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2 years ago

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Carefully consider how the accelerations a1 and a2 are related. Solve for the magnitude of the acceleration, a1, of the block of

mass m1, in meters per square second.

1 answer:

eduard2 years ago

7 0

Answer:

a1 = 3.68m/s²

Explanation:

Given values:

Mass of the block placed on the table, m1 = 12.25 kg

Mass of the block hanging vertically, m2 = 7.5 kg

Acceleration due to gravity, g = 9.8 m/s2

Tension in the string is T

Let the acceleration of mass 1 and mass 2 be a1 and a2

a1 and a2 are equal in magnitude but different in direction. This because the string does not stretch. Hence the two bodies must move equal distances in equal times, and so their speechless at any instant must be equal. When the speeds change , they change by equal amounts in a given time, so the acceleration of the two bodies must have the same magnitude a,

a = m2*g/(m1 + m2)

a = 7.5 x 9.8 / (12.5 + 7.5)

a = 3.68 m/s²

a1 = a2 = 3.68m/s²

a1 is directed to the right and a2 is directed downwards

Below is a diamonds to show the geometrical arrangements of both masses

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Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$

Explanation:

From the question we are told that

The Coulomb constant is $k = 9.0 *10^{9} \ N\cdot m^2 /C^2$

The charge on the electron/proton is $e = 1.6*10^{-19} \ C$

The mass of proton $m_{proton} = 1.67*10^{-27} \ kg$

The mass of electron is $m_{electron } = 9.11 *10^{-31} \ kg$

Generally for the electron to be held up by the force gravity

Then

Electric force on the electron = The gravitational Force

i.e

$m_{electron} * g = \frac{ k * e^2 }{r^2 }$

$\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81$

$r = \sqrt{25.78}$

$r = 5.077 \ m$

7 0

2 years ago

Mt. Asama, Japan, is an active volcano complex. In 2009, an eruption threw solid volcanic rocks that landed far from the crater.

solong [7]

Answer:u=97.41m/s

Explanation:

Given

inclination $\theta =58.7^{\circ}C$

Horizontal distance travel by Particle $d=1200 m$

Vertical height $h=780 m$

Let u be the initial velocity

calculating vertical distance

$y=u\sin \theta +\frac{at^2}{2}$

$y=u\sin \theta t-\frac{gt^2}{2}$ -------1

Calculating horizontal distance

$x=u\cos \theta \times t+0$

$t=\frac{x}{u\cos \theta }$

put value of t in equation 1

$y=u\sin \theta \times \frac{x}{u\cos \theta }-\frac{g}{2}\times (\frac{x}{u\cos \theta })^2$

$y=x\tan \theta -\frac{gx^2}{2u^2\cos ^2\theta }$

$\frac{gx^2}{2u^2\cos ^2\theta }=x\tan \theta -y$

$u^2=\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}$

$u=\sqrt{\frac{gx^2}{2cos^2\theta (x\tan \theta -y)}}$

at $y=-780\ m\ x=1200 m$

$u^2=\frac{18.154}{2753.65}\times 1200^2$

$u=97.41 m/s$

6 0

2 years ago

When Jane drives to work, she always places her purse on the passenger’s seat. By the time she gets to work, her purse has falle

LUCKY_DIMON [66]

At some time during her drive she backed up with a substantial negative. ( backwards) acceleration. Since the pocket book is not physically connected to the seat it is free to move. Upon rapid negative acceleration the pocket book remains in its position while the car accelerates backwards away from it. this demonstrates Newtons 1st law of motion. The first law is the law of inertia. Which states, an object at rest. ( pocketbook) will remain at rest and an object in motion will continue in motion at constant velocity, unless acted upon by some outside force to change its motion.

4 0

2 years ago

Read 2 more answers

A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work

SpyIntel [72]

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string $T=5 N$

$2\pi r=2$

$r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m$

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is $90^{\circ}$

$W=\int\vec{F}\cdot \vec{r}$

$W=\int Fdr\cos 90$

$W=0$

4 0

2 years ago

The brake in most cars makes use of a hydraulic system. This system consists of a fluid filled tube connected at each end to a p

Maksim231197 [3]

Answer:

The force at the brake pad = 250 N

Explanation:

The hydraulic brake system works on the Pascal's Principle for pressure transmission in fluids; the pressure applied to a fluid is transmitted undiminished in all directions.

For hydraulic systems, the pressure applied to the brake pedal is transmitted undiminished through the fluid filled tube, connected at each end to a piston, to the brake pad.

Hence, mathematically,

P(brake pedal) = P(break pad)

Pressure is given as the force applied divided by the cross sectional Area perpendicular to the direction of applied force.

P(brake pedal) = (Force applied on the brake pedal) ÷ (Cross Sectional Area of the brake pedal)

Force applied on the brake pedal = 50 N

Cross Sectional Area of the brake pedal = 3 cm²

P(brake pedal) = (50/3) = 16.67 N/cm²

P(brake pad) = P(brake pedal) = 16.67 N/cm²

P(brake pad) = (Force applied on the brake pad) ÷ (Cross Sectional Area of the brake pad)

Force applied on the brake pad = F = ?

Cross Sectional Area of the brake pad = 15 cm²

16.67 = (F/15)

F = 16.67 × 15 = 250 N

Hence, the force at the brake pad = 250 N

Hope this Helps!!!

7 0

2 years ago