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lisabon 2012 [21]
3 years ago
13

A block of mass 22 kg is sliding along the ice at constant speed 5.0 m/s just ahead of it is q block of mass 29 kg sliding in th

e same direction at constant speed 4.6 m/s when the two blocks collide,the 29-kg block travels at a new speed of 7.2 m/s what is the new speed of the 22-kg block?
Physics
1 answer:
mars1129 [50]3 years ago
8 0

Answer:

Answer:

New speed of the 22-kg block is 1.57 m/s

Explanation:

Mass of block  

Mass of another block  

Initial speed of the block  

Initial speed  of the another block  

Initial speed  of the another block  

For conservation of momentum, we have

Substitute all the values and solving for final speed of the 22kg block is

new speed of the 22-kg block is 1.57 m/s

Couldnt write the answer so check picture

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An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f
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Answer:

B_2 = \frac{B_1}{2}

Explanation:

As per Ampere's law the magnetic field at the surface of the wire is given as

\int B. dL = \mu_0 i

here we have

B . (\pi d) = \mu_0 i

so we will have

B_1 = \frac{\mu_0 i}{\pi d}

now again we use same value of current but wire with double the diameter

so the magnetic field at the surface is given as

B_2 = \frac{\mu_0 i}{2\pi d}

so we have

B_2 = \frac{B_1}{2}

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What are the issues that hinders efforts to achieve sustainability?
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3 years ago
Which of the following is an oxidation-reduction reaction? ZnS(s) + 2O2(g) mc011-1.jpg ZnSO4(s) CaO(s) + H2O(l) mc011-2.jpg Ca(O
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2 years ago
Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0
2 years ago
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