same.

c) Electric Field =?

As we know,

E = V/(K.d)

E = 9V/(2.1 x 0.2 x [tex]10^{-3}[/tex])

E = 21.42 x [tex]10^{3}[/tex] V/m

d) After the Teflon is removed

Q = CV

Q = (99.5 [tex]10^{-13}[/tex] ) ( 9)

Q = 895.5 x [tex]10^{-13}[/tex] C

e) Again the potential difference will not change it will remain same as 9 V

f) Electric Field = ?

E = [tex]\frac{V}{d}[/tex]  (Teflon is removed)

E = 9/0.2 x [tex]10^{-3}[/tex]

E = 45 x [tex]10^{3}[/tex] V/m

The increase in thermal energy of the gas sample is +14 J, and the temperature is increased.

Given information:

The heat removed from the gas sample is [tex]Q=-14\rm\;J[/tex]. The negative sign represents the heat removal.

The work done on the gas sample is [tex]W=28\rm\; J[/tex].

Work is done on the gas. So, it will be taken as positive.

According to the first law of thermodynamics, the change in thermal energy of the  gas or system will be calculated as,

[tex]\Delta E=Q+W\\\Delta E=-14+28\\\Delta E=14\rm\;J[/tex]

The change in thermal energy of the system will be 14 J. It is positive. So, the thermal energy is increased. It implies the temperature of the system or gas is also increased.

Therefore, the increase in thermal energy of the gas sample is +14 J, and the temperature is increased.

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Explanation:

p=m x v

to find the mass:

m= p/v

=47/11

=4.27 kg

Answer:

Hello your question is incomplete attached below is the missing part of the question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I

answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L

              part A = attached below

Explanation:

Part A :

Assuming that mass of swing is negligible

α = T/I

where ; T = torque, I = inertia,

hence T =  L/2*9*(M1 - M2)

also;  I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]=  ( M1 + M2) * (L/2)^2

Finally the magnitude of the angular acceleration α

α = 2*[(M1 - M2)/(M1 + M2)]*g/L

Part B attached below

Answer:

742.3N

Explanation:

Given parameters:

Distance  = 39m

Mass 1  = 9.2 x 10⁷kg

Mass 2  = 1.84 x 10⁸kg

Unknown:

Gravitational force between the ships  = ?

Solution:

To solve this problem, we apply the newton's law of universal gravitation:

            Fg  = [tex]\frac{G x mass 1 x mass 2}{r^{2} }[/tex]  

G is the universal gravitation constant  = 6.67 x 10⁻¹¹

r is the distance or separation

   Fg  = [tex]\frac{6.67 x 10^{-11} x 9.2 x 10^{7} x 1.84 x 10^{8} }{39^{2} }[/tex]   = 742.3N

Answer:

Resistance is the opposing of the flow of current through a conductor.

Answer:

E; The object is negatively charged

Explanation:

Here, we want to state the conclusion that can be drawn from a positively charged rod being attracted to an object.

Generally as we know, oppositely charged materials attract while the ones with same charges repel each other.

Thus, in this case, for the rod to attract the object, there must have been an opposite charge of negativity on the object

So we conclude that the reason why the rod attracted the object was because of the presence of opposing charges on both of them. And since the rod has taken the positive charge, it is only correct to state that the object is negatively charged

Answer:

[tex]v_f=2\:\mathrm{m/s}[/tex]

Explanation:

From the Law of Conservation of Momentum, momentum is conserved. Therefore, we can set up the following equation:

[tex]m_1v_1+m_2v_1=m_fv_f[/tex]

Since the 5kg toy car was initially and rest, [tex]m_2v_2=0[/tex].

Therefore, plugging in our values, we have:

[tex]10\cdot 3=(10+5)v_f,\\v_f=\frac{30}{15},\\v_f=\fbox{$2\:\mathrm{m/s}$}[/tex].

