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insens350 [35]
1 year ago
7

A block is suspended from a scale and then lowered into a bucket of water. The density of the water is 1 gm/cm3. The initial rea

ding on the scale is 19N and one complete revolution of the scale is a change of 10N. 1) When the block is lowered into the water, the mass of the block:
Physics
1 answer:
miv72 [106K]1 year ago
6 0

Answer:

1.94 kg

Explanation:

we have given density of water = 1 gm/cm^3

the initial reading of the scale = 19 N

change in reading of scale =10 N

we have to find the mass of the block when the block is lowered in water

mass of the block when it is lowered in the water is given by

w_b=\frac{f}{g}

here g is the gravity of acceleration which value is given by 9.8 m/sec^2

so the mass of the block =\frac{19}{9.8}=1.94\ kg

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The question with the complete options:

Look at these two sentences about Undeposited Funds. 1. By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy ___________. 2. When receiving a payment, make sure _________________. Which of the options below correctly fills in the blanks? A.)1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds

B) 1. To match your bank register with your bank statement; 2. the Deposit to account is Checking

C)1. To match your expenses with your bank statement; 2. the Deposit to account is Uncategorized asset

D)1. To match your bank register with your bank statement; 2. the Deposit to account is Uncategorized funds.

Answer: The correct option is A (1. To match your bank register with your bank statement; 2. the Deposit to account is Undeposited Funds)

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By posting to Undeposited Funds, you can create a single bank deposit for multiple payments, making it easy to match your bank register with your bank statement. When receiving a payment, make sure the Deposit to account is Undeposited Funds.

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4 0
2 years ago
On a guitar, the lowest toned string is usually strung to the E note, which produces sound at 82.4 82.4 Hz. The diameter of E gu
Vsevolod [243]

The complete and comprehensive solution is attached.

8 0
1 year ago
Find an expression for the acceleration a of the red block after it is released. use mr for the mass of the red block, mg for th
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<span>Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.</span>

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6 0
2 years ago
Read 2 more answers
Two disks with the same rotational inertia i are spinning about the same frictionless shaft, with the same angular speed ω, but
valentina_108 [34]

Answer:

3. none of these

Explanation:

The rotational kinetic energy of an object is given by:

K=\frac{1}{2}I \omega^2

where

I is the moment of inertia

\omega is the angular speed

In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

K_1 = \frac{1}{2}I (\omega)^2 = \frac{1}{2}I\omega^2

For disk 2:

K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2

so the total energy is

K=K_1 + K_2 = \frac{1}{2}I\omega^2 + \frac{1}{2}I\omega^2 = I\omega^2

So, none of the options is correct.

5 0
2 years ago
Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value.
wolverine [178]

Answer:

V

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

|I|=m(v-u)=(1000)(0-25)=25,000 Ns

(II)

u = 25 m/s

v = 0

|I|=m(v-u)=(1000)(0-25)=25,000 Ns

(III)

In this case,

F = 2000 N is the force

\Delta t = 10 s is the time

So the magnitude of the impulse is

|I| =F\Delta t = (2000N)(10)=20,000 Ns

(IV)

F = 2000 N is the force

\Delta t = 10 s is the time

So the magnitude of the impulse is

|I| =F\Delta t = (2000N)(10)=20,000 Ns

(V)

u = 25 m/s

v = -25 m/s

|I|=m(v-u)=(1000)(-25-25)=50,000 Ns

So the ranking from largest to smallest is:

V

I and II

III and IV

5 0
2 years ago
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