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3 years ago

How long would this acceleration last

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

6 0

It would last as long as the applied force continued, or until the accelerating object hit something.

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If you toss a stick into the air, it appears to wobble all over the place. specifically, about what place does it wobble?

Otrada [13]

I would assume the end?

8 0

3 years ago

A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What vol

Gwar [14]

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

$C=\frac{n}{V}$

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

$230mL=0.23L$

$n=C*V=0.240 M*0.23L=0.055 mol$

Now, we isolate the variable V to know the new volume with the new concentration given.

$V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL$

Finally, the volume of water evaporated is the difference between initial and final volume.

$V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL$

4 0

3 years ago

Suppose you take a trip that covers 180 km and takes 3 hours to make. Your average speed is A. 30 km/h. B. 60 km/h. C. 180 km/h.

Ivenika [448]

B. 60 kilometers per hour

5 0

3 years ago

A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente

inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

$F=\frac{mv^2}{r}$

Where, $m$ is the mass of the ball, $v$ is the speed and $r$ is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

$20.4 = \frac{3.1\times v^2}{2}$

and, $v^2 = 13.16$

Thus, $v = 3.63$ m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = $\frac{\text {Path length} }{\text {Speed}}$ = $\frac{12.56}{3.63}$ = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0

3 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0

3 years ago