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bearhunter [10]

2 years ago

5

An object floats on water with 80% of its volume below the surface. the same object when placed in another liquid floats on that

liquid with 72% of its volume below the surface. determine the density of the unknown fluid.

1 answer:

GarryVolchara [31]2 years ago

8 0

0.9g/cm cubed, this is my full answer however i have to write at least 20 characters so this is filler :)

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Find the mass of a sample of water if its temperature dropped 24.8 °C when it lost 870 J of

rosijanka [135]

Answer:

8.35 × $10^{-3}$ kg

Explanation:

Remember the formula for heat:

$Q=mc\Delta T$

You just need to solve for m, doing so you're left with:

$m=\dfrac{Q}{c\Delta T}$

Replace the variables with numbers and that's it.

7 0

2 years ago

Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)

Andrew [12]

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

5 0

2 years ago

A 30-g car rolls from a hill 12 cm high and is traveling at 154 cm/s as it travels along a 275 cm horizontal track. What is the

Bond [772]

Momentum of the car is 46.2 kgm/s

8 0

2 years ago

Question 2 (1 point)

Tju [1.3M]

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

7 0

2 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

3 years ago