Answer:

Let m1 = mass of big toy car=10kg

m2= mass of small toy car= 5kg

U1= initial velocity of big toy car= 3m/s

U2= initial velocity of small toy car=0

Since the big toy car moved the small one after the collision, their final velocity will be the same.

m1u1 + m2u2= (m1+m2)v

10(3)+(5)(0)=(10+5)v

30=15v

Divide both sides by 15

V=2

The final velocity is 2m/s

Explanation:

the shadows are exactly the same length in the morning as they are in the evening.

 is so obvious it’s that when the sun is low you get long shadows and when the sun is up in the sky like in the noon the shadow is shorter.

Answer:

Hydrochloric and Hydrofluoric Acids.

Answer:

v = 3.34 m/s

Explanation:

Given that,

Distance, d = 1100 m

Time, t = 330 s

We need to find the average speed of the Megan. It is equal to the total distance divided by total time taken.

[tex]v=\dfrac{1100\ m}{330\ s}\\\\v=3.34\ m/s[/tex]

So, the average speed of Megan is 3.34 m/s.

Answer:

33.3

Explanation:

Answer:

0.83 m³/s  

Explanation:

The speed of the airflow is given as;

             1.85 x 10³ ft³/min

Now we are to express this unit in m³/s  

    1ft  = 0.3m

    60s  = 1 min

So;

  1.85 x 10³  x ft³ x [tex]\frac{1}{min}[/tex] x [tex]\frac{(0.3m)^{3} }{ft^{3} }[/tex] x [tex]\frac{1min}{60s}[/tex]  

    = 0.83 m³/s  

Answer:

b) Springs oscillate with the same frequency,

Explanation:

expression for frequency of vibration of mass hanging from a spring is given as follows

f = [tex]\frac{1}{2\pi} \times \sqrt{\frac{k}{m} }[/tex]

k is force constant of spring and m is mass vibrating .

In the present case, if mass stretches the spring by x and remains balanced

mg = k x

[tex]\frac{k}{m} =\frac{g}{x}[/tex]

g and x are same  for both cases

[tex]\frac{k}{m}[/tex] will also be same for both cases .

Hence frequency of vibration will also be same for both the balls .

Answer:

They can cause sunburn. Exposure to UV rays can cause premature aging of the skin and signs of sun damage such as wrinkles, leathery skin, liver spots, actinic keratosis, and solar elastosis. UV rays can also cause eye problems.

Explanation:

Answer:

C. The sum remains the same.

Explanation:

The sum of the kinetic and potential energy remains the same as the all rolls from point A to E.

We know this based on the law of conservation of energy that is in play within the system.

The law of conservation of energy states that "energy is neither created nor destroyed within a system but transformed from one form to another".

Answer:

1. 1800 W

2. $ 17.3

Explanation:

From the question given above, the following data were obtained:

Current (I) = 15 A

Voltage (V) = 120 V

Time (t) = 20 h per day

Duration = 31 days

Cost = 15.5 cents per kWh

1. Determination of the power.

Current (I) = 15 A

Voltage (V) = 120 V

Power (P) =?

P = IV

P = 15 × 120

P = 1800 W

Thus, 1800 W of power is required.

2. Determination of the cost per month (31 days).

We'll begin by converting 1800 W to KW.

1000 W = 1 KW

Therefore,

1800 W = 1800 W × 1 KW / 1000 W

1800 W = 1.8 KW

Next, we shall determine the energy consumption for 31 days. This can be obtained as follow:

Power (P) = 1.8 KW

Time (t) = 2 h per day

Time (t) for 31 days = 2 × 31 = 62 h

Energy (E) =?

E = Pt

E = 1.8 × 62

E = 111.6 KWh

Finally, we shall determine the cost of consumption. This can be obtained as follow:

1 KWh = 15.5 cents

Therefore,

111.6 KWh = 111.6 KWh × 15.5 cents / 1 KWh

111.6 KWh = 1729.8 cents

Converting 1729.8 cents to dollar, we have:

100 cents = $ 1

Therefore,

1729.8 cents = 1729.8 cents × $ 1 / 100 cents

1729.8 cents = $ 17.3

Thus, it will cost $ 17.3 per month to run the electric heater.

