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koban [17]
1 year ago
12

A large crate is at rest on a ramp at a loading dock

Physics
1 answer:
VARVARA [1.3K]1 year ago
4 0

Answer:

yuhhh

Explanation:

You might be interested in
If c1=c2=4.00μf and c4=8.00μf, what must the capacitance c3 be if the network is to store 2.70×10−3 j of electrical energy?
Akimi4 [234]
Missing detail in the text: total voltage of the circuit \Delta V = 46.0 V
Missing figure: https://www.physicsforums.com/attachments/prob-24-68-jpg.190851/

Solution:

1) The energy stored in a circuit of capacitors is given by
U= \frac{1}{2} C_{eq} (\Delta V)^2
where C_{eq} is the equivalent capacitance of the circuit. We can find the value for C_{eq} by using \Delta V=46.0 V and the energy of the system, U=2.7\cdot 10^{-3} J
C_{eq}= \frac{2U}{(\Delta V)^2}=2.55\cdot 10^{-6} F=2.55\mu F

2) Then, let's calculate the equivalente capacitance of C1 and C2. The two capacitors are in series, so their equivalente capacitance is given by
\frac{1}{C_{12}}= \frac{1}{C_1}+ \frac{1}{C_2}= \frac{1}{4 \mu F} + \frac{1}{4 \mu F}
from which we find C_{12}=2 \mu F

3) Then let's find C_{123}, the equivalent capacitance of C_{12} and C3. C_{123} is in series with C4, therefore we can write
\frac{1}{C_{eq}}= \frac{1}{C_{123}}+ \frac{1}{C_4}
Since we already know C_4=8 \mu F and C_{eq}=2.55 \mu F, we find
C_{123}=3.70 \mu F

4) Finally, we can find C_{3}, because it is in parallel with C_{12}, and the equivalent capacitance of the two must be equal to C_{123}:
C_{123}=C_{12}+C_3
So, using C_{123}=3.70 \mu F and C_{12}=2 \mu F, we find
C_3=1.70 \mu F

7 0
2 years ago
Jocko the clown, whose mass is 60-Kg, stands on a skateboard. A 20-Kg ball is thrown at Jocko at 3m/s, and when he catches the b
Mekhanik [1.2K]

Answer:

The speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

Explanation:

Given that,

Mass if Jocko, m = 60 kg

Mass of the ball, m' = 20 kg

Speed of the ball, v = 3 m/s

Let V is the speed of Jocko and the ball move after he catches the ball. The momentum of the system remains conserved. Using the conservation of momentum as :

m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{20\times 3}{(60+20)}\\\\V=0.75\ m/s

So, the speed of the Jocko and the ball move after he catches the ball is 0.75 m/s.

7 0
2 years ago
If you are swimming upstream (i.e., against the current), at what speed does your friend on the shore see you moving?
Alex777 [14]
Could be very slow since they’re basically going against the current which is hard so will be going slow
7 0
2 years ago
The electric potential in a region that is within 2.00 mm of the origin of a rectangular coordinate system is given by V=Axl+Bym
Fynjy0 [20]

Answer:

Given the potential, V = Ax^l+By^m+Cz^n+D

The components of the electric field are:

E_x = \frac{-dV}{dx} = -Alx^l^-^1

E_y = \frac{-dV}{dy} = - Bmy^m^-^1

E_z = \frac{-dV}{dz} = - nCzn^n^-^1

Let's calculate the potential difference for all given points.

V(0, 0, 0) = 10V => Ax^l+By^m+Cz^n+D = 10

=> D = 10

V(1, 0, 0) = 4V => A + 10 = 4

Solving for A, we have:

A = 4 - 10

A = -6

V(0, 1, 0) = 6V => B + 10 = 6

Solving for B, we have:

B = 6 - 10

B = -4

V(0, 0, 1) = 8V => C + 10 = 4

Solving for C, we have:

C = 8 - 10

C = -2

For all given points, let's calculate the magnitude of electric field as follow:

E_x (1, 0, 0) = 16 => - Alx^l^-^1 = 16

Al = -16

Solving for l, we have:

l = \frac{-16}{A}

From above, A = -6

l = \frac{-16}{-6}

l = \frac{8}{3}

E_y (0, 1, 0) = 16=> Bmy^m^-^1 = 16

Bm = -16

Solving for m, we have:

m = \frac{-16}{A}

From above, B = -4

m = \frac{-16}{-4}

m = 4

E_y (0, 0, 1) = 16=> nCz^n^-^1 = 16

nC = - 16

Solving for n, we have:

n = \frac{-16}{C}

From above, C = -2

n = \frac{-16}{-2}

n = 8

4 0
1 year ago
If a metal wire is 4m long and a force of 5000n causes it to stretch by 1mm, what is the strain?
barxatty [35]

Answer:

2.5\cdot 10^{-4}

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

S=\frac{\Delta L}{L_0}

where

\Delta L is the change in length of the object

L_0 is the original length of the object

In this problem, we have L_0 = 4 m and \Delta L=1 mm=0.001 m, therefore the strain is

S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}


5 0
2 years ago
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