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3 years ago

6

Will brainlist 20 points

2 answers:

Len [333]3 years ago

3 0

Your answer for this question is the third option.

Strike441 [17]3 years ago

3 0

The 3rd one I think try that

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1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

3 years ago

Sergeu [11.5K]

Answer:

The answer is: To accelerate an object <u>the force applied to the object</u> has to increase.

Explanation:

the acceleration of an object <u>increases with increased force</u> and <u>decreases with increased mass.</u>

3 0

3 years ago

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

8090 [49]

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

8 0

4 years ago

Any help????????????

pashok25 [27]

The answer should be D

3 0

3 years ago

Read 2 more answers

A ladybug sits 10.8 cm from the center of a beatles music album spinning at 33.33 rpm. What is the maximum velocity (in m/s) of

snow_lady [41]

Answer:

The maximum velocity is 0.377 m/s

Explanation:

Please, the solution is in the Word file attached

7 0

3 years ago

Read 2 more answers