The amount of power it requires is 180 Watt, and the cost per month to operate the electric heater for 2.0 h is 5.58 cent

To calculate the electric power required by the electric heater, we use the formula below.

⇒ Formula:

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

For the cost of running the electric heater per month, if it operates 2.0 h per day. We use the relation below

⇒ Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 2

Hence, The amount of power it requires is 180 Watt, and the cost per month to operate the electric heater for 2.0h is 5.58 cent

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Answer:

[tex]v_{2'}=8.1\:\mathrm{m/s}[/tex]

Explanation:

In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:

[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]

Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].

Plugging in all given values, we have:

[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot8.4^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot 4.4\cdot {v_{2'}}^2,\\\\{v_{2'}}=\sqrt{64.31},\\\\{v_{2'}}\approx\fbox{$8.1\:\mathrm{m/s}$}[/tex]..

Answer:

reducing temperature of the surrounding

Explanation:

combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer

Answer:

A chemical reaction is a process in which one or more substances, also called reactants, are converted to one or more different substances, known as products.

Answer:

the answer is reactants i guess

Answer:

Amplitude increases with decreasing velocity.

Explanation:

At the same time, an increase in attention takes place

Answer:

F = 402.18 N

Explanation:

Given that,

A particular satellite with a mass of m is put into orbit around Ganymede (the largest moon of Jupiter) at a distance 300 km from the surface. Let the mass of the satellite is 350 kg.

We need to find the gravitational force of attraction between the satellite and the moon.

The formula for the gravitational force is given by :

[tex]F=G\dfrac{Mm}{(R+h)^2}[/tex]

M is mass of Ganymede

m is mass of satellite

R is Radius of Ganymede

h is distance = 300 km

Putting all the values,

[tex]F=6.67\times 10^{-11}\times \dfrac{1.48\times 10^{23}\times 350}{(2631\times 10^{3}+300\times 10^3)^2}\\F=402.18\ N[/tex]

So, the required force of attraction between the satellite and the moon is 402.18 N.

Answer: [tex]5.944\times 10^{23}\ kg[/tex]

Explanation:

Given

Time period [tex]T=7.08\times 10^3\ s[/tex]

Orbital speed [tex]v=3.40\times 10^3\ m/s[/tex]

mass of GS [tex]m_{GS}=930\ kg[/tex]

Radius of Mars [tex]r=3.43\times 10^6\ m[/tex]

Consider the mass of mars is M

Here, Gravitational pull will provide the centripetal force

[tex]F_G=F_c[/tex]

[tex]\dfrac{GMm_{GS}}{r^2}=\dfrac{m_{GS}v^2}{r}\\M=\dfrac{v^2\cdot r}{G}\\M=\dfrac{(3.43\times 10^3)^2\cdot 3.43\times 10^6}{6.67\times 10^{-11}}[/tex]

[tex]M=5.944\times 10^{23}\ kg[/tex]

In March 1999 the Mars Global Surveyor (GS) entered its final orbit on Mars, sending data back to Earth. The mass of Mars is approximately 6.419 × 10²³ kg.

Kepler's Third Law states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit:

T² = (4π² / GM) × a³

In a circular orbit, the semi-major axis is equal to the radius of the orbit (r).

Given:

Orbital period (T) = 7.08 × 10³ s

Orbital speed (v) = 3.40 × 10³ m/s

Mass of GS (m) = 930 kg

Radius of Mars (r) = 3.43 × 10⁶ m

The orbital speed (v) is related to the radius (r) and the gravitational constant (G) by:

v = √(GM / r)

v² = GM / r

G = (v² × r) / M

T² = (4π² / [(v² × r) / M]) × r³

T² = (4π² × M × r²) / v²

M = (T² × v²) / (4π² × r²)

M = ( (7.08 × 10³)² × (3.40 × 10³)² ) / (4π² × (3.43 × 10⁶)²)

M = 6.419 × 10²³ kg

Therefore, the mass of Mars is approximately 6.419 × 10²³ kg.

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Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.