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zloy xaker
2 months ago
10

One component of a metal sculpture consists of a solid cube with an edge of length 38.9 cm. The alloy used to make the cube has

a density of 7.15 x10^3 kg/m3. Find the cube's mass.
Physics
1 answer:
Ostrovityanka [3.2K]2 months ago
7 0

Answer:

The cube's mass is calculated to be 420.8 kg.

Explanation:

Given that,

Edge length = 38.9 cm

Density \rho= 7.15 \times10^{3}\ kg/m^3

We need to determine the cube's volume

Using the volume formula

V = 38.9^3

V=0.058863\ m^3

Next, we need to calculate the cube's mass

Employing the density formula

\rho = \dfrac{m}{V}

m = V\times\rho

m =0.058863\times7.15 \times10^{3}

m=420.8\ kg

Consequently, the cube's mass is determined to be 420.8 kg.

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Three objects of the same mass begin their motion at the same height. One object falls straight down, one slides down a low-fric
Softa [3030]

Answer:

D. All are equal

Explanation:

The only force acting on the objects is gravity, which is consistent across all items

4 0
1 month ago
A rod 12.0 cm long is uniformly charged and has a total charge of -20.0 µc. determine the magnitude and direction of the electri
Ostrovityanka [3204]
<span>Let Q be the charge, thus Q = -20.0 µC.</span>
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m, which represents the length of the rod
</span><span>To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
</span><span>E = K·Q/r²
</span>or<span>E = kQ/D(D+L), where k</span> is a constant equal to 8.99 x 10<span>9</span> N m
2/C2.<span>Consequently,[TAG_21]]E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>

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7 0
2 months ago
Read 2 more answers
3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta
ValentinkaMS [3465]

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

a.

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

b.

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

6 0
2 months ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size1.055\times 10^{- 7} is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, m_{p} = 1.67\times 10^{- 27} kg

Initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This proton energy is \simeq 250 GeV

Thus, the speed of the proton, v\simeq c

The time to cover 1 km = 1000 m of distance is calculated as:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

According to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 months ago
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