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AlexFokin
14 days ago
13

Calculate the mass (in kg) of 54.3 m³ of granite. The density of granite is 2700 kg/m³. Give your answer to 2 decimal places.

Physics
1 answer:
Ostrovityanka [2.2K]14 days ago
6 0

Answer:

Explanation:

Density is defined as d=m/v.

To find mass, the formula transforms into:

m=d*v

m=2700*54.3

m=146610

m=14.6*10^4

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A particle with mass 1.81×10−3 kg and a charge of 1.22×10−8 C has, at a given instant, a velocity v⃗ =(3.00×104m/s)j^. What are
ValentinkaMS [2425]

Answer:

The particle's acceleration magnitude and direction is a= 0.3296\ \hat{k}\ m/s^2

Explanation:

Given:

Mass m = 1.81\times10^{-3}\ kg

Velocity v = (3.00\times10^{4}\ m/s)j

Charge q = 1.22\times10^{-8}\ C

Magnetic field B= (1.63\hat{i}+0.980\hat{j})\ T

To find the particle’s acceleration

The formula for acceleration is

F = ma=q(v\times B)

a =\dfrac{q(v\times B)}{m}

We will calculate v\times B

v\times B=(3.00\times10^{4}\ m/s)j\times(1.63\hat{i}+0.980\hat{j})

v\times B=4.89\times10^{4}

Now, substitute all known values into the formula for acceleration

a =\dfrac{1.22\times10^{-8}\times(-4.89\times10^{4}\hat{k})}{1.81\times10^{-3}}

a= -0.3296\ \hat{k}\ m/s^2

The negative sign indicates the direction is opposite.

Thus, the particle’s acceleration magnitude and direction is a= 0.3296\ \hat{k}\ m/s^2

7 0
17 days ago
Read 2 more answers
(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
serg [2593]

Answer:

(a) the coefficient of friction is 0.451

This was derived using the energy conservation principle (the total energy in a closed system remains constant).

(b) No, the object stops 5.35 m away from point B. This is due to the spring's expansion only performing 43 J of work on the block, which isn't sufficient compared to the 398 J required to overcome friction.

Explanation:

For more details on how this issue was resolved, refer to the attached material. The solution for part (a) separates the body’s movement into two segments: from point A to B, and from B to C. The total system energy originates from the initial gravitational potential energy, which transforms into work against friction and into work compressing the spring. A work of 398 J is needed to counteract friction over the distance of 6.00 m. The energy used for this is lost since friction is not a conservative force, leaving only 43 J for spring compression. When the spring expands, it exerts a work of 43 J back on the block, which is only sufficient to move it through a distance of 0.65 m, stopping 5.35 m short of point B.

Thank you for your attention; I trust this is beneficial to you.

4 0
20 days ago
Evaluate the final kinetic energy of the supply spacecraft for the actual tractor beam force, F(x)=αx3+βF(x)=αx3+β.
Sav [2230]

Answer:

K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x

Explanation:

To calculate the kinetic energy variation, we can utilize the work-energy theorem.

W = ΔK

∫ F .dx = K - K₀

If the object starts from rest, then K₀ = 0.

So, ∫ F dx cos θ = K.

As the force and displacement directions align, the angle is zero, and hence the cosine is 1.

Now we can substitute and perform integration:

α ∫ x³ dx + β ∫ dx = K.

Thus, α x⁴ / 4 + β x = K.

Next, we evaluate from the limits F = 0 to F:

α (x⁴ / 4 - 0) + β (x - 0) = K.

Consequently, K = αX⁴ / 4 + β x.

This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.

To finalize the computation, we need to ascertain the displacement.

8 0
1 month ago
Consider a 4-mg raindrop that falls from a cloud at a height of 2 km. When the raindrop reaches the ground, it won't kill you or
inna [2210]

Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

Hello!

According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = gravitational acceleration (9.8 m/s²)

h = height.

Let's determine the initial potential energy of the raindrop:

(4 mg should be converted into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

As the raindrop descends, some of its potential energy converts into kinetic energy while the rest is lost to the air resistance. Upon reaching the ground, all initial potential energy has been either turned into kinetic energy or spent overcoming air resistance:

initial PE = final KE + Work by air

Where:

KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

<pThus:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can find the work done by air resistance:

initial PE = final KE + Work by air

0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

Since work is performed in the opposite direction to movement, this results in a negative value. Therefore, the work done by air resistance is -0.0782 J.

5 0
29 days ago
Potential energy matter has a result of its ____ or ____.
Yuliya22 [2438]
Position or composition
3 0
15 days ago
Read 2 more answers
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