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3 years ago

Your bike is too big if your can't touch the ground A.Knees B.Hands C.Feet

Physics

1 answer:

SOVA2 [1]3 years ago

6 0

Answer: C: Feet

Explanation: Why...

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The answer is A) First law of motion.

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Which of the following is a vector quantity?

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A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m

zzz [600]

Answer:

$310.38\times 10^{-4}j$

Explanation:

We have given mass m=0.155 kg

Force constant K = 305 N/m

Distance $X=1.25\times 10^{-2}m$

Velocity $v=.305 m/sec$

The total energy at any position of the motion is give by $E=\frac{1}{2}mv^2+KX^2$ here $\frac{1}{2}mv^2$ is energy due to motion and $KX^2$ is energy due to spring elongation

So total energy $E=\frac{1}{2}\times 0.155\times 0.305^2+305\times 0.0125^2=310.38\times 10^{-4}j$

4 0

3 years ago

Shalnov [3]

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

$36=\frac{1}{2}m(12)^2$ and isolating the m:

$m=\frac{2(36)}{12^2}$ which gives us

m = .50 kg

3 0

3 years ago

0.3-L glass of water at 20C is to be cooled with ice to 5C. Determine how much ice needs to be added to the water, in grams, i

abruzzese [7]

Answer:

$a. m_i_c_e=54.6g\\b. m_i_c_e=48.7g\\m_c_o_l_d_w_a_t_e_r=900g$

Explanation:

First we need to state our assumptions:

Thermal properties of ice and water are constant, heat transfer to the glass is negligible, Heat of ice $h_i_f=333.7KJkg$

Mass of water, $m_w=\rho V =1\times0.3=0.3Kg$ .

Energy balance for the ice-water system is defined as

$E_i_n-E_o_u_t=\bigtriangleup E_s_y_s\\0=\bigtriangleup U=\bigtriangleup U_i_c_e+\bigtriangleup U_w$

a.The mass of ice at $0\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[0+333.7+418\times(5-0)]+0.3\times4.18\times(5-20)=0\\m_i_c_e=0.0546Kg=54.6g$

b.Mass of ice at $20\textdegree C$ is defined as:

$[mc(0-T_1)+mh_i_f+mc(T_2-0)]_i_c_e+[mc(T_2-T_1)]_w=0\\\\m_i_c_e[2.11\times(0-(20))+333.7+4.18\times(5-0)]+0.3\times4.18\times(5-20)=0\\\\m_i_c_e=0.0487Kg=48.7g$

c.Mass of cooled water at $T_c_w=0\textdegree C$

$\bigtriangleup U_c_w+\bigtriangleup U_w=0$

$[mc(T_2-T_1)]_c_w+[mc(T_2-T_1)]_w=0\\m_c_w\times4.18\times(5-0)+0.3\times4.18\times(5-20)\\m_c_w=0.9kg=900g$

8 0

3 years ago

Your bike is too big if your can't touch the ground A.Knees B.Hands C.Feet​

Your bike is too big if your can't touch the ground A.Knees B.Hands C.Feet