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2 years ago

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A basketball is tossed upwards with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction. We can ignore air resistance. What is the maximum height reached by the basketball from its release point?

1 answer:

Hatshy [7]2 years ago

4 0

Answer:

The last one

Explanation:

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A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

2 years ago

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

= surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ / 2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀

= 6.78 X 10³ N/C

4 0

1 year ago

The image on the left shows the charges on a balloon after it’s been rubbed with a wool cloth. The image on the right is a piece

taurus [48]

When wool is rubbed with a balloon, the wool is left with a positive charge as electrons have travelled from the wool to the balloon which means the balloon now has a negative charge.

Now that the balloon has a negative charge, you need to know:
The tissue paper originally contains electrons and protons
The fact that the balloon has a negative charge, it will ATTRACT protons because protons are POSITIVE and electrons are NEGATIVE.
So once they are attracted, they will move closer to one another.

3 0

1 year ago

Read 2 more answers

A plane wall with constant properties is initially at a uniform temperature To. Suddenly, the surface at x = L is exposed to a c

Rzqust [24]

Answer:

The distribution is as depicted in the attached figure.

Explanation:

From the given data

The plane wall is initially with constant properties is initially at a uniform temperature, To.
Suddenly the surface x=L is exposed to convection process such that T∞>To.
The other surface x=0 is maintained at To

Uniform volumetric heating q' such that the steady state temperature exceeds T∞.

Assumptions which are valid are

There is only conduction in 1-D.
The system bears constant properties.
The volumetric heat generation is uniform

From the given data, the condition are as follows

<u>Initial Condition</u>

At t≤0

$T(x,0)=T_o$

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.

<u>Boundary Conditions</u>

<u>At x=0</u>

<u /> $T(0,t)=T_o$ <u />

This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.

<u>At x=L</u>

<u /> $-k\frac{\partial T}{\partial x}]_{x=L}=h[T(L,t)-T_{\infty}]$ <u />

This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.

The temperature distribution along with the schematics are given in the attached figure.

Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.

It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.

3 0

2 years ago

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th

Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

Explanation:

Given that :

The internal shear force V = 80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress $\tau$ can be calculated by using the equation:

$\tau = \dfrac{VQ}{Ib}$

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10 mm

From the centroid ;

Q = $Q_1 + Q_{2}$

Q = $A_1y_1 + A_{2}y_{2}$

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³

Q = ( 38500 + 307125 ) mm³

Q = 345625 mm³

$\tau_H = \dfrac{VQ}{Ib}$

$\tau_H = \dfrac{80*10^3 * 345625}{64900000*10 }$

$\tau_H = \dfrac{2.765*10^{10}}{649000000 }$

$\tau_H = 42.60400616 \ N/mm^2$

$\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

7 0

2 years ago