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Anuta_ua [19.1K]
3 years ago
8

A 1.20 g sample of an unknown has a volume of 1.73 cm what is the density of the unknown

Physics
1 answer:
mariarad [96]3 years ago
7 0
1.73 divide by 1.20=1.4416
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;-; please help me....​
sineoko [7]
The answer is 24N. Since the body is moving with constant velocity all the forces must balance (equal & opposite)
5 0
2 years ago
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
Which of the following is typically NOT a method scientist use to determine astronomic distances
tangare [24]
I think the answer is A
5 0
3 years ago
The auto in the sketch moves forward as the brakes are applied. A bystander says that during the interval of braking, the auto's
Ivan

Answer:

The statement is true: velocity and acceleration have opposite directions in the interval of braking.

Explanation:

Let's say we have a velocity v>0.

The acceleration a is the rate of change of the velocity v. This means that if v is <em>increasing during</em> time, then a must be positive. But if v is <em>decreasing over</em> time, then a will be negative (even though the velocity is positive).

Mathematically:

a=\frac{dv}{dt}

v decreases ⇒\frac{dv}{dt}

⇒a.

Example:

v(t)=e^{-t}>0 \\\\\frac{dv}{dt}=-te^{-t}

3 0
3 years ago
A constant magnetic field passes through a single rectangular loop whose dimensions are 0.35 m × 0.55 m. The magnetic field has
Pani-rosa [81]

Answer:

Part a)

EMF = 0.38 V

Part b)

\frac{dA}{dt} = 0.43 m^2/s

Explanation:

Part a)

Initial value of magnetic flux is given as

\phi_1 = BAcos\theta

\phi_1 = (2.1)(0.35 \times 0.55) cos65

so we have

\phi_1 = 0.17 Wb

Final flux through the loop is given as

\phi_2 = 0

now EMF is given as

EMF = \frac{\phi_1 - \phi_2}{\Delta t}

EMF = \frac{0.17 - 0}{0.45}

EMF = 0.38 V

Part b)

If magnetic field is constant while Area is changing

So EMF is given as

E = Bcos65 \times \frac{dA}{dt}

0.38 = 2.1 cos65(\frac{dA}{dt})

\frac{dA}{dt} = 0.43 m^2/s

5 0
2 years ago
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