The velocity of a motor car moving along a road increases from 10m/s to 50{1} ms in 8s. Find its avarege acceleration. RESULT

25POINTS please answer this

WARRIOR [948]

i know for sure the water one is correct with

A. Water is important to some organisms.

I'm 50/50 about the circulatory one BUT

A. carrying heat around the body.

that is your best bet

4 0

3 years ago

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17,874,000 what is the value of 1

kotegsom [21]

Salutations!

17,874,000 what is the value of 1?

The value of 1 is 10 million. The place value would be 10,000,000.

Hope I helped.

3 0

3 years ago

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A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W

Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)$

$V^{2} = 24.5 m/s$

$V = \sqrt{24.5} \ m/s$

$V = 4.95 \ m/s$

V = ± 4.95 m/s

V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is V = - 4.95 m/s

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

$V^{2} = U^{2} + 2gh$

Substituting the values,

$0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)$

$0 = U^{2} - 16.072 m/s$

$U^{2} = 16.072 m/s$

$U = \sqrt{16.072} \ m/s$

U = ± 4.01 m/s

U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is U = + 4.01 m/s

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

$F = \frac{mv - mu}{t}$

$F.t = m(v - u)$

⇒ Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,

∴ F.t = 0.120 kg(4.01 m/s - (-4.95 m/s) )

F.t = 0.120 kg(4.01 m/s + 4.95 m/s) )

F.t = 0.120 kg × 8.96 m/s

Impulse = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

6 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

3 years ago

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Mass m, moving at speed 2v, approaches mass 4m, moving at speed v. The two collide elastically head-on. Part A Find the subseque

meriva

Thank you for posting your question here at brainly. A mass of m moves with 2V towards in the opposite direction of a mass, 4m moving at a speed of V, the speed of m was 2/5V and the mass of 4m was 7.5V. I hope it helps.

4 0

3 years ago

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