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1 year ago

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A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W

hat is the magnitude of the acceleration of the ball when it is at its highest point?

1 answer:

dalvyx [7]1 year ago

5 0

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

$F_g = mg$

now the acceleration of ball is given as

$a = \frac{F_g}{m}$

$a = \frac{mg}{m} = g$

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

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The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym

Fynjy0 [20]

Answer:

Given the potential, $V = Ax^l+By^m+Cz^n+D$

The components of the electric field are:

$E_x = \frac{-dV}{dx} = -Alx^l^-^1$

$E_y = \frac{-dV}{dy} = - Bmy^m^-^1$

$E_z = \frac{-dV}{dz} = - nCzn^n^-^1$

Let's calculate the potential difference for all given points.

$V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10$

$=> D = 10$

$V(1, 0, 0) = 4V => A + 10 = 4$

Solving for A, we have:

$A = 4 - 10$

$A = -6$

$V(0, 1, 0) = 6V => B + 10 = 6$

Solving for B, we have:

$B = 6 - 10$

$B = -4$

$V(0, 0, 1) = 8V => C + 10 = 4$

Solving for C, we have:

$C = 8 - 10$

$C = -2$

For all given points, let's calculate the magnitude of electric field as follow:

$E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16$

$Al = -16$

Solving for l, we have:

$l = \frac{-16}{A}$

From above, A = -6

$l = \frac{-16}{-6}$

$l = \frac{8}{3}$

$E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16$

$Bm = -16$

Solving for m, we have:

$m = \frac{-16}{A}$

From above, B = -4

$m = \frac{-16}{-4}$

$m = 4$

$E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16$

$nC = - 16$

Solving for n, we have:

$n = \frac{-16}{C}$

From above, C = -2

$n = \frac{-16}{-2}$

$n = 8$

4 0

1 year ago

A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th

mr Goodwill [35]

Answer:

$T=4985.5^{\circ}K$

Explanation:

The equation that relates heat Q with the temperature change $T-T_0$ of a substance of mass <em>m </em>and specific heat <em>c </em>is $Q=mc(T-T_0)$ .

We want to calculate the final temperature <em>T, </em>so we have:

$T=\frac{Q}{mc}+T_0$

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add $273^{\circ}$ to our temperature in $^{\circ}C$ to have it in $^{\circ}K$ as it must be):

$T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K$

3 0

1 year ago

Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co

mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copper $V=8.0\times10^{-6}\ m^3$

Volume of liquid $V=7.6\times10^{-6}\ m^3$

Latent heat of fusion $L= 13.05 kJ$

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

$\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}$

Put the value into the formula

$\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}$

$\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}$

$\Delta T=4.162\ K$

Hence, The change in the equilibrium melting point is 4.162 K.

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1 year ago

(a) Two point charges totaling 8.00 μC exert a repulsive force of 0.150 N on one another when separated by 0.500 m. What is the

Anastaziya [24]

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1 year ago

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When two objects are in contact, moving together, which of the following statements must be true? Choose all that apply. When tw

Setler [38]

Answer:

The objects must have the same acceleration and the objects must exert the same magnitude force on each other.

Explanation:

The objects must have the same weight: FALSE. This is not needed, any two object can move together in contact no matter their mass.

The objects must have the same acceleration: TRUE. If they have different accelerations, they will separate since the distance each of them travel at a given time will be different.

The objects must have the same net force acting on them: FALSE. This is not needed, since what matters is acceleration, and a=F/m, so if both objects have different net force acting on them, they could have different masses also to compensate and result in the same acceleration.

The objects must exert the same magnitude force on each other: TRUE, this is the 3rd Newton Law, an action must follow the same reaction.

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2 years ago