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4 years ago

Two objects of different mass have equal, non-zero kinetic energies. Which object has the greater momentum?

1 answer:

4 0

Momentum= mass*velocity thus if you keep velocity constant an increase in mass increases momentum so the heavier object (assuming constant gravity) has a higher momentum than a lighter object

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Explain how the formula for the force and the force between charged particles is similar to the formula for the gravitational fo

sashaice [31]

Formula for gravitational force shows it is inversely proportional to the distance between two masses squared this is similar to the electrostatic force thay changes according to the separation squared.
The difference is that gravity is proportional to the product of the two masses. Electro static force is proportional to the product of the two charges.

7 0

3 years ago

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

4 years ago

How many moles of gas must be forced into a 4.6 l tire to give it a gauge pressure of 31.2 psi at 26 âc? the gauge pressure is r

Bess [88]

Use the Ideal Gas Law to the air in the tire :

( P ) ( V ) = ( n ) ( R ) ( T )
n = ( P ) ( V ) / ( R ) ( T )
P = P gauge + P baro = 31.2 psig + 14.8 psia = 46 psia
P = ( 46 psia ) ( 1 atm / 14.696 psia ) = 3.13 atm
n = ( P ) ( V ) / ( R ) ( T )
n = ( 3.13 atm ) ( 4.6 L ) / ( 0.08206 atm - L / mol - K ) ( 26.0 + 273.2 K )
n = 0.5864 moles
m = ( n ) ( M )
m = ( 0.5864 mol ) ( 28.96 g/ mol ) = 16.98 g

7 0

3 years ago

A piece of zinc was lying outside and had a temperature of 26 degrees C. Later that day the sun came out and the temperature inc

pickupchik [31]

Answer:

570 J

Explanation:

im sorry im not sure about the rest but i hope this helps

3 0

3 years ago

Consider the diagram for the following questions.

Ksivusya [100]

Answer:

1) Charges

2) Vectors

4 0

3 years ago

Read 2 more answers