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Shtirlitz [24]
3 years ago
10

I'd like some help in these two questions please, thankyou so much. have a great day! stay safe and stay happy. (there are two p

arts of one question so uh yea.

Physics
1 answer:
Andrews [41]3 years ago
6 0

Answer:

The resultant force is 25 N in the negative x-direction.

Explanation:

We need to find the total force in the x and y-direction and then the resultant force will be the magnitude of the sum of Fx and Fy.

<u>Total force in the x-direction:</u>

F_{x}=12-37=-25\: N

<u>Total force in the y-direction:</u>

F_{y}=10-10=0\: N

Therefore, the magnitude of the resultant force will be:

F_{R}=\sqrt{F_{x}^{2}+F_{y}^{2}}

F_{R}=\sqrt{(-25)^{2}+0}

|F_{R}|=25\: N

The resultant force is 25 N in the negative x-direction, in other words, the car will move to the left.

I hope it helps you!

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Use Newton’s second law: F=ma
m= 250kg. a= 750ms-2
F= 250 x 750 = 187 500N
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g A ray of light is incident on a flat reflecting surface and is reflected. If the incident ray makes an angle of 28.7° with the
Alex_Xolod [135]

Answer:

\theta_2 = 15.8degree

Explanation:

given data:

\theta_1 = 28.7 degree

accoding to the snell's law

n_1*sin\theta_1=n_2*sin\theta_2

where n1 is refracting index of air = 1

n2 is refracting index of glass = 1.55

putting all value to get angle mad by incident ray with normal

\theta_2 = \frac{1*sin28.7^{o}}{1.55}

\theta_2 = 15.8degree

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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
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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

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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

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Is equal to 2486 milliliters
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