Ask question

Ask question

All categories

4 years ago

15

A man climbs on to a wall that is 3.6m high and gains 2268J of potential energy. What is the mass of the man? *

1 answer:

Ilia_Sergeevich [38]4 years ago

3 0

The mass of the man is 63 kilograms
There is it written down properly. Its hard to type this stuff in here

You might be interested in

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

True [87]

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:

Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².

For a constant acceleration, distance = 402.3

m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2

µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

5 0

3 years ago

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

barxatty [35]

Answer:

a

The mass of blood is $m= 5.7876kg$

b

The number of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the question we are told that

The volume of blood is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of the blood is $\rho_b = 1060 kg/m^3$

% of blood that is cell is = 45.0%

% of the blood that is plasma is = 55.0%

density of blood cell is $\rho_d = 1125kg/m^3$

% of cell that are white is = 1%

% of cell that is red is = 99%

The diameter of the red blood cell is = $7.5 \mu m = 7.5*10^{-6}m$

The radius of the red blood cell is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

Generally the mass is mathematically represented as

$m = \rho_b * V_b$

Substituting value

$m = 1060 * 0.00546$

$m= 5.7876kg$

Mass of cell is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The volume of white blood cell is $V_w$ = 1% of volume of cells

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The volume of red blood cells is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The number of red blood cell is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The Number of white blood cell is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The total number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

4 years ago

Read 2 more answers

Determine the volume displaced and then calculate the density of this 54 g sample of brass.

inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density $d_{w}$ of water is : $1000kgm^{-3}=1gcm^{-3}$

Buoyant force exerted by a liquid on an object with $V_{imm}$ of it's volume immersed is :

$F_{B}=V_{imm}*d_{l}*g$

where ,

$F_{B}$ is the buoyant force
$d_{l}$ is the density of the liquid
$g$ is the acceleration due to gravity

Thus at equilibrium:

$m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}$

from these , we get the density of brass to be $1gcm^{-3}$

which is not possible

7 0

3 years ago

If you took a white piece of plastic to a depth of 180 m, what color would it appear?.

SashulF [63]

Answer:

It would appear blue

Explanation:

3 0

2 years ago

How much force is required to accelerate a 9.0-g object at 10000 g's?

Yuki888 [10]

Hey give us m = 9.0 g = 9.0 x 10-3 kg, and a = 10,000 "g's" = 98000 m/s/s so:F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N

6 0

3 years ago