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kaheart [24]
2 years ago
8

7. Four perpendicular forces, 2 N, 3 N, 9 N and 10 N act on object P as shown in Diagram 2. 10 N A B 9 N 3 N 2 N P Diagram 2 Wha

t is the magnitude of the resultant force? 10 N C 18 N 14 N D 20 N​

Physics
1 answer:
Anvisha [2.4K]2 years ago
6 0

Answer: R=

A

2

+B

2

+2ABcosθ

​

θ=90

o

cos90

o

=0

R=

A

2

+B

2

​

=

3N

2

+4N

2

​

=5N

Explanation:

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With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
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Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

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