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3 years ago

5. Layer of Earth consisting of crust & upper layer of mantle ________

Physics

2 answers:

elixir [45]3 years ago

6 0

Answer:

lithosphere

Explanation:

hope this helps you!!

dedylja [7]3 years ago

6 0

Answer:

the answer is lithosphere

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A sailboat weighing 980 lb with its occupants is running downwind at 8 mi/h when its spinnaker is raised to increase its speed.

Effectus [21]

Answer:

78.498N

Explanation:

The Net force provided by the spinnaker can be obtained from Newton's second law of motion as follows;

$F=\frac{m(v-u)}{t}................(1)$

where m is the mass, v is the final velocity, u is the initial velocity and t is the time interval for which the force acted.

Given;

m =980lb

v = 12mi/h

u =8mi/hr

t = 10s.

It is important to convert all quantities to their SI units where necessary, so we do that as follows;

1lb = 0.45kg,

hence 980lb = 980 x 0.45kg = 441kg.

1mile = 1609.34m

1hour = 3600s,

therefore;

$8mi/h=\frac{8*1609.34m}{3600s}=3.58m/s$

$12mi/h=\frac{12*1609.34m}{3600s}=5.36m/s$

Substituting all values into equation (1), we obtain the following;

$F=\frac{441(5.36-3.58)}{10}\\F=\frac{441*1.78}{10}\\F=\frac{784.98}{10}\\F=78.498N$

4 0

3 years ago

If a diode at 300°K with a constant bias current of 100μA has a forward voltage of 700mV across it, what will the voltage drop a

MAVERICK [17]

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current $I_1$ = 100 μA

current $I_2$ = 1 mA

forward voltage $V_r$ = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.

Using the formula:

$I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

$I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}$

where;

$V_r$ = 0.7

$I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}$

$I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}$

$\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$

$\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }$