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julia-pushkina [17]
3 years ago
11

29. Jorge is conducting an investigation into perfectly inelastic collisions using equipment where two carts collide with

Physics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

He used the probability assumption

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When the temperature of ice-cold water is increased slightly, does it undergo a net expansion or net contraction?
miss Akunina [59]

From the freezing temperature up to about 4°C (39°F) the water CONTRACTS.  That is, it becomes MORE dense.  I think I read that water is the ONLY known substance whose solid phase floats in its liquid phase.  That's why the cubes float in your soda and bergs float in the ocean.  And if weren't so, then life on Earth would not be possible !  Oceans and lakes would freeze from the bottom up, ONE TIME, and then never thaw again.

4 0
3 years ago
a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v
solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

                             S = ut + ½ gt²

Let tₓ be the time where the two stones  meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

                            x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

                S - x = U tₓ - ½ gtₓ²  (∵ the motion of the stone is in opposite direction)

Adding these two equations,

                      x + (S - x) = U tₓ

                               S = U tₓ

                               200 = 50 tₓ

∴                                  tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s   

6 0
3 years ago
A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student
Kruka [31]

Answer:

B. 6.6%

Explanation:

The percentage error of a measurement can be calculated using the formula;

Percent error = (experimental value - accepted value / accepted value) × 100

In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.

Hence, the percentage error can be calculated thus;

Percent error = (266.5-250.0/250.0) × 100

Percent error = 16.5/250 × 100

Percent error = 0.066 × 100

Percent error = 6.6%

7 0
3 years ago
A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done
NeX [460]

Answer: The correct answer is option B.

Explanation:

Mass of the sled = 10 kg

Initial speed of the sled = 2 m/s

Kinetic energy of the sled = \frac{1}{2}mv^2

\frac{1}{2}\times 10 kg\times (2 m/s)^2=20 Joules

Work done by the sled = 20 joules

The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.

Hence, correct answer is option B.

6 0
3 years ago
Read 2 more answers
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
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