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3 years ago

Physics When attempting to increase the distance traveled by a projectile, what must occur

2 answers:

4 0

To increase the distance of a projectile you must increase the power behind the projectile. However you must also change the angle if you are aiming for a specific point. By increasing the power the momentum transfers to the object and the projectile gains more velocity and more distance. Also remember that energy transfers. For more distance you must have more velocity.

ryzh [129]3 years ago

3 0

Energy or power has to be increased to increase the distance travelled by the projectile.

<u>Explanation: </u>

Projectile motion is defined as when an object is thrown or projected at a particular angle and it reaches the ground after sometime due to acceleration and gravity.

For example, throwing a ball in air. To increase the distance travelled by the object due to projectile motion, one must increase the power at which the object to set.

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You live on a planet far from ours. "Based on extensive communication with a physicist on earth", you have determined that all l

Ugo [173]

Answer:

8.56 m/s2

Explanation:

Using law of energy conservation while taking into account of the rotational and translation kinetic energy, when the solid cylinder rolls down the incline we have the potential energy converted to kinetic energy:

$E_p = E_{kv} + E_{k\omega}$

$mgh = mv^2/2 + I\omega^2/2$

where m is the mass, $I = mr^2/2$ is the moments of inertia of the solid cylinder $\omega = v / r$ is the angular speed of the cylinder

$mgh = mv^2/2 + \frac{mr^2}{2}\frac{(v/r)^2}{2}$

$mgh = mv^2/2 + mv^2/4 = 3mv^2/4$

$h = 3v^2/(4g)$

So if you plot a liner chart of h vs $v^2$ and get a slope of 6.42 then that means 3/(4g) = 6.42 so $g = 6.42*4/3 = 8.56 m/s^2$

The gravitational acceleration on this planet is 8.56 m/s2

3 0

4 years ago

An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following pre

Montano1993 [528]

Answer:

A) and B) are correct.

Explanation:

If the object is at rest, it means that no net force is exerted on it.

As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.

This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).

At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.

So, the statements A) and B) are true.

6 0

3 years ago

Read 2 more answers

Which tool would be most useful to identify areas with frequent earthquakes to predict future earthquakes?

enyata [817]

It is a seismograph that would be most useful.

6 0

3 years ago

A bucket weighing 5 lbs is lifted at a constant rate from the bottom of a 100 ft well by a rope which weighs 5 lbs. The bucket h

8_murik_8 [283]

Answer:

$W_{bucket} = 24934.85\,lbf\cdot ft$

Explanation:

The system is modelled after the Principle of Energy Conservation and the Work-Energy Theorem:

$W_{bucket} = U_{g,B, bucket} - U_{g,A,bucket} +U_{g, B, water}-U_{g,A,water} +U_{g, B, rope} -U_{g,A,rope}$

$W_{bucket} = (5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-0\,ft) + (25\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft) - (30\,lb)\cdot (32.174\,\frac{ft}{s^{2}})\cdot (0\,ft)+(5\,lb) (32.174\,\frac{ft}{s^{2}})\cdot (100\,ft-50\,ft)$

$W_{bucket} = 24934.85\,lbf\cdot ft$

6 0

4 years ago

Describe using examples how objects can be at rest and in motion simultaneously

elena-14-01-66 [18.8K]

An object can be at rest and still be in motion because the earth is always in motion.

5 0

3 years ago