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Mumz [18]
2 years ago
12

A 15-g block of aluminum has a volume of 5.5 cm. what is the density

Physics
1 answer:
Mnenie [13.5K]2 years ago
4 0

Answer: 2.72727272......g/cm3^

Explanation:

D = g/cm^3

soo d = 15/5.5

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Consider three planets. All have the same mass as Earth, but with different radii (from largest to smallest: Planet 1, 2, 3). Fo
LuckyWell [14K]

Answer:

option C

Explanation:

given,

mass of the three planet is same

radius of the planets are

R₁ > R₂ > R₃

expression of escape velocity

v = \sqrt{\dfrac{2GM}{R}}

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

from the above expression we can clearly conclude that the escape velocity is inversely proportional to the radius of the Planet.

radius of planet increases escape velocity decreases.

Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.

The correct answer is option C

3 0
3 years ago
Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force
marusya05 [52]

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

E=W\frac{a}{b}

From the figure, you can observe that a/b < 1, thus E < W

8 0
3 years ago
Plzzz help will mark the brainliest
ANEK [815]
Ciara is winging....etc
The answer is : 0.60 N, toward the center of the circle


A satellite....etc
The Answer is : 7400 m/s


What is the .....etc
The Answer is : 2.60 m/s
8 0
3 years ago
Find the image position for a picture placed 3.0 cm outside the focal point of a converging lens with a 4.0 cm focal length. a.
horrorfan [7]
<span>Answer: Using 1/f = 1/d' + 1/d ...(where d' object distance and d is image distance) 1/4 = 1/7 + 1/d 1/4 - 1/7 = 1/d 3/28 = 1/d d = 28/3 d = 9.33 cm</span>
5 0
3 years ago
A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
kari74 [83]

Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

4.9\times 9.8=4.9a+\frac{3.3a}{2}

Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

a=\frac{48.02}{6.55}=7.33m/s^2

Angular acceleration,\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2

7 0
3 years ago
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