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brilliants [131]

3 years ago

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Can somebody please help me.

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stira [4]3 years ago

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Answer:

what is the image in question

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Best Answer will receive BRAINLIEST One consequence of Newton's third law of motion is that __________. A. every object that has

In-s [12.5K]

One consequence of Newton's third law of motion is that all actions have equal and opposite reactions. <em>(C)</em>

In fact, that's pretty much what the law itself says in so many words.

7 0

3 years ago

Read 2 more answers

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

Simple Circuit and Ohm's Law Check-for-Understanding

Vanyuwa [196]

Answer:

current, only

Explanation:

current:I

voltage:U

resistance:R

formula: I=U/R

Increasing the battery cause the increasing in the voltage. Resistance does not normally change. And the current would increase.

4 0

3 years ago

Read 2 more answers

An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y

Xelga [282]

Answer:

The fraction of the protons would have no electrons $=1.88\times 10^{-10}$

Explanation:

We are given that

Amoeba has total number of protons= $1.00\times 10^{16}$

Net charge, Q=0.300pC

Electrons are fewer than protons= $1.88\times 10^6$

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

= $\frac{Fewer\;electrons}{Total\;protons}$

The fraction of the protons would have no electrons

= $\frac{1.88\times 10^{6}}{1.00\times 10^{16}}$

$=1.88\times 10^{-10}$

Hence, the fraction of the protons would have no electrons $=1.88\times 10^{-10}$

6 0

3 years ago

A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?

musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where we have:

$p_1 = 1.01\cdot 10^5 Pa$ (initial pressure is stp pressure)

$V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3$ is the initial volume

$T_1 = 0^{\circ}=273 K$ is the initial temperature (stp temperature)

$p_2 = 368 torr = 4.9\cdot 10^4 Pa$ is the final pressure

$V_2 = ?$ is the final volume

$T=155^{\circ}=428 K$ is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3$

which corresponds to 2.41 L.

7 0

3 years ago