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a) Điện trở của dây là:
\(R=\dfrac{U}{I}=\dfrac{50}{1,2}=\dfrac{125}{3}\left(\Omega\right)\)
b)
Ta có: \(\rho_1=\rho_2 \)
\(\Leftrightarrow\dfrac{R_1S}{l_1}=\dfrac{R_2S}{l_2}\\ \Leftrightarrow R_1Sl_2=R_2Sl_1\\ \Leftrightarrow\dfrac{125}{3}.S.300=R_2.S.200\\ \Leftrightarrow12500=200R_2\Leftrightarrow R_2=62,5\left(\Omega\right)\)
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11)11) 3x(x-5)2-(x+2)3+2(x-1)3-(2x+1)(4x2-2x+1)=3x(x2-10x+25)-(x3+6x2+12x+8)+2(x3-3x2+3x-1)-(8x3+1)=3x3-30x2+75x-x3-6x2-12x-8+2x3-6x2+6x-2-8x3-1=-4x3-42x2+63x-11
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Bài 1
Tóm tắt
\(I=3A\\ U=30V\\ U'=20\)
_________
\(a.R=?\Omega\\ b.I'=?A\)
Giải
\(a.R=\dfrac{U}{I}=\dfrac{30}{3}=10\Omega\\ b.I'=\dfrac{U'}{R}=\dfrac{20}{10}=2A\)
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Câu 3
Tóm tắt:
\(m_{MgO}=8g\\ m_{ddH_2SO_4}=200g\)
________
\(a)C_{\%H_2SO_4}=?\%\\
b)m_{MgSO_4}=?g\)
Giải
\(a)n_{MgO}=\dfrac{8}{40}=0,2mol\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ n_{MgO}=n_{H_2SO_4}=n_{MgSO_4}=0,2mol\\ C_{\%H_2SO_4}=\dfrac{0,2.98}{200}\cdot100=9,8\%\\ b)m_{MgSO_4}=0,2.120=24g\)
Câu 4
Tóm tắt:
\(V_{SO_2}=3,36l\\ V_{Ba\left(OH\right)_2}=100ml\)
________
\(a)C_M\) \(_{Ba\left(OH\right)_2}=?M\)
\(b)m_{BaSO_3}=?g\)
Giải
\(a)n_{SO_2}=\dfrac{3,36}{22,4}=0,15mol\\ SO_2+Ba\left(OH\right)_2\rightarrow BaSO_3+H_2O\)
0,15 0,15 0,15 0,15
\(C_M\) \(_{Ba\left(OH\right)_2}=\dfrac{0,15}{0,1}=1,5M\)
\(b)m_{BaSO_3}=0,15.217=32,55g\)
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8 How much do these apples cost?
9 THis is a blue car
11 Are there 40 classrooms in Phong's school?
13 How wide if the Great wall
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1 1/4 tấn=5/4 tấn=0,25 tấn=250kg
2 2/5 tạ=12/5 tạ=2,4 tạ=240 kg
Tổng của 1 1/4 tấn+2 2/5 tạ là:
250+240=490(kg)
Đáp số:490 kg
Chúc bạn học giỏi!!!