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EastWind [94]
2 years ago
8

A 0.40 kg bead slides on a straight frictionless wire with a velocity of 3.50 cm/s to the right. The

Physics
1 answer:
tensa zangetsu [6.8K]2 years ago
5 0

Answer:

Total momentum before collision

P1 =.4 * 3.5 = 1.4       ignoring units here

Total momentum after collision

P2 = .6 * V - .4 * .7 = .6 V - .28

.6 V = 1.4 + .28   momentum before = momentum after

V = 2.8 cm/sec

In 5 sec V moves 2.8 cm/sec * 5 sec = 14 cm

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Cuál es el significado del diálogo intercultural​
bagirrra123 [75]

Answer:

El diálogo intercultural es un intercambio de opiniones abierto y respetuoso entre individuos y grupos pertenecientes a diferentes culturas que conduce a una comprensión más profunda de la percepción global del otro

Explanation:

“Intercultural dialogue is an open and respectful exchange of views between individuals and groups belonging to different cultures that leads to a deeper understanding of the other's global perception.”

5 0
3 years ago
An 800 N box is pushed across a level floor for a distance of 5.0 m with a force of 400 N. How much work was done on this box.
AveGali [126]

Answer: 2000 J.

Explanation: Since work is force*displacement, we just have to multiply the force by the distance: w = f*d = 400 N*5.0 m = 2000 J.

4 0
3 years ago
A spring has a spring constant of 53N/m. How much elastic potential energy is stored in the spring in the spring when it is comp
IRISSAK [1]

Answer:11686.5 joules

Explanation:

elastic constant(k)=53N/m

extension(e)=21m

Elastic potential energy=(k x e^2)/2

Elastic potential energy=(53 x (21)^2)/2

Elastic potential energy=(53x21x21)/2

Elastic potential energy=23373/2

Elastic potential energy=11686.5

Elastic potential energy is 11686.5 joules

5 0
3 years ago
A cannonball with a mass of 1.0 kilogram is fired horizontally from a 500.-kilogram cannon, initially at rest, on a horizontal,
vfiekz [6]
The impulse J is equal to the magnitude of the force applied to the cannonball times the time it is applied:
J=F \Delta t
But the impulse is also equal to the change in momentum of the cannonball:
J=\Delta p
If we put the two equations together, we find
F \Delta t= \Delta p
And since we know the magnitude of the average force and the time, we can calculate the change in momentum:
\Delta p= F \Delta t=(8.0 \cdot 10^3 N)(1.0 \cdot 10^{-1} s)=800 kg m/s
7 0
3 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C   → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C   → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C   → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first  charge , 2nd to 5 th charges can be  written as 2e,  3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of  1.6 x 10⁻¹⁹ C  exists.

3 0
3 years ago
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