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xxTIMURxx [149]

2 years ago

12

Can someone pls do this for me for 20 pts

2 answers:

Ann [662]2 years ago

6 0

You need to delete this question ASAP! You put out your first and last name.

m_a_m_a [10]2 years ago

6 0

Answer:you got your name in that

Explanation:

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If an object has a mass of 10 kilograms, how much does it weigh in newtons?

schepotkina [342]

10 kilograms of mass weighs 98.1 newtons on Earth,
16.2 newtons on the Moon, 37.1 newtons on Mars,
and other weights in other places.

6 0

3 years ago

Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t

Nataliya [291]

Answer:

Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

Q = cos^-1 ( sqrt (0.464))

Q = cos^-1 (0.6811754546)

Q = 47.06 degrees

8 0

3 years ago

If the mass of a material is 42 grams and the volume of the material is 15 cm^3, what would the density of the material be?

Fantom [35]

Density = mass / volume

Density = 42g / 15cm^3

Density = 2.8g/cm^3

3 0

3 years ago

In the Sun, fusion reactions create helium nuclei. To form each helium

seraphim [82]

D is the answer. Bc the pointy

6 0

3 years ago

Read 2 more answers

uniform disk with mass 40.0 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is station

Alex787 [66]

Answer:

The magnitude of the tangential velocity is $v= 0.868 m/s$

The magnitude of the resultant acceleration at that point is $a = 4.057 m/s^2$

Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

Generally the angular speed i mathematically represented as

$w = \sqrt{2 \alpha \theta}$

Where $\theta$ is the angular displacement given from the question as

$\theta = 0.2000 rev = 0.2000 rev * \frac{2 \pi \ rad }{1 rev}$

$=1.257\ rad$

$\alpha$ is the angular acceleration which is mathematically represented as

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

The moment of inertial is mathematically represented as

$I = \frac{1}{2} m_dR^2_d$

Substituting values

$I = 0.5 * 40 * 0.200^2$

$= 0.8kg \cdot m^2$

Considering the equation for angular acceleration

$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

Substituting values

$\alph$ $\alpha = \frac{(30.0)(0.200)}{0.8}$

$= 7.5 rad/s^2$

Considering the equation for angular velocity

$w = \sqrt{2 \alpha \theta}$

Substituting values

$w =\sqrt{2 * (7.5) * 1.257}$

$= 4.34 \ rad/s$

The tangential velocity of a given point on the rim is mathematically represented as

$v = R_d w$

Substituting values

$= (0.200)(4.34)$

$v= 0.868 m/s$

The radial acceleration at hat point is mathematically represented as

$\alpha_r = \frac{v^2}{R}$

$= \frac{0.868^2}{0.200^2}$

$= 3.7699 \ m/s^2$

The tangential acceleration at that point is mathematically represented as

$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

$a = 4.057 m/s^2$

7 0

3 years ago