The bonds of the products store 22kJ more energy than the bonds of the reactants. How is energy conserved during this reaction?

mr_godi [17]

<span> Learning new things produces physical changes in your brain structure</span>

5 0

3 years ago

light incident on a polarizer is then passed through a second polarizer. if the polarizer and the analyzer are perpendicular to

Lerok [7]

<span>Indeed, this is one of the odd results in physics. A system of two polarizing filters arranged as shown below trasmits no light.</span>

8 0

3 years ago

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A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago

Europa, a satellite of Jupiter, is believed to have a liquid ocean of water (with a possibility of life) beneath its icy surface

Goryan [66]

Answer:

so maximum velocity for walk on the surface of europa is 0.950999 m/s

Explanation:

Given data

legs of length r = 0.68 m

diameter = 3100 km

mass = 4.8×10^22 kg

to find out

maximum velocity for walk on the surface of europa

solution

first we calculate radius that is

radius = d/2 = 3100 /2 = 1550 km

radius = 1550 × 10³ m

so we calculate no maximum velocity that is

max velocity = √(gr) ...............1

here r is length of leg

we know g = GM/r² from universal gravitational law

so G we know 6.67 × $10^{-11}$ N-m²/kg²

g = 6.67 × $10^{-11}$ ( 4.8×10^22 ) / ( 1550 × 10³ )

g = 1.33 m/s²

now

we put all value in equation 1

max velocity = √(1.33 × 0.68)

max velocity = 0.950999 m/s

so maximum velocity for walk on the surface of europa is 0.950999 m/s

3 0

3 years ago

A wind turbine is an example of what kind of device

Liula [17]

Answer:

it’s an example of a generator.

Explanation:

3 0

2 years ago