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3 years ago

A 45-gram object that has a volume of 9 mL has a density of

Physics

1 answer:

yanalaym [24]3 years ago

7 0

Answer:

<h2> The answer is </h2><h2>a. 5g/mL</h2>

Explanation:

Given data

mass m= 45g

volume v= 9mL

we know that density=m/v

substituting our given data we have

$density=\frac{45}{9} =5g/mL$

What is Density?

The Density of a body can be defined as the ratio of mass to volume,

Density, mass of a unit volume of a material substance. The formula for density is $Density= \frac{mass}{volume}$ ,

where d is density,

M is mass, and

V is volume.

Density is commonly expressed in units of grams per cubic centimetre.

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a 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they

Otrada [13]

According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^

<h3>What is collision and momentum?</h3>

The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).

<h3>Briefing:</h3>

From conservation of momentum

Initial momentum = final momentum

m u +M U =(m+M) V

2000×25 i^ +1500×30 j^ =(2000+1500) V

V = 8.735i^ + 11.25j^

Final velocity when they stick together is 8.735i^ + 11.25j^

To know more about Collide visit:

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The complete question is -

A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

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1 year ago

Lemon trees receiving the most water produced the most lemons. What is the independent variable, the dependent variable, and the

Veseljchak [2.6K]

The lemon trees will always be the same so they are the constants. The independent variable is the water amount because it is changing. The dependent variable is the amount of lemons because it is the variable being tested. (If we change the amount of water how many lemons will be produced?)

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3 years ago

What was Laurentia and what lands was it composed of? What happened to it?

snow_lady [41]

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4 years ago

A student's calculation for the work done on an object is shown.

MissTica

The statement that describes the error in the work is that the distance must be converted to meters (m).

<h3>FORMULA FOR WORK:</h3>

Work can be calculated by using the following formula:

W = F × d

Where;

W = work done
F = force (N)
d = distance (m)

According to this question, the force is given as 140N and the distance is given as 30cm. The force is calculated as follows:

F = 140N × 30cm = 4200J

This calculation is erroneous because the unit of distance should be converted from cm to meters.

Learn more about work done at: brainly.com/question/3902440

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2 years ago

Read 2 more answers

A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm. Both

Dafna11 [192]

Answer:

a. i. 1350 V ii 0 V iii -450 V b. 6.75 kV. The inner shell is at a higher potential.

Explanation:

The formula for electric potential is given by V = Σkq/r, where k = 9 × 10⁹ Nm²/C², q = charge and r = distance.

q₁ = charge on smaller shell = +6.00 nC = +6.00 × 10⁻⁹ C, r₁ = radius of smaller shell = 2.00 cm = 2.00 × 10⁻² m.

q₂ = charge on larger shell = -9.00 nC = -9.00 × 10⁻⁹ C, r₂ = radius of larger shell = 6.00 cm = 6.00 × 10⁻² m.

a. At r = 0, inside both spheres V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 1350 V

ii. At r = 4.00 cm, the point outside of smaller shell but inside larger shell. r₁ = 4.00 cm = 4.00 × 10⁻² m and r₂ = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/4.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 0 V.

iii. At r = 6.00 cm, the point outside both shells. r₁ = r₂ = r = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁ + q₂)/r = 9 × 10⁹ [+6.00 × 10⁻⁹+ (-9.00 × 10⁻⁹)]/6.00 × 10⁻² = -450 V.

b. The potential of the surface of the smaller shell is V₁ = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻²] = 2700 V = 2.7 kV.

The potential of the surface of the larger shell is V₂ = 9 × 10⁹ [-9.00 × 10⁻⁹/2.00 × 10⁻²] = -4050 V = -4.050 kV. The potential difference V₁ - V₂ = 2700 - (-4050) V = 6750 V = 6.75 kV. Since the potential difference is positive, V₁ is higher. So, the inner shell is at a higher potential.

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3 years ago