PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3

V = 6.307 x 10^-5 m^3

By density = m/V

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

6 0

3 years ago

A car battery has a rating of 170 ampere-hours. This rating is one indication of the total charge that the battery can provide t

Pavlova-9 [17]

Answer:

612000 C

Explanation:

Current, I, is given as the rate of flow of charge, that is:

I = Δq / Δt

where q = electric charge

t = time taken

This implies that:

Δq = I * Δt

The battery rating is 170 Ampere-hours, therefore:

Δq = 170 * 1 hour

But 1 hour = 3600 seconds;

=> Δq = 170 * 3600 = 612000 C

The total charge that the battery can provide is 612000 C.

8 0

3 years ago

(a) A submarine descends to a depth of 70 m below the surface of water. The density of the water is 1050 kg/m3. Atmospheric pres

timurjin [86]

Answer:

sttouyietETwe2e664yrwtwwteuwtrwruwuuwwuwtwuw7w7w5w7w772253536464647

5 0

3 years ago

A. How many calories are needed to raise the temperature of 1 gram of water by 1 °C?

sweet-ann [11.9K]

A) 1 cal

B) 80 cal

C) 540 cal

Explanation:

A)

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

$\Delta T$ is the change in temperature

In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

B)

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

- The ice is already at melting point, 0 °C

- Mass of the ice: $m=1g$

- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

$Q=(1)(80)=80 cal$

C)

For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

5 0

3 years ago

A stone is dropped from the upper observation deck of a tower, 250 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the dista

Vitek1552 [10]

(a) $y(t)=250 - 4.9 t^2$

For an object in free-fall, the vertical position at time t is given by:

$y(t) = h + ut - \frac{1}{2}gt^2$

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

$y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2$

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

$y(t) = h - \frac{1}{2}gt^2=0$

So for the stone in the problem, we have

$250 - 4.9 t^2 = 0$

Solving for t, we find:

$t=\sqrt{\frac{250}{4.9}}=7.14 s$

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

$v(t) = u - gt$

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

$v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s$

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

$y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2$

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

6 0

3 years ago