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soldi70 [24.7K]

3 years ago

12

A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati

on for the dropped ball.

2 answers:

Elza [17]3 years ago

7 0

Explanation:

The given data is as follows.

Initial velocity; u = 0, Final velocity; v = 19.6 m/s

time; t = 2 seconds

As the relation between initial velocity, final velocity and acceleration is as follows.

v = u + at

Hence, putting the given values into the above formula as follows.

v = u + at

19.6 m/s = 0 + $a \times 2 sec$

a = 9.8 $m/s^{2}$

Thus, we can conclude that acceleration of the dropped ball is 9.8 $m/s^{2}$ .

zlopas [31]3 years ago

6 0

Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

19.6 = 0 + 2a

a = 9.8 m/s²

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It goes sun moon earth the moon is blocking us from seeing the sun.

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3 years ago

Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.

aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

6 = (20 / 32.2 + 50 / 32.2)a

2.173a = 6

A = 2.76 ft/s^2

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

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3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

The height of the cylinder is $h = 0.588\ m$

The face Area is $A = 4.19 \ m^2$

The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

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3 years ago

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bogdanovich [222]

Answer:

im not really good at explaining, but i found this website url:

https://www.numerade.com/questions/a-cyclist-travels-from-point-a-to-point-b-in-10-min-during-the-first-20-min-of-her-trip-she-maintain/

same question just with the explanation

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2 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy of 0.013 J.

Aleksandr [31]

answer

v = 4.2 $\frac{m}{s}$

set up an equation

we can use the formula for kinetic energy since we know mass and kinetic energy

K = $\frac{mv^{2}}{2}$

v = $\sqrt{\frac{2K}{m} }$

values

K = 0.013 J

m = 1.5g = 0.0015 kg

plug in values

v = $\sqrt{\frac{2*0.013}{0.0015} }$

v = 4.2 $\frac{m}{s}$

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3 years ago