All categories

1 month ago

The length of the side of a cube having a density of 12.6 g/ml and a mass of 7.65 g is __________ cm.

1 answer:

4 0

Density serves as a key characteristic of any substance, defined as the mass of the substance divided by its volume (density = mass/volume). Through manipulation of this formula, we determine volume. So volume equals the mass of a substance divided by its density (Vol = mass/density). With density given as 12.6 g/ml and mass as 7.65 g, we find that volume equals 0.60714 ml; since 1 ml is equivalent to 1 cm³, volume also equals 0.60714 cm³. Finally, taking the cube root of the volume provides the side length of the cube in cm, which is 0.84677 cm.

You might be interested in

A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav

kicyunya [3294]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

The locomotive's acceleration is 1.6 $m/s^2$

The duration taken to pass the crossing is 2.4 seconds.

We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 $m/s^2$ .

32 = 0 + 1.6 * t

t = 20 seconds.

Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0

1 month ago

Read 2 more answers

A model of a spring/mass system is 4x'' + e−0.1tx = 0. By inspection of the differential equation only, discuss the behavior of

Keith_Richards [3271]

Displacement stabilizes over time. It is known that exponentials raised to infinity approach zero, hence the system model will yield as time approaches infinity, resulting in 4x'' + e−0.1tx = 0. As time approaches infinity, we deduce that 4x'' equals zero. Consequently, upon integrating, we derive 4x' = c, and further integration leads to the conclusion 4x = cx + d.

4 0

20 days ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

serg [3582]

Answer:

$4.1\cdot 10^8 N$

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ denotes the atmospheric pressure

$\rho = 1000 kg/m^3$ represents the density of the water

$g=9.8 m/s^2$ signifies the acceleration due to gravity

$h=11,000 m$ indicates the depth

By substituting these values,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

Consequently, the inward force acting on the sphere equals

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

28 days ago

Read 2 more answers

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

2 months ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

serg [3582]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: The speed of a wave on a string under tension can be determined using the following:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ denotes tension (N)

μ refers to linear density (kg/m)

Calculating the velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

Distance a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With a tension of 47.8N, the distance a pulse will cover is Δx = 11.5× $10^{-6}$ m.

When tension is doubled:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Distance in the same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With the increased tension, it moves $\Delta x =$ 15.4× $10^{-5}$ m

4 0

1 month ago