To determin the density of an empty plastic jug what would you measure

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

$80=5(50-x)$

$80=250-5x$

$5x=170$

$x=\frac{170}{5}$

$x=34 cm$

4 0

2 years ago

Two resistors of 5.0 and 9.0 ohm are connected in parallel. A 4.0 ohm resistor is then connected in series with the parallel com

rewona [7]

Answer:

I1 = 0.772 A

Explanation:

Given: R1 = 5.0 ohm, R2 = 9.0 ohm, R3 = 4.0 ohm, V = 6.0 Volts

To find: current I = ? A

Solution:

Ohm's law V= I R

⇒ I = V / R

In order to find R (total) we first find R (p) fro parallel combination. so

1 / R (p) = 1 / R1 + 1/ R2 ∴(P) stand for parallel

R (p) = R1R2 / ( R1 + R2)

R (p) = (5.0 × 9.0) / (5.0 + 9.0)

R (p) = 3.214 ohm

Now R (total) = R (p) + R3 (as R3 is connected in series)

R (total) = 3.214 ohm + 4.0 Ohm

R (total) = 7.214 ohm

now I (total) = 7.214 ohm / 6.0 Volts

I (total) = 1.202 A

This the total current supplied by 6 volts battery.

as voltage drop across R (p) = V = R (p) × I (total)

V (p) = 3.214 ohm × 1.202 A = 3.864 volts

Now current through 5 ohms resister is I1 = V (P) / R1

I1 = 3.864 volts / 5 ohm

I1 = 0.772 A

3 0

2 years ago

If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it

Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

$\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

Explanation:

From the question we are told that

The mass of the rocket is m = 6300 kg

The velocity at gas is being ejected is u = 2000 m/s

The initial acceleration desired is $a = 27.0 \ m/s$

The time taken for the gas to be ejected is t = 1 s

Generally this desired acceleration is mathematically represented as

$a = \frac{u * \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}$

Here $\frac{\Delta m}{\Delta t }$ is the rate at which gas is being ejected with respect to time

Substituting values

$27 = \frac{2000 * \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}$

=> $170100 -27* \frac{\Delta m}{\Delta t} = 2000 * \frac{\Delta m}{\Delta t}$

=> $170100 = 2027 * \frac{\Delta m}{\Delta t}$

=> $\frac{\Delta m}{\Delta t} = \frac{170100}{2027}$

=> $\frac{\Delta m}{\Delta t} = 83.92 \ Kg/s$

3 0

2 years ago

You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car

gogolik [260]

B. The sound of the engine will get louder and the pitch higher.

4 0

2 years ago

Read 2 more answers

The position of a 50 g oscillating mass is given by x(t)=(2.0cm)cos(10t−π/4), where t is in s. If necessary, round your answers

Leona [35]

For the answer to the question above, this is the maximum displacement, the spring has only elastic potential energy.
spring is constant @ 5 N/m
maximum displacement = 2 cm = 0.02 m
elastic potential energy = 1/2 kx²
= 0.5 x 5 x 0.02²
So the answer would be
= 0.001 Joules

5 0

2 years ago