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2 years ago

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A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

illates up and down in simple harmonic motion. Calculate the period of motion.

1 answer:

marin [14]2 years ago

4 0

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

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2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the

Lubov Fominskaja [6]

Answer:

$P=39.2205\, watt$

$E=374.948 \,cal$

Explanation:

Given that:

force applied, $F=66\,N$

displacement, $s=23.77\,m$ (length of a tennis court)

time taken for pushing, t = 40 s

Since, work is given by:

$W=F.s$

$W=66\times 23.77$

$W=1568.82\,J$

Now, power is given as:

$P=\frac{W}{t}$

$P=\frac{1568.82}{40}$

$P=39.2205 \,watt$

Calories consumed is:

$E= 1568.82\times 0.239$

$E=374.948\, cal$

3 0

3 years ago

A rural mail carrier logged the mileage on the odometer at the start of the mail route as 31,500 miles. After 6.5 hours, the car

dusya [7]

Answer: $32.77\ mi/hr$

Explanation:

Given

Initially Reading on the odometer is $31,500\ miles$

Final reading on the odometer is $31,713\ miles$

Time taken is $t=6.5\ hr$

average velocity $=\dfrac{\text{Displacement}}{\text{time}}$

$v_{avg}=\dfrac{31713-31500}{6.5}$

$v_{avg}=\dfrac{213}{6.5}$

$v_{avg}=32.77\ mi/hr$

Thus the average velocity of mail truck is $32.77\ mi/hr$

8 0

2 years ago

Three cannons are located at the top of a cliff above a level plain. Cannon A is aimed at an angle of 25° above the horizontal a

vodka [1.7K]

Answer:

The canon B hits the ground fast.

Explanation:

Given that,

Speed of cannon A = 85 m/s

Speed of cannon B= 100 m/s

Speed of cannon C = 75 m/s

We need to calculate the cannonballs will hit the ground with the greatest speed

Using conservation of energy

The final kinetic energy of canon depends on initial kinetic energy and potential energy.

The final velocity depends upon initial velocity and initial height.

So, the initial velocity of canon B is high.

Hence, The canon B hits the ground fast.

3 0

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>

6 0

2 years ago

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Alex777 [14]

The resultant speed of the plane is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Hence, we have that the resultant speed of the plane is (3) 226 m/s

Have a nice day!

5 0

2 years ago