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3 years ago

You throw a rock straight up into the air with a speed of 14.2 m/s. how long does it take the rock to reach its highest point?

1 answer:

4 0

The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ? I don't hear any objection, so I'll assume that it is.

That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed. (The same thing.)

If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)

At that point, since it has no more upward speed, it can't go any higher. Right ?

(crickets . . .)

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Two forces act on a 1250 kg sailboat as it moves through the water with an initial velocity of 11 m/s. The forward force of the

Naddik [55]

Answer:

The velocity of the boat 15 seconds later is 5.6 meters per second.

Explanation:

We assume that sailboat can be modelled as particle, so that we use solely translations equations. It is noticed that the sailboat is move by action of the wind and drag force of the water is opposed to such force, but the last force has a greater magnitude than the first one, meaning that net force is less than zero.

From Newton's Laws we have the following equation of equilibrium for the sailboat:

$\Sigma F = F - f = m\cdot a$ (Eq. 1)

Where:

$F$ - Force from the wind exerted on the sailboat, measured in newtons.

$f$ - Drag force of the water, measured in newtons.

$m$ - Mass of the sailboat, measured in kilograms.

$a$ - Net acceleration of the sailboat, measured in meters per square second.

If we know that $F = 3.90\times 10^{3}\,N$ , $f = 4.35\times 10^{3}\,N$ and $m = 1250\,kg$ , then the net acceleration of the sailboat is:

$a = \frac{F-f}{m}$

$a = \frac{3.90\times 10^{3}\,N-4.35\times 10^{3}\,N}{1250\,kg}$

$a = -\frac{9}{25}\,\frac{m}{s^{2}}$

If sailboat decelerates uniformly, then we can get the final velocity of the boat by using this equation of motion:

$v = v_{o}+a\cdot t$ (Eq. 2)

Where:

$v_{o}$ , $v$ - Initial and final velocities of the sailboat, measured in meters per second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 11\,\frac{m}{s}$ , $a = -\frac{9}{25}\,\frac{m}{s^{2}}$ and $t = 15\,s$ , then the final velocity of the sailboat is:

$v = 11\,\frac{m}{s} +\left(-\frac{9}{25}\,\frac{m}{s^{2}} \right)\cdot (15\,s)$

$v = 5.6\,\frac{m}{s}$

The velocity of the boat 15 seconds later is 5.6 meters per second.

3 0

3 years ago

Which mass is undergoing to the greatest amount of acceleration ??

lina2011 [118]

Answer:

Option (3)

Explanation:

Formula used to calculate acceleration is,

F = ma

Where F = force exerted on a mass

m = mass

a = acceleration due to force exerted on the mass

Option (1),

When F = 100 N and m = 100 kg

100 = 100a

a = 1 m per sec²

Option (2)

For F = 1 N and m = 100 kg

1 = 100a

a = $\frac{1}{100}$

a = 0.01 m per sec²

Option (3)

For F = 100 N and m = 1 kg

100 = 1(a)

a = 100 m per sec²

Option (4)

For F = 1 N and m = 1 kg

1 = 1(a)

a = 1 m per sec²

Therefore. acceleration in Option (3) is the maximum.

4 0

3 years ago

A man is using a fishing rod to catch fish in figure 1.

ryzh [129]

Answer:

Explanation:

Remark

In general, these 3rd class levers are very inefficient. Because the force distance is smaller than the load distance, you need to pull upward with more force that the weight of the load. So whatever the load is, the force is going to be much greater.

The distances are always measured to the pivot unless you are asked something specific otherwise.

Givens

F = ?

weight = 6N

Force Distance = F*d = 0.5 m

Weight Distance =W*d1 = 2 m

Formula

F*Fd = W*Wd

Solution

F*0.5 = 6 * 2 Divide by 0.5

F = 12/0.5

F = 24 N upwards

5 0

2 years ago

A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup

ohaa [14]

Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) = 7.5 mol

As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)

When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
$V= \frac{(7.5 \, mol)*(8.314 \, J/(mol-K))*(300 \, K)}{(92205.75 \, Pa)} =0.2029 \, m^{3}$

V = (0.2029 m³)*(10³ L/m³) = 202.9 L

Answer: 203 liters

3 0

3 years ago

What is the resistance of resistor R3?

Vsevolod [243]

Answer: 2.0

Explanation: I KNOW !

5 0

3 years ago