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pshichka [43]
2 years ago
14

if you have 10,000 red blood cells, how many red blood cells and white blood cells do you have altogether

Physics
1 answer:
ivolga24 [154]2 years ago
6 0
I believe its 17,000 blood cells in your body 

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A small boat coasts at constant speed under a bridge. A heavy sack of sand is dropped from the bridge onto the boat. The speed o
Tju [1.3M]

Answer:

d. decreases

Explanation:

The law of conservation of momentum tells us that the sum of momenta before the collision is equal to the sum of momenta after the collision. The bag has no momentum as it falls onto the boat because its velocity is zero in the horizontal direction. But after it hits the boat, it's momentum increases while the momentum of the system remains the same. That means a component of the system must decrease somewhere else. And that component is the velocity, not the mass, of the boat.

7 0
3 years ago
Give reason Pascal is a derived unit​
Stells [14]

Answer:

Pascal is a derived unit because <u>it</u><u> </u><u>cannot</u><u> </u><u>be</u><u> </u><u>expressed</u><u> </u><u>in</u><u> </u><u>any</u><u> </u><u>physics</u><u> </u><u>terms</u><u>,</u><u> </u><u>but</u><u> </u><u>it</u><u> </u><u>is</u><u> </u><u>an</u><u> </u><u>expression</u><u> </u><u>of</u><u> </u><u>fundamental</u><u> </u><u>quantities</u><u>.</u>

Explanation:

{ \sf{Pasacal \: ( Pa) =  \frac{newtons}{metres {}^{2} } }} \\  \\ { \sf{Pasacal  \: (Pa) =  \frac{kg \times  {ms}^{ - 2} }{ {m}^{2} } }}

4 0
3 years ago
The size of the gravitational force between two objects depends on their__. frictional force, inertia, masses and the distance b
Inessa [10]

Answer:

Masses and distance between them

Explanation:

The gravitational force between two objects can be calculated using Newton's Gravitational Law.

However, using logic, we can already dictate what the answer will be, for example. We know that the bigger an object is, the stronger its gravity is. This can be seen with how the moon is much smaller, and also has much less gravity.

Also, the distance between two objects also influences the gravity. This can be seen the further an object gets from Earth, the less of a pull the gravitational field has on it. Another example is that Pluto (being very far from the sun) has less of a gravitational effect from the sun, in comparison to Mercury (the closest plant to the sun).

3 0
3 years ago
A container of rocks is brought back from the Moon's surface where the acceleration due to gravity is 162 meters per second if t
spin [16.1K]

Answer:

6318 N

Explanation:

From the question given above, the following data were obtained:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Mass (m) of container = 650 kg

Weight (W) of container on the earth =.?

Next, we shall determine the acceleration due to gravity of the earth. This can be obtained as follow:

Acceleration due to gravity of the moon (gₘ) = 1.62 m/s²

Acceleration due to gravity of the earth (gₑ) =.?

gₘ = 1/6 × gₑ

1.62 = 1/6 × gₑ

1.62 = gₑ /6

Cross multiply

gₑ = 1.62 × 6

gₑ = 9.72 m/s²

Finally, we shall determine the weight of the container on the earth as follow:

Mass (m) of container = 650 kg

Acceleration due to gravity of the earth (gₑ) = 9.72 m/s²

Weight (W) of container on the earth =.?

W = m × gₑ

W = 650 × 9.72

W = 6318 N

Therefore, the weight of the container on earth is 6318 N

6 0
3 years ago
Coulomb's Law: Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrost
Ahat [919]

Answer:

The magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force. ( ¹/₄ F)

Explanation:

The electrostatic force between two charges is given by Coulomb's law;

F = \frac{kQ_1Q_2}{r^2}

where;

Q₁ and Q₂ are the magnitude of the charges

r is the distance between the charges

k is Coulomb's constant

Since the charges are identical;

Q₁ = Q

Q₂ = Q

the electrostatic force experienced by each charge is given by;

F =  \frac{kQ^2}{r^2}

When each of the spheres has lost half of its initial charge;

Q₁ = Q/2

Q₂ = Q/2

F_2 = \frac{k(Q/2)(Q/2)}{r^2}\\\\ F_2 = \frac{k(Q)(Q)}{4r^2}\\\\F_2 = \frac{1}{4} (\frac{kQ^2}{r^2} )\\\\F_2 = \frac{1}{4} (F)

Therefore, the magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force.

6 0
2 years ago
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