All categories

4 years ago

How much work is done to move 3c of charge through a potential difference of 1.5v

Physics

1 answer:

kaheart [24]4 years ago

4 0

Answer:4.5 joules

Explanation:

charge=3c

Potential difference=1.5v

Work=potential difference x charge

Work=1.5 x 3

Work=4.5

Work=4.5 joules

You might be interested in

a ferry which runs at 12m/s in still water travels between towns a and b on a river which flows south at 9 m/s. the ferry leavs

Reil [10]

Answer:

= 15 m/s

Explanation:

Considering right side(west) as positive x-axis and south as negative y-axis.

velocity of boat in still water $v_b=12\hat{i}$

velocity of stream $v_s=-9\hat{j}$

now relative velocity of boat w.r.t. stream $v_{b/s}=12\hat{i}+9\hat{j}$

this velocity with which ac distance will be covered.

therefore magnitude of $v_{b/s} =\sqrt{12^2+9^2}$

= 15 m/s

6 0

3 years ago

Please help me with thus question (picture) D:

IrinaVladis [17]

well thanks for the Information

4 0

3 years ago

Read 2 more answers

The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i

Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

$T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1$

Block 2

$T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2$

Block 3

$- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3$

Solving Eq1,2,3 simultaneously

Divide 1 and 2

$\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4$

Put Eq 4 into Eq3

$T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5$

Put Eq 5 into Eq2 and solve for a

$a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6$

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

$T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\$

5 0

4 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

Answer: The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

Explanation:

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

4 years ago

Theories have both an explanatory and a predictive function. a. True b. False

Simora [160]

Theories have both an explanatory an a predictive function. True

3 0

3 years ago

Read 2 more answers