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alexandr1967 [171]

2 years ago

12

Which one ? need to know asap

1 answer:

Snezhnost [94]2 years ago

8 0

Answer:

D or A ( I think)

Explanation:

Cause you know how people say the smaller the chips the more the chips its kind of the same with amplitude.

( hope it helps if I am wrong I am sorry)

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Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

A six pack of your favorite beverage (12 fl. oz. cans, assumed to have the same characteristics as water) is placed in a cooler

nasty-shy [4]

Answer:

$Q = 169 BTU$

Explanation:

As we know that volume is given as

$V = 12 Fl oz$

so it is given in liter as

$V = 12 fl oz = 0.355 Ltr$

now we have six pack of such volume

so total volume is given as

$V = 6\times 0.355 Ltr$

$V = 2.13 ltr$

so its mass is given as

$m = 2.13 kg$

now the change in temperature is given as

$\Delta T = 70 - 34 = 36 ^oF$

$\Delta T = 20^oC$

now the heat given to the liquid is given as

$Q = ms\Delta T$

$Q = 2.13(4186)(20)$

$Q = 1.78 \times 10^5 J$

$Q = 169 BTU$

4 0

2 years ago

When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When

Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

or

$L=\dfrac{\epsilon}{(di/dt)}$ ........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$ .........(2)

From equation (1) and (2) we get :

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0

2 years ago

Please please help I'm stuck!!!

Korvikt [17]

Answer: iDc im in 3grade

Explanation:sorry

6 0

2 years ago

Read 2 more answers

a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a

Whitepunk [10]

Answer:

$W = -2.76\times 10^{-9}~J$

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

$U = \frac{1}{2}CV^2$

where C can be calculated using the plate separation and area.

$C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon$

Therefore, the potential energy in the first case is

$U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon$

In the second case:

$C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon$

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

$W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J$

6 0

2 years ago