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3 years ago

7

I’ll name u BRAINLIEST if u get this right

1 answer:

nekit [7.7K]3 years ago

7 0

V = m1 u1 - m2 u2 / m1
v = 0.01 * 500 - 2 * 1.4 / 0.01
v = 220 m/s

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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

Soapy water between surfaces

Inessa [10]

Answer:

<em>A</em><em>.</em><em>increases</em><em> </em><em>friction</em>

Explanation:

6 0

2 years ago

Read 2 more answers

Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass

Artist 52 [7]

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by $\frac{1}{2} mv^2$

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = $\frac{1}{2} (4m)(v^2) = 2mv^2$

B) KE = $\frac{1}{2} (3m)(2v)^2 = 6mv^2$

C) KE = $\frac{1}{2} (2m)(3v)^2 = 9mv^2$

D) KE = $\frac{1}{2} (3)(4v)^2 = 8mv^2$

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

4 0

3 years ago

What is the impulse when a test car traveling at 15 m/s strikes a cement wall at a force of 1,900,000 N if the impact lasts for

svp [43]

Answer:

impulse = force x time

impulse = 1900000 x 0.005 = 9500Ns

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3 years ago

Do electric field lines actually exist?

konstantin123 [22]

Answer: No

Explanation:they do not exist electric field lines are only imaginary

8 0

3 years ago